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Transcript of M2 Exam Questions
8/2/2019 M2 Exam Questions
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Mechanics M2 Exam
Questions
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Click to go straight to a particular topic
Moments
Centre of Mass
Collisions
Work Energy Power
Kinematics (Vectors)
Work Energy Power 2
Projectiles
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Question 1
• A uniform rod AB , of length 8a and weight W , is free torotate in a vertical plane about a smooth pivot at A . One endof a light inextensible string is attached to B . The other endis attached to point C which is vertically above A , with AC =6a . The rod is in equilibrium with AB horizontal, as shownbelow.
6a
A B
C
8a
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W
T
X
Y
(a ) By taking moments about A , or otherwise, show that thetension in the string is 5/6W . (4)
6 3in
10 5
a S B
a
34 8
5aW aT
6a
A B
C
8a
Add the forces to diagram.
By Pythag CB = 10a
Taking moments about A
4aW=8aTSinB
5
6T W
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(b ) Calculate the magnitude of the horizontal component of theforce exerted by the pivot on the rod.
(3)
8 4
10 5
a CosB
a
5
6T W
Add the forces to diagram.
Resolving forces horizontally.
X = TCosB
Therefore2
3X W
W
T
X
Y6a
A B
C
8a
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Question 2Figure 2 shows a metal plate that is made by removing a circleof centre O and radius 3 cm from a uniform rectangular laminaABCD , where AB = 20 cm and BC = 10 cm. The point O is 5 cmfrom both AB and CD and is 6 cm from AD .
A
5 cm
D
6 cm
20 cm
C
B
10 cm
O
3 cm
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(a ) Calculate, to 3 significant figures, the distance of thecentre of mass of the plate from AD . (5)
Circle Rectangle Plate
Masses 9∏ 200 200-9∏
Centreof mass
6 10 x
A
5
cm
D
6
cm
20 cm
C
B
10
cm
O
3
cm
Centre of mass will lie onmirror line.
X = 10.7cm
9 6 (200 9 ) 200 10x
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The plate is freely suspended from A and hangs in equilibrium.(b ) Calculate, to the nearest degree, the angle between AB
and the vertical. (3)
G will be directly below A. A B
D C
G
Therefore angle GAB is given by:
1 525
10.7Tan
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Question 3A small package P is modelled as a particle of mass 0.6 kg. Thepackage slides down a rough plane from a point S to a point T ,where ST = 12 m. The plane is inclined at an angle of 30 to thehorizontal and ST is a line of greatest slope of the plane, asshown in Figure 3. The speed of P at S is 10 m s–1 and the speedof P at T is 9 m s–1.
12 m
S
P
T
30
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Calculate(a ) the total loss of energy of P in moving from S to T , (4)
12
m
S
P
T
30
KE at S = ½ × 0.6 × 100 = 30Js
KE at T = ½ × 0.6 × 81 = 24.3Js
KE lost = 5.7Js
PE Lost = mgh = 0.6 × 9.8 × 12Sin30º
= 35.28Js
Total loss of energy = 41.0Js
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Calculate(b ) the coefficient of friction between P and the plane. (5)
12
m
S
P
T
30
Add forces to diagram, and resolveperpendicular to the plane.
R = 0.6 × 9.8 × Sin30º = 5.09
Using F = μR F = μ × 5.09
Work done against F = loss of energy
μ × 5.09 × 12 = 40.98
R
F
μ = 0.67
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Question 4A particle P of mass 0.4 kg is moving under the action of a singleforce F newtons. At time t seconds, the velocity of P , v m s–1, isgiven by
v = (6t + 4)i + (t 2 + 3t ) j.When t = 0, P is at the point with position vector (–3i + 4 j) m.
(a ) Calculate the magnitude of F when t = 4. (4)
Using F = ma and the fact that the acceleration is differentialof the velocity vector.
v = (6t + 4)i + (t 2 + 3t ) j.
a = 6i + (2t + 3) j
When t = 4, a = 6i + 11 j therefore F = 0.4 × (6i + 11 j)
Magnitude of F =√(2.42 + 4.42) = 5.0
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When t = 4, P is at the point S .(b ) Calculate the distance OS . (5)
32 2
1 2
t 3r=(3t + 4t)i+( + t )j+C i C j
3 2
2r = i (6t + 4)dt + j (t + 3t)dt
The position vector is found by integrating the velocity vector
When t = 0, P has position vector (-3i + 4 j).
