M1 Topics Part 1

7/24/2019 M1 Topics Part 1

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M1 Collisions - Direct impact PhysicsAndMathsTutor.com

1. Particle Phas mass mkg and particle Qhas mass 3mkg. The particles are moving in opposite

directions along a smooth horizontal plane when they collide directly. Immediately before the

collision Phas speed 4ums1and Qhas speed kums1, where kis a constant. As a result of the

collision the direction of motion of each particle is reversed and the speed of each particle is

halved.

(a) Find the value of k.(4)

(b) Find, in terms of mand u, the magnitude of the impulse exerted on Pby Q.(3)

(Total7marks)

2. A particleAof mass 2 kg is moving along a straight horizontal line with speed 12 m s1.

Another particleBof mass mkg is moving along the same straight line, in the opposite direction

toA, with speed 8 m s1. The particles collide. The direction of motion ofAis unchanged by the

collision. Immediately after the collision,Ais moving with speed 3 m s1andBis moving with

speed 4 m s1. Find

(a) the magnitude of the impulse exerted byBonAin the collision,(2)

(b) the value of m.(4)

(Total6marks)

3. Two particlesAandBare moving on a smooth horizontal plane. The mass ofAis 2mand themass of B is m. The particles are moving along the same straight line but in opposite directions

and they collide directly. Immediately before they collide the speed ofAis 2uand the speed of

Bis 3u. The magnitude of the impulse received by each particle in the collision is2

7mu.

Edexcel Internal Review 1


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Find

(a) the speed ofAimmediately after the collision,

(3)

(b) the speed ofBimmediately after the collision.(3)

(Total 6 marks)

4. Two particlesAandBare moving on a smooth horizontal plane. The mass ofAis km, where

2 < k< 3, and the mass ofBis m. The particles are moving along the same straight line, but in

opposite directions, and they collide directly. Immediately before they collide the speed ofAis

2uand the speed ofBis 4u. As a result of the collision the speed of Ais halved and its direction

of motion is reversed

(a) Find, in terms of kand u, the speed ofBimmediately after the collision.(3)

(b) State whether the direction of motion ofBchanges as a result of the collision, explaining

your answer.(3)

Given that ,3

7=k

(c) find, in terms of mand u, the magnitude of the impulse thatAexerts onBin the collision.(3)

(Total 9 marks)

5. Two particlesAandB, of mass 0.3 kg and m kg respectively, are moving in opposite directions

along the same straight horizontal line so that the particles collide directly. Immediately before

the collision, the speeds ofAandBare 8 m s1and 4 m s1respectively. In the collision the

direction of motion of each particle is reversed and, immediately after the collision, the speed of

each particle is 2 m s1. Find




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(Total 7 marks)

6. A particle P of mass 0.3 kg is moving with speed u m s1in a straight line on a smooth

horizontal table. The particle P collides directly with a particle Q of mass 0.6 kg, which is at rest

on the table. Immediately after the particles collide, P has speed 2 m s1and Q has speed 5 m

s1. The direction of motion of P is reversed by the collision. Find

(a) the value of u,(4)

(b) the magnitude of the impulse exerted by P on Q.(2)

Immediately after the collision, a constant force of magnitudeR newtons is applied to Q in the

direction directly opposite to the direction of motion of Q. As a result Q is brought to rest in

1.5s.

(c) Find the value ofR.(4)

(Total 10 marks)



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1. (a)

4 3 2 32

umu mku mu mk = + M1 A1

4

3k = M1 A1cso 4

(b) For P, I= m(2u 4u) M1 A1

= 6mu A1 3

OR

For Q,I= 3m )2

( kuku

(M1 A1)

[7]

2. (a) I=21223=18 (N s) M1 A1 2

(b) LM 2128m = 23+4m M1 A1

Solving to m= 1.5 DM1 A1 4

Alternative

I= m(4(8)) = 18 M1 A1

Solving to m= 1.5 DM1 A1[6]

3. (a) ForA: ( )uvmmu A 222

7 = M1 A1

4

uvA = A1 3

(b) ForB: ( )uvmmu

B3

2

7= M1 A1

2

uvB = A1 3

OR CLM:



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4mu 3mu=B

mvu

m +4

2 M1 A1

2

uvB = A1

[6]

4. (a) 2u 4u km2u 4mu= kmu+ mv M1 A1

km m u(3k 4)= v

u v A1 3

(b) k> 2 v> 0 dirnof motion reversed M1A1A1

cso 3

(c) ForB, m(u(3k 4) 4u)

= 7 mu M1 A1 f.t.

A1 3[9]

5.

0.3

2 m s1

4 m s1

m

2 m s1

(a) A:I= 0.3(8 + 2) M1A1

= 3 (Ns) A1 3

(b) LM 0.3 8 4m= 0.3 (2) + 2m M1A1

m= 0.5 DM1A1 4

Alternative

B: m(4 + 2) = 3 M1A1

m= 0.5 DM1A1 4

The two parts of this question may be done in either order.[7]

6 (a) CLM 0.3u = 0.3 (2) + 0.6 5 M1A1

u = 8 M1A1 4

(b) I = 0.6 5 = 3 (Ns) M1A1 2



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(c) v =u + at 5=a 1.5(a =3

10) M1A1

N2LR = 0.6 3

10 = 2 M1 A1 4

[10]



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1. This question produced very many correct responses. In part (a) most candidates were able to

apply the conservation of momentum principle with few problems, with many candidates

achieving all four marks. As usual a significant number, maybe fewer than in previous years,

made sign errors, with the occasional candidate missing the odd ms or us. Very few put gs

into the equation while others had difficulty in manipulating the fractions Arithmetic errors in

working out the value of kwere not uncommon and negative values obtained for kseldomalerted the candidates to a possible error in their work. In the second part, the majority of

candidates chose to use the change in momentum of Pwith many correct answers being

obtained. However there were the inevitable errors with signs, more than in part (a), with too

many candidates thinking that a negative answer was acceptable, misunderstanding the meaning

of magnitude.

2. This proved to be a good starter and was well-answered by the majority of candidates. In part

(a), most knew that impulse = change in momentum and almost all errors were with the signs.

Candidates would be well-advised to put impulses, with arrows, on their diagrams as well asvelocities. There are still some candidates giving a negative answer for a magnitude which

always loses a mark. Most used conservation of momentum in part (b) which was preferable

since it did not rely on their answer from the previous part. Those who used impulse = change in

momentum again, applied to the other particle, could lose two marks if their answer to part (a)

was wrong.

3. Impulses continue to cause problems and a correct solution to part (a) was rarely seen. Most

candidates know that impulse = change in momentum but few can cope with the signs correctly

and the impulse in the first part almost always had the wrong sign. The second part producedmore success and if the impulse-momentum principle was used again, this part was independent

of part (a) and so full marks could be scored. Some tried to use the conservation of momentum

principle in part (b), but this relied on using their possibly incorrect answer to part (a).

4. Almost all candidates attempted to use a conservation of momentum equation in part (a) but

there were many who either did not draw a diagram at all or else drew one which did not show

the directions of motion of each particle after the collision. This lead to problems in all three

parts of the question. Few realised the significance of the question asking for the speed of B,

and gave a negative answer u(4 3k). There were also sign errors in the momentum equation

and general problems dealing with the algebra. The second part required the significance of therange of values of kto be explicitly referred to in the identification of direction and there were a

number of fully correct and often well-expressed solutions. However, many did not mention kat

all and scored little. In part (c), many knew the relevant impulse-momentum equation and

attempted to apply it to one of the particles but there was often confusion over direction and

substitution of values and some gave a negative answer, losing the final mark.

5. Notwithstanding the usual sign errors, this question was more successfully answered than in

some previous years.

(a) Sign errors were frequent for this part withI = 0.3(8 2) = 1.8 being frequently seen. An

extra g was not often seen but knowledge of the units for impulse although not

separately marked this time was tenuous, with N or even N/s being proffered. The



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requirement for magnitudedemands a positive answer [()3 was marked down, as

being an attempt to have it both ways].

(b) Almost all candidates were able to write down a momentum equation (even if with sign

errors)but 6m = 3 far too often led to the deduction m = 2. Sign errors could lead to a

negative mass, an outcome which should have alerted candidates to a problem; just

dropping the negative sign should not be an option!

6. This was also a good source of marks for many candidates. Most candidates knew to use

conservation of momentum in the first part, but there were often sign errors leading to an

incorrect value of 12 for u. In part (b), diagrams are advisable so that candidates can clearly

define direction when using the Impulsemomentum equation. Some candidates threw away a

mark for not giving a positive answer, as a magnitude was required. For the final part, most

were able to find the acceleration, but often an extra force appeared when applying F = ma.



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M1 Collisions PhysicsAndMathsTutor.com

1. Two particles Pand Qhave mass 0.4 kg and 0.6 kg respectively. The particles are initially at

rest on a smooth horizontal table. Particle Pis given an impulse of magnitude 3 N s in the

direction PQ.

(a) Find the speed of Pimmediately before it collides with Q.(3)

Immediately after the collision between Pand Q, the speed of Qis 5 m s1.

(b) Show that immediately after the collision Pis at rest.(3)

(Total 6 marks)

2. Two particlesAandBhave masses 4 kg and mkg respectively. They are moving towards each

other in opposite directions on a smooth horizontal table when they collide directly.

Immediately before the collision, the speed ofAis 5 m s1and the speed ofBis 3 m s1.

Immediately after the collision, the direction of motion ofAis unchanged and the speed ofA is

1 m s1.

(a) Find the magnitude of the impulse exerted onAin the collision.(2)

Immediately after the collision, the speed ofBis 2 m s1.

(b) Find the value of m.(4)

(Total 6 marks)



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1. (a) I =mv

3 = 0.4 v M1 A1

v = 7.5(m s1) A1 3

(b)

0.4 0.6

v 5

7.5

LM 0.4 7.5 = 0.4v + 0.6 5 M1A1

0 = 0.4vv = 0 * cso A1 3[6]

2. (a)

A B

I I

1 2

4

5 3

I = 4( 5 1) = 16 Ns M1 A1 2

(b) CLM: 4 5 m 3 = 4 1 + m 2 M1A1

m =3.2 DM1A1 4or

16 = m(3 + 2) M1A1

m =3.2 DM1A1 4[6]



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1. This was done well by the majority of candidates. Part (a) was a straightforward opening

question, almost always correctly answered. A few candidates wrote 3 = 0.4(0 v), thus only

gaining the method mark. In the second part most knew and could apply appropriately the

conservation of momentum principle, with only occasional sign errors. Drawing a clear velocitydiagram would have helped candidates who confused before and after velocities. Since it

was a show that question it was important that full working was seen in order to achieve full

marks. Wordy explanations involving impulses with no equation, tended to achieve no marks.

2. A good starter question enabling most candidates to obtain marks. A significant number of

candidates gave an answer of 16 in part (a) rather than giving the magnitude of the impulse and

lost a mark.

In part (b) 16 was a common incorrect answer resulting from an incorrect direction of motion

for particleBi.e. 4 5 m 3 = 4 1 m 2. A few candidates seemed unconcerned with a

negative mass obtained from using (+ m 3) on the L.H.S. and there were also a few instancesof candidates quoting and using the formula m1u1+ m1v1=m2u2+ m2v2. It was rare to see

correct solutions using Impulse and many included g in their Impulse-Momentum equation.



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M1 Dynamics - Analysis of force systems PhysicsAndMathsTutor.com

1. Two forces Pand Qact on a particle. The force Phas magnitude 7 N and acts due north. The

resultant of Pand Qis a force of magnitude 10 N acting in a direction with bearing 120. Find

(i) the magnitude of Q,

(ii) the direction of Q, giving your answer as a bearing.(Total 9 marks)



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1. R= 103/2 i 5j M1 A1

Using P= 7jand Q= R Pto obtain Q = 53i 12j M1 A1

Magnitude = [(53)2+ 122] 14.8 N (AWRT) M1 A1

angle with i = arctan (12/53) 64.2 M1 A1

bearing 144 (AWRT) A1 9

Alternative method

120

PQ

R

Vector triangle correct B1

Q2= 102+ 72+ 2 10 7 cos 60 M1 A1

Q 14.8 N (AWRT) A1

sin

10

120sin

8.14= M1 A1

= 35.8, bearing 144 (AWRT) M1 A1, A1[9]



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1. This question proved to be the most demanding on the paper. The majority attempted it by

trying to draw a vector triangle, but the triangles drawn were often unclear and rarely correct

(with quite a few right-angled triangles drawn or assumed). Others attempted to use coordinates,

though often made mistakes in using the implied equation P+ Q= R(instead simply adding the

two given vectors, i.e. assuming P+ R= Q). The presentation of work here was also very poor,

with calculations or numbers often splayed all over the page with no clear justification for whatwas being attempted. Fully correct solutions were seen, but only occasionally!



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M1 Dynamics - Connected particles PhysicsAndMathsTutor.com

1.

Two particlesAandBhave mass 0.4 kg and 0.3 kg respectively. The particles are attached to

the ends of a light inextensible string. The string passes over a small smooth pulley which is

fixed above a horizontal floor. Both particles are held, with the string taut, at a height of 1m

above the floor, as shown in the diagram above. The particles are released from rest and in the

subsequent motionBdoes not reach the pulley.

(a) Find the tension in the string immediately after the particles are released.(6)

(b) Find the acceleration ofAimmediately after the particles are released.(2)

When the particles have been moving for 0.5 s, the string breaks.

(c) Find the further time that elapses untilBhits the floor.(9)

(Total17marks)



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2.

Two particlesAandBhave masses 5mand kmrespectively, where k< 5. The particles areconnected by a light inextensible string which passes over a smooth light fixed pulley. The

system is held at rest with the string taut, the hanging parts of the string vertical and with Aand

Bat the same height above a horizontal plane, as shown in Figure 4. The system is released

from rest. After release,Adescends with acceleration .4

1g

(a) Show that the tension in the string asAdescends is .4

15mg

(3)

(b) Find the value of k.(3)

(c) State how you have used the information that the pulley is smooth.(1)

After descending for 1.2 s, the particleAreaches the plane. It is immediately brought to rest by

the impact with the plane. The initial distance betweenBand the pulley is such that, in the

subsequent motion,Bdoes not reach the pulley.

(d) Find the greatest height reached byBabove the plane.(7)

(Total14marks)



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3. A car of mass 800 kg pulls a trailer of mass 200 kg along a straight horizontal road using a light

towbar which is parallel to the road. The horizontal resistances to motion of the car and the

trailer have magnitudes 400 N and 200 N respectively. The engine of the car produces a

constant horizontal driving force on the car of magnitude 1200 N. Find

(a) the acceleration of the car and trailer,(3)

(b) the magnitude of the tension in the towbar.(3)

The car is moving along the road when the driver sees a hazard ahead. He reduces the forceproduced by the engine to zero and applies the brakes. The brakes produce a force on the car of

magnitudeFnewtons and the car and trailer decelerate. Given that the resistances to motion are

unchanged and the magnitude of the thrust in the towbar is 100 N,

(c) find the value ofF.(7)

(Total 13 marks)



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4.

One end of a light inextensible string is attached to a block Pof mass 5 kg. The blockPis held

at rest on a smooth fixed plane which is inclined to the horizontal at an angle , where

.5

3sin = The string lies along a line of greatest slope of the plane and passes over a smooth

light pulley which is fixed at the top of the plane. The other end of the string is attached to a

light scale pan which carries two blocks QandR, with block Qon top of blockR, as shown in

Figure 3. The mass of block Qis 5 kg and the mass of blockRis 10 kg. The scale pan hangs at

rest and the system is released from rest. By modelling the blocks as particles, ignoring air

resistance and assuming the motion is uninterrupted, find

(a) (i) the acceleration of the scale pan,

(ii) the tension in the string,(8)

(b) the magnitude of the force exerted on block Qby blockR,(3)

(c) the magnitude of the force exerted on the pulley by the string.(5)

(Total 16 marks)



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5.

A m( ) P

B(2m)

Two particlesA andB, of mass m and 2m respectively, are attached to the ends of a light

inextensible string. The particleA lies on a rough horizontal table. The string passes over a

small smooth pulleyP fixed on the edge of the table. The particleB hangs freely below the

pulley, as shown in the diagram above. The coefficient of friction betweenA and the table is .

The particles are released from rest with the string taut. Immediately after release, the

magnitude of the acceleration ofA andB is g94 . By writing down separate equations of motion

forA andB,

(a) find the tension in the string immediately after the particles begin to move,(3)

(b) show that3

2= .

(5)

WhenB has fallen a distance h, it hits the ground and does not rebound. ParticleA is then a

distance h3

1 fromP.

(c) Find the speed ofA as it reachesP.(6)

(d) State how you have used the information that the string is light.(1)

(Total 15 marks)



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6.

P(0.5 kg)

Q m( kg)

3.15 m

Two particlesP and Q have mass 0.5 kg and m kg respectively, where m < 0.5. The particles are

connected by a light inextensible string which passes over a smooth, fixed pulley. InitiallyP is

3.15 m above horizontal ground. The particles are released from rest with the string taut and the

hanging parts of the string vertical, as shown in the diagram above. AfterP has been descending

for 1.5 s, it strikes the ground. ParticleP reaches the ground before Q has reached the pulley.

(a) Show that the acceleration ofP as it descends is 2.8 m s2.(3)

(b) Find the tension in the string asP descends.(3)

(c) Show that m =18

5.

(4)

(d) State how you have used the information that the string is inextensible.(1)

WhenP strikes the ground,P does not rebound and the string becomes slack. Particle Q then

moves freely under gravity, without reaching the pulley, until the string becomes tautagain.

(e) Find the time between the instant whenP strikes the ground and the instant when the

string becomes taut again.(6)

(Total 17 marks)



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7.

0.8 m

Q(2 kg)P(3 kg)

30

A

The diagram above shows two particlesP and Q, of mass 3 kg and 2 kg respectively, connected

by a light inextensible string. InitiallyP is held at rest on a fixed smooth plane inclined at 30 to

the horizontal. The string passes over a small smooth light pulley A fixed at the top of the plane.

The part of the string fromP toA is parallel to a line of greatest slope of the plane. The particle

Q hangs freely belowA. The system is released from rest with the string taut.

(a) Write down an equation of motion forP and an equation of motion for Q.(4)

(b) Hence show that the acceleration of Q is 0.98 m s2.(2)

(c) Find the tension in the string.(2)

(d) State where in your calculations you have used the information that the string is

inextensible.(1)

On release, Q is at a height of 0.8 m above the ground. When Q reaches the ground, it is brought

to rest immediately by the impact with the ground and does not rebound. The initial distance of

P fromA is such that in the subsequent motionP does not reachA. Find

(e) the speed of Q as it reaches the ground,(2)



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(f) the time between the instant when Q reaches the ground and the instant when the string

becomes taut again.(5)

(Total 16 marks)

8. A car is towing a trailer along a straight horizontal road by means of a horizontal tow-rope. The

mass of the car is 1400 kg. The mass of the trailer is 700 kg. The car and the trailer are modelled

as particles and the tow-rope as a light inextensible string. The resistances to motion of the car

and the trailer are assumed to be constant and of magnitude 630 N and 280 N respectively. The

driving force on the car, due to its engine, is 2380 N. Find

(a) the acceleration of the car,(3)

(b) the tension in the tow-rope.(3)

When the car and trailer are moving at 12 m s1, the tow-rope breaks. Assuming that the driving

force on the car and the resistances to motion are unchanged,

(c) find the distance moved by the car in the first 4 s after the tow-rope breaks.

(6)

(d) State how you have used the modelling assumption that the tow-rope is inextensible.(1)

(Total 13 marks)

9.

(smooth) (rough)

30 30

A m(3 ) B m( )

A fixed wedge has two plane faces, each inclined at 30 to the horizontal. Two particles Aand

B, of mass 3mand mrespectively, are attached to the ends of a light inextensible string. Each

particle moves on one of the plane faces of the wedge. The string passes over a small smooth

light pulley fixed at the top of the wedge. The face on whichAmoves is smooth. The face on

whichBmoves is rough. The coefficient of friction betweenBand this face is .ParticleAis

held at rest with the string taut. The string lies in the same vertical plane as lines of greatest

slope on each plane face of the wedge, as shown in the figure above.



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The particles are released from rest and start to move. ParticleAmoves downwards andB

moves upwards. The accelerations ofAandBeach have magnitude g101 .

(a) By considering the motion ofA, find, in terms of mandg, the tension in the string.(3)

(b) By considering the motion ofB, find the value of .(8)

(c) Find the resultant force exerted by the string on the pulley, giving its magnitude and

direction. (3)(Total 14 marks)

10.

15

This figure shows a lorry of mass 1600 kg towing a car of mass 900 kg along a straight

horizontal road. The two vehicles are joined by a light towbar which is at an angle of 15to theroad. The lorry and the car experience constant resistances to motion of magnitude 600 N and

300 N respectively. The lorrys engine produces a constant horizontal force on the lorry of

magnitude 1500 N. Find

(a) the acceleration of the lorry and the car,(3)

(b) the tension in the towbar.(4)

When the speed of the vehicles is 6 m s1, the towbar breaks. Assuming that the resistance to the

motion of the car remains of constant magnitude 300 N,

(c) find the distance moved by the car from the moment the towbar breaks to the moment

when the car comes to rest.

(4)



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(d) State whether, when the towbar breaks, the normal reaction of the road on the car is

increased, decreased or remains constant. Give a reason for your answer.(2)

(Total 13 marks)

11.

A(0.5 kg)P

B(0.8 kg)

A block of woodAof mass 0.5 kg rests on a rough horizontal table and is attached to one end of

a light inextensible string. The string passes over a small smooth pulley Pfixed at the edge of

the table. The other end of the string is attached to a ballBof mass 0.8 kg which hangs freely

below the pulley, as shown in the diagram above. The coefficient of friction betweenAand the

table is . The system is released from rest with the string taut. After release, Bdescends a

distance of 0.4 m in 0.5 s. ModellingAandBas particles, calculate

(a) the acceleration ofB,(3)

(b) the tension in the string,(4)

(c) the value of .(5)



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(d) State how in your calculations you have used the information that the string is

inextensible.(1)

(Total 13 marks)

12.

3 kg m kg

The particles have mass 3 kg and mkg, where m< 3. They are attached to the ends of a light

inextensible string. The string passes over a smooth fixed pulley. The particles are held in

position with the string taut and the hanging parts of the string vertical, as shown in the diagram

above. The particles are then released from rest. The initial acceleration of each particle has

magnitude73g. Find

(a) the tension in the string immediately after the particles are released,(3)


(Total 7 marks)

13.

P (4 kg) Q(6 kg) 40 N



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Two particlesPand Q, of mass 4 kg and 6 kg respectively, are joined by a light inextensible

string. Initially the particles are at rest on a rough horizontal plane with the string taut. The

coefficient of friction between each particle and the plane is7

2 . A constant force of magnitude

40 N is then applied to Qin the directionPQ, as shown in the diagram above.

(a) Show that the acceleration of Qis 1.2 m s2.(4)

(b) Calculate the tension in the string when the system is moving.(3)

(c) State how you have used the information that the string is inextensible.(1)

After the particles have been moving for 7 s, the string breaks. The particle Qremains under the

action of the force of magnitude 40 N.

(d) Show thatPcontinues to move for a further 3 seconds.(5)

(e) Calculate the speed of Qat the instant whenPcomes to rest. (4)(Total 17 marks)

14.

30

A(4 kg)

B(3 kg)



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A particleAof mass 4 kg moves on the inclined face of a smooth wedge. This face is inclined at

30 to the horizontal. The wedge is fixed on horizontal ground. ParticleAis connected to a

particleB, of mass 3 kg, by a light inextensible string. The string passes over a small light

smooth pulley which is fixed at the top of the plane. The section of the string fromAto thepulley lies in a line of greatest slope of the wedge. The particleB hangs freely below the pulley,

as shown in the diagram above. The system is released from rest with the string taut. For the

motion beforeAreaches the pulley and beforeBhits the ground, find

(a) the tension in the string,(6)

(b) the magnitude of the resultant force exerted by the string on the pulley.(3)

(c) The string in this question is described as being light.

(i) Write down what you understand by this description.

(ii) State how you have used the fact that the string is light in your answer to part (a).

(2)(Total 11 marks)

15.

1 m

1.4 m

P

30

A m( kg)

B (0.4kg)



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The diagram above shows two particlesAandB, of mass mkg and 0.4 kg respectively,

connected by a light inextensible string. InitiallyAis held at rest on a fixed smooth plane

inclined at 30 to the horizontal. The string passes over a small light smooth pulley Pfixed at

the top of the plane. The section of the string fromAtoPis parallel to a line of greatest slope ofthe plane. The particleBhangs freely belowP. The system is released from rest with the string

taut andBdescends with acceleration51g.

(a) Write down an equation of motion forB.(2)

(b) Find the tension in the string.(2)

(c) Prove that m=3516 .

(4)

(d) State where in the calculations you have used the information thatPis a light smooth

pulley.(1)

On release,Bis at a height of one metre above the ground andAP = 1.4 m. The particleB

strikes the ground and does not rebound.

(e) Calculate the speed ofB as it reaches the ground.(2)

(f) Show thatAcomes to rest as it reachesP.(5)

(Total 16 marks)



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16. A car which has run out of petrol is being towed by a breakdown truck along a straight

horizontal road. The truck has mass 1200 kg and the car has mass 800 kg. The truck is

connected to the car by a horizontal rope which is modelled as light and inextensible. The

trucks engine provides a constant driving force of 2400 N. The resistances to motion of thetruck and the car are modelled as constant and of magnitude 600 N and 400 N respectively. Find

(a) the acceleration of the truck and the car,(3)

(b) the tension in the rope.(3)

When the truck and car are moving at 20 m s1, the rope breaks. The engine of the truck

provides the same driving force as before. The magnitude of the resistance to the motion of the

truck remains 600 N.

(c) Show that the truck reaches a speed of 28 m s1approximately 6 s earlier than it would

have done if the rope had not broken.(7)

(Total 13 marks)



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17.

0.6 m

AP

B

A particleA of mass 0.8 kg rests on a horizontal table and is attached to one end of a light

inextensible string. The string passes over a small smooth pulley Pfixed at the edge of the table.

The other end of the string is attached to a particleBof mass 1.2 kg which hangs freely below

the pulley, as shown in the diagram above. The system is released from rest with the string taut

and withBat a height of 0.6 m above the ground. In the subsequent motionAdoes not reachP

beforeBreaches the ground. In an initial model of the situation, the table is assumed to be

smooth. Using this model, find

(a) the tension in the string beforeBreaches the ground,(5)

(b) the time taken byBto reach the ground.(3)

In a refinement of the model, it is assumed that the table is rough and that the coefficient of

friction betweenAand the table is51 . Using this refined model,

(c) find the time taken byBto reach the ground.

(8)(Total 16 marks)



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1. (a) ( )0.4 0.4g T a = M1 A1

( ) 0.3 0.3T g a = M1 A1

solving for T DM1

T= 3.36 or 3.4 or 12g/35 (N) A1 6

(b) 0.4 0.3 0.7g g a = DM1-21.4 m sa= ,g/7 A1 2

(c) ( )v u at = +

0.5 x 1.4v= M1

= 0.7 A1 ft on a

212

( )s ut at = + 2

0.5 x 1.4 x 0.5s= M10.175= A1 ft on a

212

( )s ut at = + 21.175 = 0.7 4.9t t + DM1 A1 ft

24.9 0.7 1.175 = 0t t 20.7 0.7 19.6 x 1.175

9.8t

+= DM1 A1 cao

0.5663..or ...=

Ans 0.57 or 0.566 s A1 cao 9

[17]

2. (a) N2L A: gmTmg4

155 = M1 A1

mgT4

15= * cso A1 3

(b) N2L B: T kmg= km g41 M1 A1

k= 3 A1 3

(c) The tensions in the two parts of the string are the same B1 1



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(d) Distance of A above ground ( )764.118.02.14

1

2

1 21 == ggs M1 A1

Speed on reaching ground ( )94.23.02.141

== ggv M1 A1

ForBunder gravity ( ) ( )

( )441.02

3.023.0

2

22

2=== gsgsg M1 A1

S= 2s1 +s2=3.969 4.0 (m) A1 7

[14]

3. (a) For whole system: 1200 400 200 = 1000a M1 A1

a= 0.6 m s2 A1 3

(b) For trailer: T 200 = 200 0.6 M1 A1 ft

T= 320 N A1

OR:

OR: For car: 1200 400 T= 800 0.6 M1 A1 ft

T= 320 N A1 3

(c) For trailer: 200 + 100 = 200for 200f M1 A1

f= 1.5 m s2(1.5) A1

For car: 400 +F 100 = 800for 800f M1 A2

F= 900 A1 7

(N.B. For both: 400 + 200 +F= 1000f)[13]

4. (a) T 5gsin= 5a M1 A1

15g T= 15a M1 A1

solving for a M1

a= 0.6g A1

solving for T M1

T= 6g A1 8

(b) For Q: 5g N= 5a M1 A1

N= 2g A1 f.t. 3



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(c)

)2

90cos(2

TF= M1 A2

= 12gcos 26.56.. A1 f.t.

= 105 N A1 5[16]

5. (a) B: 2mg T= 2m 4g/9 M1A1

T= 10mg/9 A1 3

(b) A: T mg= m 4g/9 M1B1A1Sub for Tand solve: = 2/3* DM1A1 5

(c) WhenBhits: v2= 2 4g/9 h M1A1

Deceleration ofAafterBhits: ma= mga= 2g/3 M1A1ftSpeed ofAatP: V2= 8gh/9 2 2g/3 h/3 DM1

V=3

2(gh) A1 6

(d) Same tension onA andB B1 1[15]

6. (a) s= ut+4

9

2

115.32

1 2 = aat M1A1

a= 2.8 (m s2) * cso A1 3

(b) N2L forP: 0.5g T= 0.5 2.8 M1A1

T= 3.5 (N) A1 3

(c) N2L for Q: T mg= 2.8m M1A1

m= *

18

5

6.12

5.3= cso DM1A1 4



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(d) The acceleration ofPis equal to the acceleration of Q. B1 1

(e) v= u+ atv= 2.8 1.5 M1A1(or v2= u2+ 2asv2= 2 2.8 3.15)(v2= 17.64, v= 4.2)

v= u+ at4.2 = 4.2 + 9.8t DM1A1

t=7

6, 0.86, 0.857 (s) DM1A1 6

[17]

7. (a) N2L Q 2g T =2a M1 A1

N2LP T 3g sin 30 = 3a M1 A1 4

(b) 2g 3g sin 30 = 5a M1

a = 0.98 (m s2) * cso A1 2

(c) T =2(g a) or equivalent M1

18 (N) accept 17.6 A1 2

(d) The (magnitudes of the) accelerations ofP and Q are equal B1 1

(e) v2= u2+ 2as v2=2 0.98 0.8 (=1.568) M1v1.3 (m s1) accept 1.25 A1 2

(f) N2L forP 3g sin 30 = 3a

a = g2

1)( M1 A1

s =ut +

2

1at20= 1.568t 29.4

2

1t or equivalent M1 A1

t = 0.51 (s) accept 0.511 A1 5[16]

A maximum of one mark can be lost for giving too great accuracy.

8. (a) Car + trailer: 2100a= 2380 280 630 M1 A1

= 1470 a= 0.7 m s2 A1 3

M1 for a complete (potential) valid method to get a



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(b) e.g. trailer 700 0.7 = T280 M1 A1ft

T = 770 N A1 3If consider car: then get 1400a = 2380 630 T.

Allow M1 A1 for equn of motion for car or trailer whereverseen (e.g. in (a)).

So if consider two separately in (a), can get M1 A1 from (b) for

one equation; then M1 A1 from (a) for second equation, and

then A1 [(a)] for a and A1 [(b)] for T.

In equations of motion, M1 requires no missing or extra terms

and dimensionally correct (e.g. extra force, or missing mass, is

M0).

If unclear which body is being considered, assume that the body

is determined by the mass used. Hence if 1400a used, assume

it is the car and mark forces etc accordingly.

But allow e.g. 630/280 confused as an A error.

(c) Car: 1400a = 2380 630 M1 A1

a= 1.25 ms2 A1

distance = 12 4 + 1.25 42 M1 A1ft

= 58 m A1 6

Must be finding a new acceleration here. (If they get 1.25

erroneously in (a), and then simply assume it is the same acceln

here, it is M0).

(d) Same acceleration for car and trailer B1 1

Allow o.e. but you must be convinced they are saying that it is

same acceleration for both bodies.

E.g. acceleration constant on its own is B0

Ignore extras, but acceleration and tension same at A and B is

B0

[13]

9. (a)

3mg

T A: 3mg sin 30 T = 3m. 101 g M1 A1

T =56 mg A1 3



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(b)T R

mg

F: R(perp): R = mgcos 30 M1 A1

R(//): T mg sin 30 F= m.101 g M1 A2, 1, 0

UsingF= R M1

mgmgmgmg101

2

3

21

56 = M1

= 0.693 or 0.69 or5

32 A1 8

(c) T T

Magn of force on pulley = 2Tcos60 =56 mg M1 A1ft

Direction is vertically downwards B1 (cso) 3[14]

10. (a)

T T600

1500300

Lorry + Car: 2500a= 1500 300 600 M1 A1

a= 0.24 m s2 A1 3

(b) Car: Tcos 15 300 = 900a

ORLorry: 1500 Tcos 15 600 = 1600a M1 A1

Sub and solve: T534 N M1 A1 4

(c) 300 Deceleration of car = 300/900 = 1/3 m s1 M1 A1

Hence 62= 2 1/3 ss= 54 m M1 A1 4

(d) Vertical component of Tnow removed M1

Hence normal reaction is increased A1cso 2[13]



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11. (a) s= ut+ at2 forB: 0.4 = a(0.5)2 M1 A1

a = 3.2 m s2 A1 3

(b) N2L forB: 0.8g T = 0.8 3.2 M1 A1ftT = 5.28 or 5.3 N M1 A1 4

(c) A: F = 0.5g B1

N2L for A: T F = 0.5a M1 A1

Sub and solve = 0.75 or 0.751 M1 A1 5

(d) Same acceleration for A and B. B1 1[13]

12. (a) 3 kg: 3g T= 3 7

3g M1 A1

T=7

12g or 16.8 N or 17 N A1 3

(b) mkg: T mg= m.7

3g M1 A1

7

12g = mg+

7

3mg(Sub for Tand solve) M1

m= 1.2 A1 4[7]

13. R1

4g 6g

F2 R240

(a) F1= 72 4g(= 11.2) orF2= 7

2 6g(= 16.8) B1

System: 40 72 4g

72 6g= 10a(equn in aand not T) M1 A1

a= 1.2 m s2(*) A1 4



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(b) P: T78g= 4 1.2 orQ:40 T

712g= 6 1.2 M1 A1

T= 16 N A1 3

(c) Accelerations ofPand Qare same B1 1

(d) v= 1.2 7 = 8.4 B1

P: ()7

8g= 4aa= ()

7

2g= 2.8 M1 A1

0 = 8.4 2.8tt= 3 s (*) M1 A1 5

(e) Q: 40 7

12 g= 6a(a3.867) M1 A1

v= 8.4 + 3.867 3 = 20 m s1 M1 A1 4

[17]

14. (a) T

T

3mg

4mg

a

aR

A: T 4gsin 30 = 4a M1 A1

B: 3g T= 3a M1 A1

T=7

18g= 25.2 N M1 A1 6

(b) T

TR

R= 2Tcos 30 M1 A1N6.43or44 A1 3

(c) (i) String has no weight/mass B1

(ii) Tension in string constant, i.e. same at A and B B1 2[11]



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15. (a)T

B

0.4g0.2g

0.4g T= 0.4 5

1g M1 A1 2

(b) T=25

8g or 3.14 or 3.1 N M1 A1 2

(c)

mg

A

T

T mg sin 30 = m5

1g M1 A1

m=35

16 (*) M1 A1 4

(d) Same Tfor A & B B1 1

(e) v2= 2 5

1g 1 M1

v=5

2g1.98 or 2 ms1 A1 2

(f) A: 2

1mg= maa=

2

1g M1 A1

v2=

5

2g 2

2

1g 0.4 M1 A1

v= 0 A1 5[16]

16. (a) Car + truck: 2000a= 2400 600 400 M1 A1

a= 0.7 m s2 A1 3



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(b) Car only: T 400 = 800 0.7 M1 A1 ft

[ortruck only: 2400 T 600 = 1200 0.7]

T = 960 N A1 3

(c) New acceleration of truck agiven by1200 a= 2400 600 M1a= 2400 600 = 1.5 m s1 A1

Time to reach 28 m s1=5.1

2028 = 5.33 s M1 A1

Time to reach 28 m s1if rope had not broken =7.0

2028 = 11.43 s M1 A1

Difference = 6.1 s 6 s (*) A1 7[13]

17. (a)A

RT

0.8g B

T

1.2g

a

A:T= 0.8a B1

B: 1.2g T= 1.2a M1 A1

Solve: T= 0.48g = 4.7 N M1 A1 5

(b) a= 0.6g = 5.88 M1

Hence 0.6 = 0.6g t2 M1

t= 0.45 or 0.452 s A1 3

F

RT

0.8g

T

1.2g

,

,

F= R=5

1 0.8g B1

A: TF= 0.8a M1 A1B: 1.2g T= 1.2a B1

Solve: a= 0.52g M1 A10.6 = 0.52g t2 M1

t= 0.49 or 0.485 s A1 8[16]



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1. In parts (a) and (b), most were able to make a reasonable attempt at two equations of motion,

but there were errors in signs and solutions. This was not helped by the fact that Twas asked for

first rather than aand some candidates lost marks due to trying to solve for T first rather than

the easier route of solving for a. A few attempted the whole system equation and these solutions

were in general less successful than those who used two separate equations to start with. In the

last part, too many candidates were unable to visualise the situation clearly and then deal with it

in a methodical fashion. If they failed to find both the velocity ofAon impact with the ground

and the distance that it had travelled they were unable to progress any further. Only the more

able students managed correct solutions. Of those that managed to progress in part (c), there

were sign errors which caused problems. Many chose to split the motion ofBinto two parts and

these were usually quite successful provided that the extra distance travelled byBin the upward

direction was taken into account.

2. Part (a) was reasonably well done by the majority of students, with good use of the printedanswer to correct sign errors etc. but there was less success in the second part, with omission of

m and/or g from some terms. The mark in part (c) was very rarely scored and candidates should

be aware that if they give a list of answers they will not be awarded the mark, even if the

correct answer appears in their list. The final part was a good discriminator and led to this

question being the worst answered question on the paper. Consideration of two stages to the

motion was required, with two distinct accelerations. Many completely omitted the motion

under gravity and found the distance moved byA and either gave that as their answer or else just

doubled it.

3. Part (a) was well done by the majority of candidates and a good number went on to use the

answer correctly in part (b). If mistakes were made they were the usual sign errors or more

seriously, in terms of marks lost, missing terms.

The third part was poorly done. There was confusion over the direction of the forces and the

concept of thrust. A few candidates halved the thrust and used 50N in each equation. Some used

the values of the acceleration and tension from previous parts.

4. In part (a), most candidates were able to set up the two equations of motion, one for each of the

two particles and most then went on to solve these correctly to find values for both Tand a. Afew persist in trying to use a whole system equation to find a, usually with limited success. In

the second part the vast majority of candidates were unable to select the correct particle, forces

or equation to score any of the marks. Part (c) also proved to be discriminating, with some

weaker candidates not attempting it. Only a minority of candidates managed to produce a

correct solution. Of those who did, many used the cosine rule applied to a vector triangle, or a

resolution into two perpendicular components. Common misconceptions involved using just T+

Tsin/cos alpha or answers involving components of 5g and 15g. Many had difficulty in

identifying the correct size for the angle whichever method was attempted. A few very good

candidates realised that the force acted along the angle bisector and scored five quick marks.



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10. Candidates found this question more challenging and fully correct answers to all parts were not

so common, though many could make good progress with some parts of the question. In part

(a), many could do this correctly, and many this time did adopt the approach of considering the

whole system (car + trailer together) to obtain the acceleration correctly. Considerably more

problems were found with part (b) and only the best candidates seemed able to work out

correctly the relevant forces acting on the car or the trailer. Many simply ignored the fact that

the tension was acting at an angle and so failed to make any attempt to resolve it. However,

many managed to recover well and produce correct answers to part (c) where it was pleasing to

see so many correctly coping with the changed situation in terms of the forces. A common error

was to assume that the acceleration found in (a) was now the deceleration of the car on its own.

In part (d) a number of well argued answers were given, explaining that the normal reaction

increased. Some however gave totally irrelevant reasons (e.g. that the friction changed, or that

the weight remained constant and hence the normal reaction remained constant). This part again

proved to be a good discriminator for more able candidates.

11. This question was generally well answered and it was pleasing to see candidates being able to

write down equations of motion for the two particles separately. Mistakes from weaker

candidates arose from sometime including the weight ofAin the (horizontal) equation of

motion forA, or confusing the two particles and the forces acting on them. Most realised that

they had to use the given data to solve part (a) though a few launched straight into writing down

the equations of motion and then floundering when they did not have enough information to

solve these. Answers to part (d) were almost uniformly incorrect: the vast majority stated that

the inextensibility of the string meant that the tensions were the same (or constant throughout

the string).

12. Again this was generally well done with candidates showing themselves able to write down

separate equations of motion for each particle with very attempting the (mathematically

spurious) equation for the whole system approach. However, some mistakes occurred with

missinggs. Also some failed to realise the mechanical significance of the condition m< 3 and

assumed that acceleration was in the wrong direction.

13. The question was generally well done by some candidates, but some evidently failed tounderstand the mechanics of connected bodies. For those who knew what they were doing,

setting up and solving the relevant equations of motion was relatively easy, though quite a few

failed to take the easiest route for part (a) by considering the motion of the whole system. The

existence of the given answer in part (a) also led to a number of fudged solutions here. It was

disappointing to see virtually no correct answers to part (c): the vast majority stated that the

inextensibility of the string meant that the tension was the same at both ends (or constant). In

parts (d) and (e), terms were sometimes omitted in calculations of the accelerations, and weaker

candidates assumed that the acceleration was still 1.2 m s2. Again, somewhat messy working

often hindered many candidates in this question. Standard of presentation was often quite poor

with working sometimes apparently distributed randomly over the paper.



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14. This type of connected particle question seemed to be much more familiar to many, and hence

part (a) was generally well done: most could write down two equations of motion and solve

them successfully. Some however failed to realise that the weight ofAneeded resolving when

considering the motion up/down the slope. Very few however realised what was required in part

(b). Again the fact that the tensions at the different ends of the string were the same in

magnitude did not seem to be appreciated, so that quite a few appeared to think that the resultant

force on the pulley was made up of the components of the two weights; others simply assumed

that the two tensions were acting perpendicular to each other. A number who realised what to do

also lost a mark by failing to round their answer to no more than 3 significant figures. In part

(c), most realised that a light string is one that has no weight/mass; but very few realised what

the implications of this were for the equations they had written down earlier, viz. that the

tension in the string remained constant throughout its length.

15. In part (a), several evidently did not understand what was meant by an equation of motion,though many managed to recover by actually producing the relevant equation in their working

for part (b). In part (b), most could produce an equation for the tension, but several again lost a

mark by failing to give their answers to an appropriate degree of accuracy (again 2 or 3 s.f.).

In part (c), most could write down the equation of motion forA, but several fudged the working

which required proof that the answer was exactly the given fraction: several produced a

decimal and verified that the given fraction was (approximately) the same. It was though

disappointing to see a number of candidates failing to make any attempt to resolve the weight

when considering the motion up the inclined plane. In part (d), few realised the significance of

the modelling assumption as implying that the tension was constant throughout. Part (e) was

generally well done, but in part (f) the majority failed to find a new acceleration and simplyassumed that the acceleration was stillg/5. For those who did, there were still some problems

with the signs given to the various quantities and fully correct solutions here characterised the

stronger candidates.

16. It was pleasing to be able to report good overall performance on this question. Candidates

showed general confidence in dealing with equations of motion. Most difficulty occurred in part

(b) where one vehicle alone had to be considered, and some failed to put the tension correctly

into their equations. Part (c) was also generally well done with most showing that they realised

that they had to find a new acceleration for the new situation; several fully correct solutionswere seen here.



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17. This proved to be a popular question, and nearly all seemed to have been taught to use separate

equations for the motion of each particle, which was pleasing to see. Perhaps surprisingly,

candidates found the first part (assuming no friction) more challenging than the last part

(bringing friction in). Several felt compelled to bring in the weight ofAin the equation of

horizontal motion forA. Several too failed to round their answer for Tto an appropriate degree

of accuracy, leaving their answer to 4 significant figures despite having used a value ofgas 9.8.

Part (c) was generally well done, though weaker candidates failed to realise that the tension

could not be assumed to have the same value as before and so unjustifiably used the value from

the earlier part of the question.



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M1 Dynamics - F = ma horizontally PhysicsAndMathsTutor.com

1. A car of mass 800 kg pulls a trailer of mass 200 kg along a straight horizontal road using a lighttowbar which is parallel to the road. The horizontal resistances to motion of the car and the

trailer have magnitudes 400 N and 200 N respectively. The engine of the car produces aconstant horizontal driving force on the car of magnitude 1200 N. Find

(a) the acceleration of the car and trailer,(3)

(b) the magnitude of the tension in the towbar.(3)

The car is moving along the road when the driver sees a hazard ahead. He reduces the forceproduced by the engine to zero and applies the brakes. The brakes produce a force on the car ofmagnitudeFnewtons and the car and trailer decelerate. Given that the resistances to motion areunchanged and the magnitude of the thrust in the towbar is 100 N,

(c) find the value ofF.(7)

(Total 13 marks)

2.

P (2kg) Q (3kg)

F

Two particlesP and Q, of mass 2 kg and 3 kg respectively, are joined by a light inextensiblestring. Initially the particles are at rest on a rough horizontal plane with the string taut. Aconstant force F of magnitude 30 N is applied to Q in the directionPQ, as shown in the diagramabove. The force is applied for 3 s and during this time Q travels a distance of 6 m. Thecoefficient of friction between each particle and the plane is. Find

(a) the acceleration of Q,(2)

(b) the value of,(4)

(c) the tension in the string.(4)



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(d) State how in your calculation you have used the information that the string is

inextensible. (1)

When the particles have moved for 3 s, the force F is removed.

(e) Find the time between the instant that the force is removed and the instant that Q comesto rest.

(4)

(Total 15 marks)

3. A particleP of mass 2 kg is moving under the action of a constant force F newtons. When t = 0,

P has velocity (3i + 2j) m s1and at time t = 4 s,P has velocity (15i 4j) m s1. Find

(a) the acceleration ofP in terms of i andj,(2)

(b) the magnitude of F, (4)

(c) the velocity ofP at time t =6s.(3)

(Total 9 marks)

4.

20

PN

A box of mass 30 kg is being pulled along rough horizontal ground at a constant speed using arope. The rope makes an angle of 20 with the ground, as shown in the diagram above. Thecoefficient of friction between the box and the ground is 0.4. The box is modelled as a particleand the rope as a light, inextensible string. The tension in the rope isP newtons.

(a) Find the value ofP.(8)



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The tension in the rope is now increased to 150 N.

(b) Find the acceleration of the box.(6)

(Total 14 marks)

5. A stone Sis sliding on ice. The stone is moving along a straight horizontal lineABC, whereAB= 24 m andAC= 30 m. The stone is subject to a constant resistance to motion of magnitude 0.3

N. AtAthe speed of Sis 20 m s1, and atBthe speed of Sis 16 m s1. Calculate

(a) the deceleration of S,(2)

(b) the speed of Sat C.(3)

(c) Show that the mass of Sis 0.1 kg.(2)

At C, the stone Shits a vertical wall, rebounds from the wall and then slides back along the lineCA. The magnitude of the impulse of the wall on Sis 2.4 Ns and the stone continues to moveagainst a constant resistance of 0.3 N.

(d) Calculate the time between the instant that Srebounds from the wall and the instant that Scomes to rest.

(6)

(Total 13 marks)

6.

20



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A sledge has mass 30 kg. The sledge is pulled in a straight line along horizontal ground by

means of a rope. The rope makes an angle 20 with the horizontal, as shown in the diagramabove. The coefficient of friction between the sledge and the ground is 0.2. The sledge ismodelled as a particle and the rope as a light inextensible string. The tension in the rope is 150N. Find, to 3 significant figures,

(a) the normal reaction of the ground on the sledge,(3)

(b) the acceleration of the sledge.(3)

When the sledge is moving at 12 m s1, the rope is released from the sledge.

(c) Find, to 3 significant figures, the distance travelled by the sledge from the moment whenthe rope is released to the moment when the sledge comes to rest.

(6)

(Total 12 marks)

7. A particlePof mass 0.4 kg is moving under the action of a constant force Fnewtons. Initially

the velocity ofPis (6i 27j) m s1and 4 s later the velocity ofPis (14i+ 21j) m s1.

(a) Find, in terms of iandj, the acceleration ofP.(3)

(b) Calculate the magnitude of F.(3)

(Total 6 marks)



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1. (a) For whole system: 1200 400 200 = 1000a M1 A1

a= 0.6 m s2 A1 3

(b) For trailer: T 200 = 200 0.6 M1 A1 ft

T= 320 N A1

OR:

OR: For car: 1200 400 T= 800 0.6 M1 A1 ft

T= 320 N A1 3

(c) For trailer: 200 + 100 = 200for 200f M1 A1

f= 1.5 m s2(1.5) A1

For car: 400 +F 100 = 800for 800f M1 A2

F= 900 A1 7

(N.B. For both: 400 + 200 +F= 1000f)[13]

2. (a)

T T 30

2g 3g

s =ut +2

1at

26=

2

1a 9 M1

a = 13

1 (m s2) A1 2

(b) N2L for system 30 5g = 5a ft their a, accept symbol M1A1ft

21

10

3

14==

g or awrt 0.48 DM1A1 4

(c) N2L forP T 2g =2a ft their , their a,accept symbols M1A1ft

3

422

3

14= g

gT

Leading to T= 12 (n) awrt 12 DM1A1 4

AlternativelyN2L for Q30 T 3g =3a M1A1Leading to T = 12 (n) awrt 12 DM1A1



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(d) The acceleration ofP and Q (or the whole of the system)is the same. B1 1

(e) v =u +at v = 433

4= B1ft on a

N2L (for system or either particle)5g =5a or equivalent M1a =gv =u +at 0 = 4 gt DM1

Leading to t =7

6 (s) accept 0.86, 0.857 A1 4

[15]

3. (a) a =4

)23()415( jiji + =3i 15j M1A1 2

(b) N2L F = ma = 6i 3j ft their a M1A1

F=(62+ 32) 6.71 (N) accept 45, awrt 6.7 M1A1 4

(c) v6= (3i + 2j) + (3i 1.5j)6 ft their a M1A1ft

= 21i 7j (m s1) A1 1[9]

4. (a)

R

P

20

30g

Use ofF =R B1P cos20 =R M1A1

R + Psin20 =30g M1 A1P cos20 =(30g P sin 20) M1

P=+

20sin4.020cos

304.0 g M1

110 (N) accept 109 A1 8



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(b) i R+ 150 sin 20 = 30g M1A1(R242.7)

N2L 150 cos 20 R= 30a M1A1

a30

7.2424.020cos150 M1

= 1.5 (m s2) accept 1.46 A1 6[14]

5. (a) 162 = 202 2 a 24 a = 3 m s2 M1 A1 2

(b) v2 = 202 2 3 30 M1 A1ft

v = 220 or 14.8 m s1 A1 3

(c) 0.3 = m 3 m = 0.1 kg (*) M1 A1 2

(d) 0.1(w+ 220) = 2.4 M1 A1ft

w = 9.17 A1

0 = 9,17 3 t M1 A1ft

t 3.06 s A1 6[13]

6. (a)

30g

R 150

0.2R

R()R+ 150 sin 20 = 30g M1 A1R243 N A1 3

(b) R(): 150 cos 20 0.2R= 30a M1 A1

a3.08 m s2 A1 3



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1. Part (a) was well done by the majority of candidates and a good number went on to use theanswer correctly in part (b). If mistakes were made they were the usual sign errors or moreseriously, in terms of marks lost, missing terms.

The third part was poorly done. There was confusion over the direction of the forces and theconcept of thrust. A few candidates halved the thrust and used 50N in each equation. Some usedthe values of the acceleration and tension from previous parts.

2. Full marks were rarely achieved in this question. Some made a poor start by usingF=main part(a) rather than an appropriate constant acceleration formula. In the second part many usedseparate equations of motion for the two particles (sometimes with extra or omitted terms) butthen not uncommonly solved them as simultaneous equations with the sameF(friction term),showing a lack of understanding of the problem. Only a minority used the more straightforwardwhole system approach. There was some recovery in part (c) where follow through markswere available as long as the appropriate terms were included in the equation of motion of oneparticle. A significant number of candidates knew that an inextensible string implied that theaccelerations of the two particles were the same in part (d), but some of those went on toincorrectly mention the tension as well and so lost the mark. Many candidates who reached part(e) seemed to know they had to find the new deceleration but lost marks by including a tensionor the 30N in their equation of motion.

3. In part (a) most candidates knew the method and it was often fully correct but a number failed to

find the magnitude of the force in the second part, with some, subtracting the squares of thecomponents instead of adding them. Part (c) was well answered.

4. Good candidates found this question reasonably straightforward, but many of the weaker oneslost significant numbers of marks because they thought that R = 30g. It was odd that manycandidates could get part (a) completely correct but then were unable to make any progress at allwith part (b) and didnt appreciate the similarity between them. Some marks were again lost dueto overaccurate answers. A clearly labelled diagram in each part made a huge difference.

5. Most could make good attempts at the first three parts of the question, though a misreading ofthe information (confusing AC and BC was not uncommon). In part (d) the most commonmistake was to confuse signs again (similar to qu.1) in writing down the impulse-momentumequation, but most could then go on to use their result in an appropriate way to get a value forthe time.

6. Parts (a) and (b) were well done by some with few writing down that the normal reaction in (a)

was equal to the weight; in part (b) several missed out a force in trying to find the accelerationHowever, many failed to observe the specific demand to give their answers to 3 significantfigures. In part (c), many failed to realise that the normal reaction would change: hence they



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used their working from the earlier parts of the question to find the frictional force; others usedthe answer form part (b) (for acceleration) in their working for part (c) (where the sledge isdecelerating!).

7. This was generally very well done. A few errors occurred in part (a) with candidates using u v,instead of v u, and there were some arithmetical slips. In part (b), a few simply found themagnitude of their acceleration, failing to multiply by the mass; quite a few simply found theforce as a vector, failing to understand the significance of the word magnitude here.



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M1 Dynamics - F = ma on a slope PhysicsAndMathsTutor.com

(b) show that 2p+ q+ 3 = 0.

(4)

Given also that q= 1 and thatPmoves with an acceleration of magnitude 85 m s2,

(c) find the value of m.(7)

(Total 14 marks)

4.

4 N

P

A particlePof mass 0.5 kg is on a rough plane inclined at an angle to the horizontal,

where43tan = . The particle is held at rest on the plane by the action of a force of

magnitude 4 N acting up the plane in a direction parallel to a line of greatest slope of the

plane, as shown in the figure above. The particle is on the point of slipping up the plane.

(a) Find the coefficient of friction betweenPand the plane.(7)

The force of magnitude 4 N is removed.

(b) Find the acceleration ofPdown the plane.(4)

(Total 11 marks)



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5.

30

PN

A parcel of weight 10N lies on a rough plane inclined at an angle of 30 to the horizontal. A

horizontal force of magnitudePnewtons acts on the parcel, as shown in the figure above. The

parcel is in equilibrium and on the point of slipping up the plane. The normal reaction of the

plane on the parcel is 18N. The coefficient of friction between the parcel and the plane is .

Find

(a) the value ofP,(4)

(b) the value of .(5)

The horizontal force is removed.

(c) Determine whether or not the parcel moves.(5)

(Total 14 marks)

6.

20

18 N

A box of mass 2 kg is pulled up a rough plane face by means of a light rope. The plane is

inclined at an angle of 20to the horizontal, as shown in the diagram. The rope is parallel to aline of greatest slope of the plane. The tension in the rope is 18 N. The coefficient of friction

between the box and the plane is 0.6. By modelling the box as a particle, find

(a) the normal reaction of the plane on the box,(3)



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(b) the acceleration of the box.

(5)(Total 8 marks)

7.

50 m

15

B

The diagram above shows a boatBof mass 400 kg held at rest on a slipway by a rope. The boat

is modelled as a particle and the slipway as a rough plane inclined at 15 to the horizontal. The

coefficient of friction betweenBand the slipway is 0.2. The rope is modelled as a light,

inextensible string, parallel to a line of greatest slope of the plane. The boat is in equilibrium

and on the point of sliding down the slipway.

(a) Calculate the tension in the rope.(6)

The boat is 50 m from the bottom of the slipway. The rope is detached from the boat and the

boat slides down the slipway.

(b) Calculate the time taken for the boat to slide to the bottom of the slipway.(6)

(Total 12 marks)

8. A tile on a roof becomes loose and slides from rest down the roof. The roof is modelled as a

plane surface inclined at 30 to the horizontal. The coefficient of friction between the tile and

the roof is 0.4. The tile is modelled as a particle of mass mkg.

(a) Find the acceleration of the tile as it slides down the roof.(7)



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(a) find, to 2 significant figures, the acceleration of the parcel.

(8)

The rope now breaks. The parcel slows down and comes to rest.

(b) Show that, when the parcel comes to this position of rest, it immediately starts to move

down the plane again.(4)

(c) Find, to 2 significant figures, the acceleration of the parcel as it moves down the plane

after it has come to this position of instantaneous rest.(3)

(Total 15 marks)



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1. (a) s= ut + 92

17.2

2

1 2= aat M1 A1

a= 0.6 (m s2) A1 3

(b)

R= 0.8gcos30(6.79) B1

Use ofF=R B1

0.8gsin 30 R = 0.8 a M1 A1(0.8gsin 30 0.8gcos30 = 0.8 0.6)

0.51 accept 0.507 A1 5

(c)

Rcos30 =Rcos60 + 0.8g M1 A2 (1,0)

(R12.8)

X=Rsin 30+Rsin 60 M1 A1

Solving forX, X12 accept 12.0 DM1 A1 7

Alternative

R=Xsin 30 + 0.8 9.8sin 60 M1 A2 (1,0)

R+ 0.8g cos60 =Xcos30 M1 A1

+=

30sin30cos

60cos8.060sin8.0

ggX

Solving forX, X12 accept 12.0 DM1 A1[15]

2. 0.5gsinF= 0.5a M1 A1 A1

RF31

= seen B1

R= 0.5gcos M1 A1

Use of sin=54 or cos=

53 or decimal equiv or



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decimal angle e.g 53.1 or 53 B1

or5

3ga = 5.88

m s2

or 5.9 m s2

DM1 A1[9]

3. (a)

4.6312tan == M1 A1

angle is 153.4 A1 3

(b) (4 +p)i+ (q 5)j B1

(q 5) = 2(4 +p) M1 A1

2p+ q+ 3 = 0 * A1 4

(c) 21 == pq B1

R= 2i 4j M1

20)4(2 22

=+=R M1 A1 f.t.

5820 m= M1 A1 f.t.

4

1=m A1 cao 7

[14]

4. (a)R

F

0.5g

4

R = 0.5g cos = 0.4g M1 A1

4 =F + 0.5g sin M1 A1

F=Rused M1

4 = 0.4g. + 0.3g

0.27(0) M1 A1 7

1sttwo M1s require correct number of the

correct terms,with valid attempt to resolve

the correct relevant term (valid resolve = x sin / cos).



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6. (a)

F

R

2g

18

R (perp to plane):R= 2gcos 20 M1 A1

18.4 or 18 N A1 3

(b) R (// to plane): 18 2gsin 20 F= 2a M1 A1

F= 0.6Rused B1

Sub and solve: a= 0.123 or 0.12 m s2 M1 A1 5

[8]

7. (a)

400g

T

FR

R= 400gcos 15(3786 N) B1F= 0.2Rused B1

T+ 0.2R= 400gsin 15 M1 A1

T257 or 260 N M1 A1 6

(b) 400gsin 15 0.2 400gcos 15= 400a M1 A1a= 0.643() A1

50 =21 0.643 t2 M1 A1f.t.

t= 12.5 or 12 s A1 6[12]



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8. (a)

amg

RF

R( ):R= mgcos 30 B1

R( ): ma= mgsin 30 F M1A1

F= 0.4Rused B1

EliminateR ma= mgsin 30 0.4 mgcos 30 M1

Solve: a= 4.9 0.4 9.8 3 /2 M1

1.5 or 1.51 m s2 A1 7

(b) v2= 2 1.51 3 v= 3 or 3.01 m s1 M1A1 2

(c) 1.5/1.51 m s2(same as (a)) B1 ft 1[10]

9. (a)

F

R a

3g

R():R= 3gcos 30(= 25.46 N) M1 A1F = 0.4R10.2 N (accept 10 N) M1 A1 4

(b) R(): F + 3gsin 30= 3a M1 A2

(1 eeoo)

a8.3 m s2 M1 A1

v

2

= u

2

+ 2as : 6

2

= 2 as M1s2.17 m (accept 2.2 m) A1 7[11]



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10. (a)

30

30

a 24

2g

F

N

R( )N + 24 cos 60 = 2gcos 30 M1 A1 A1

N= 16.97 12 = 4.97 N M1 A1

F= 0.4 . 4.97 = 1.99 N M1 A1

R( ) 2a = 24 cos 30 2gcos 60 1.99 A1 8

(b) a 4.5 m s2

2g

NF

R( ) N= 2gcos 30= 16.97 M1 A1

Fmax = 0.4 . 16.97 = 6.79 NComponent of weight down plane = 2gsin 30= 9.8 N M1

(c) 9.8 >Fmax net force down plane parcel moves A1 4

2f = 9.8 6.79, f 1.5 m s2 M1 A1, A1[12]



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1. Part (a) had a very high success rate and all three marks were regularly scored but the second

part was found to be more challenging. Most were able to resolve perpendicular to the plane to

find the reaction and use it to find the limiting friction. However, all too often there were

omissions from the equation of motion parallel to the plane, either the mass x acceleration term

and/or the weight component or else g was missing. Part (c) was a good discriminator andcandidates needed to realise that this was a new system and that there was no acceleration.

Those who failed to appreciate this and used their friction force from part (b) scored no marks.

The majority of successful candidates resolved parallel and perpendicular to the plane (although

a sizeable minority resolved vertically and horizontally) but even then a correct final answer

was rarely seen due to premature approximation or else it was given to too many figures.

2. This question was well done by the majority of candidates. Most made valid attempts at

resolving parallel and perpendicular to the plane. The most common error was where candidates

obtained the sin/cos of the complementary angle. Others used sin(4/5) or cos(3/5). Many

successful candidates used the actual angle 53.1 rather than working with fractions for the trig.

ratios. Some thought that the friction force was 1/3. A few managed to obtain the correct

answer fortuitously by usingR= 0.5a.

3. Many were able, in the first part, to use tan to find an acute angle, scoring two of the three

marks, but were then unable to identify and find the required angle. In part (b), the first mark

was for adding the two vectors together but many students then stated that this sum was equal to

(i 2j) rather than a multiple of it and were unable to make any progress. In the final part, many

who failed in (b), obtainedp= 2 from the printed equation and, even if their Rwas wrong,

were able to benefit from follow-through marks. It was amazing to see so many arrive correctlyat 20 = m85 then correctly write m= 25 / 85 but then give m= 5/4!

4. This was generally well done and many fully correct solutions were seen to part (a). However, a

number of weaker candidates could not handle the angle in question (e.g using 0.75 degrees);

also some weaker candidates were evidently confused about what precisely F was in the

equationsF= RandF= ma. In part (b) a number of candidates also lost marks by effectively

omitting one of the two terms in the equation of motion, forgetting about either the friction or

(more commonly) the component of the weight acting down the plane.

5. In parts (a) and (b) candidates managed to recover and most could make good attempts here.

Most could resolve perpendicular to the plane and parallel to the plane, with a correct use of the

frictional force. Some lost marks by omitting forces, but several gained at least all the method

marks here. Part (c) proved to be more discriminating, at least for gaining all 5 marks. Some

could make little progress since they did not realise that the normal reaction had now changed.

Others did realise this and could get to the stage of setting up the value of the component of the

weight down the plane and the value ofR. However it proved to be very difficult for

candidates to understand clearly that R was not necessarily the actual frictional force acting

(except in limiting equilibrium): hence there were many final answers to part (c) stating that thebox remained in equilibrium because the frictional force was greater than the component of the

weight.



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6. In part (a), most obtainedR= 2gcos 20 correctly; however, a large number failed to give their

answer to an appropriate degree of accuracy (here at most 3 significant figures), thus losing a

mark. In part (b), most realised that the frictional force was equal to R; but there was asurprisingly high number of errors in attempting to write down the equation of motion with

several omitting a force (e.g. the component of the weight) in their equation; others appeared to

be very confused with what F was, using the same F in both F = ma and F =R. The

figures involved were also quite sensitive to the accuracy presumed, and premature

approximation led to a number of incorrect answers. Candidates should be aware that, if they

give a final answer to 3 s.f., then they must work with previous values which are accurate to at

least 4 s.f.

7. Some very good answers were seen to this question with many fully correct (or all but correct)answers. However, as with Q4, many again lost a mark by giving their answers (especially in

part (a)) to 4 or more significant figures. The most common other mistake in (a) was to have the

wrong sign with the friction in the equilibrium equation. It was however pleasing to note the

very high standard generally of accuracy in processing the resulting equation here with

awkward figures involved. In part (b), most realised that they had to find the acceleration, but

several omitted one or other of the relevant forces in doing so. Most however could use their

resulting acceleration to find a value of the time appropriately.

8. This proved to be a little more demanding for the weaker candidates. Some omitted forces when

trying to write down the equation of motion of the tile down the slope. Others too failed to makeclear which direction they were assuming was positive, and then produced some rather unclear

working to come up with a positive answer at the end. Some too lost a mark by failing to give

their answer to 2 or 3 significant figures. Most realised how to tackle part (b) correctly.

In part (c) very few ignored the hint in the instruction to write down their answer (and in the

fact that there was only one mark available) and proceeded to repeat the whole calculation; even

when doing so, not many came out with the same answer as part (a).

9. This question tended to be handled very well or rather poorly. In part (a) there was sometimes

some confusion between the frictional force and the resultant force, though candidates then

often proceeded correctly in part (b). There were also a number who lost a mark for failing to

given their answer to part (a) to an appropriate degree of accuracy again (i.e. 2 or 3 significant

figures). In part (b), some correct answers were seen, but frequently candidates missed out a

force (e.g. the weight) in writing down the equation of motion to find the acceleration of the

particle. A clear diagram with the forces clearly marked would have helped candidates here to

sort out the Mechanics of the situation better.

10. No Report available for this question.



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M1 Dynamics - Momentum and impulse PhysicsAndMathsTutor.com

1. Particle Phas mass mkg and particle Qhas mass 3mkg. The particles are moving in oppositedirections along a smooth horizontal plane when they collide directly. Immediately before the

collision Phas speed 4ums1and Qhas speed kums1, where kis a constant. As a result of thecollision the direction of motion of each particle is reversed and the speed of each particle ishalved.

(a) Find the value of k.(4)

(b) Find, in terms of mand u, the magnitude of the impulse exerted on Pby Q.(3)

(Total7marks)

2. A particleAof mass 2 kg is moving along a straight horizontal line with speed 12 m s1.Another particleBof mass mkg is moving along the same straight line, in the opposite direction

toA, with speed 8 m s1. The particles collide. The direction of motion ofAis unchanged by the

collision. Immediately after the collision,Ais moving with speed 3 m s1andBis moving with

speed 4 m s1. Find



(Total6marks)

3. Two particlesAandBhave mass 0.4 kg and 0.3 kg respectively. They are moving in oppositedirections on a smooth horizontal table and collide directly. Immediately before the collision,

the speed ofAis 6 m s1 and the speed ofBis 2 m s1. As a result of the collision, the direction

of motion ofBis reversed and its speed immediately after the collision is 3 m s1. Find

(a) the speed ofAimmediately after the collision, stating clearly whether the direction ofmotion ofAis changed by the collision,

(4)



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6. A particle Pof mass 1.5 kg is moving along a straight horizontal line with speed 3 m s1.Another particle Qof mass 2.5 kg is moving, in the opposite direction, along the same straight

line with speed 4 m s1. The particles collide. Immediately after the collision the direction of

motion of Pis reversed and its speed is 2.5 m s1.

(a) Calculate the speed of Qimmediately after the impact.(3)

(b) State whether or not the direction of motion of Qis changed by the collision.(1)

(c) Calculate the magnitude of the impulse exerted by Qon P, giving the units of youranswer.

(3)

(Total 7 marks)

7. A stone Sis sliding on ice. The stone is moving along a straight horizontal lineABC, whereAB= 24 m andAC= 30 m. The stone is subject to a constant resistance to motion of magnitude 0.3

N. AtAthe speed of Sis 20 m s1, and atBthe speed of Sis 16 m s1. Calculate

(a) the deceleration of S,(2)

(b) the speed of Sat C.(3)

(c) Show that the mass of Sis 0.1 kg.

(2)

At C, the stone Shits a vertical wall, rebounds from the wall and then slides back along the lineCA. The magnitude of the impulse of the wall on Sis 2.4 Ns and the stone continues to moveagainst a constant resistance of 0.3 N.

(d) Calculate the time between the instant that Srebounds from the wall and the instant that Scomes to rest.

(6)

(Total 13 marks)



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8. A tent peg is driven into soft ground by a blow from a hammer. The tent peg has mass 0.2 kgand the hammer has mass 3 kg. The hammer strikes the peg vertically.

Immediately before the impact, the speed of the hammer is 16 m s1

. It is assumed that,immediately after the impact, the hammer and the peg move together vertically downwards.

(a) Find the common speed of the peg and the hammer immediately after the impact.(3)

Until the peg and hammer come to rest, the resistance exerted by the ground is assumed to beconstant and of magnitudeRnewtons. The hammer and peg are brought to rest 0.05 s after theimpact.

(b) Find, to 3 significant figures, the value ofR.(5)

(Total 8 marks)

9. A particle Pof mass 2 kg is moving with speed um s1in a straight line on a smooth horizontalplane. The particle Pcollides directly with a particle Qof mass 4 kg which is at rest on the samehorizontal plane. Immediately after the collision, Pand Qare moving in opposite directions andthe speed of Pis one-third the speed of Q.

(a) Show that the speed of Pimmediately after the collision is51 um s1.

(4)

After the collision Pcontinues to move in the same straight line and is brought to rest by aconstant resistive force of magnitude 10 N. The distance between the poin

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