When t = 4, r = 61i +49⅓j
32 2t 3
r=(3t + 4t-3)i+( + t +4)j
3 2
22 1
OS= 61 49 78m3
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Question 5A car of mass 1000 kg is towing a trailer of mass 1500 kg along astraight horizontal road. The tow-bar joining the car to the
trailer is modelled as a light rod parallel to the road. The totalresistance to motion of the car is modelled as having constantmagnitude 750 N. The total resistance to motion of the trailer ismodelled as of magnitude R newtons, where R is a constant.When the engine of the car is working at a rate of 50 kW, thecar and the trailer travel at a constant speed of 25 m s–1.
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(a ) Show that R = 1250. (3)
Power = Force × Velocity
Force = (50000/25) = 2000N
Constant velocity implies that F = 750 + R
Therefore R = 1250N
1000kg1500kg
750NRNF
T T
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When travelling at 25 m s–1 the driver of the car disengages theengine and applies the brakes. The brakes provide a constantbraking force of magnitude 1500 N to the car. The resistingforces of magnitude 750 N and 1250 N are assumed to remainunchanged. Calculate
(b ) the deceleration of the car while braking, (3)
Using F = ma 2000N + 1500N = 2500a Therefore a = 1.4ms-2
1000kg1500kg
750N1250N
T T
1500N
Why have the arrows changed?
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When travelling at 25 m s–1 the driver of the car disengages theengine and applies the brakes. The brakes provide a constantbraking force of magnitude 1500 N to the car. The resistingforces of magnitude 750 N and 1250 N are assumed to remain
unchanged. Calculate
(c ) the thrust in the tow-bar while braking, (2)
Equation of motion of car is T – 750 – 1500N = 1000 × -1.4
1000kg1500kg
750N1250N
T T
1500N
Thrust = 850N
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When travelling at 25 m s–1 the driver of the car disengagesthe engine and applies the brakes. The brakes provide aconstant braking force of magnitude 1500 N to the car. Theresisting forces of magnitude 750 N and 1250 N are assumed
to remain unchanged. Calculate
(d ) the work done, in kJ, by the braking force in bringing the carand the trailer to rest. (4)
u = 25ms-1 a = -1.4ms-2 v = 0, therefore s =223.2m
Work done = force x distance
Work done = 1500 x 223.2 = 335KJ
Using 2 2v u 2as
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When travelling at 25 m s–1 the driver of the car disengages theengine and applies the brakes. The brakes provide a constantbraking force of magnitude 1500 N to the car. The resistingforces of magnitude 750 N and 1250 N are assumed to remainunchanged. Calculate
(e ) Suggest how the modelling assumption that the resistances tomotion are constant could be refined to be more realistic. (1)
Resistance varies with respect to speed.
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Question 6A particle P of mass 3m is moving with speed 2u in a straight line on asmooth horizontal table. The particle P collides with a particle Q ofmass 2m moving with speed u in the opposite direction to P . The
coefficient of restitution between P and Q is e .
(a ) Show that the speed of Q after the collision is 0.2u (9e + 4).
By conservation of momentum 6mu – 2mu = 3mv1 + 2mv2
4u = 3v1
+ 2v2
(1)
Coefficient of restitution = speed of separation/speed of approach
Therefore 3eu = v2 - v1
Remember to always draw a diagram
2u u
2m3mBefore
P Q
v2
2m3m
v1
After QP
and v1 = v2 - 3eu (2)
Sub (2) into (1) 4u = 3v2 - 9eu +2v2 Hence v2 = 0.2u(9e+ 4)
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As a result of the collision, the direction of motion of P isreversed.
(b ) Find the range of possible values of e . (5)
From a) v2 = 0.2u(9e+ 4) and v1 = v2 - 3eu
Therefore v1 = 0.2u(9e+4) - 3eu
Hence v1 = 0.4u(2 – 3e)
But v1<0 Therefore (2 – 3e) < 0
So e>(2/3) and e<1
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Given that the magnitude of the impulse of P on Q is 6.4mu ,(c ) find the value of e . (4)
Impulse = change in momentum
6.4mu = 2m(0.2u(9e+4) +u)
6.4u = 3.6eu + 1.6u + 2u
2.8 = 3.6e
e=7/9
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Question 7
A particle P is projected from a point A with speed 32 m s–1 at an angle of elevation , where sin = 3/5
The point O is on horizontal ground, with O vertically belowA and OA = 20 m. The particle P moves freely under gravityand passes through a point B , which is 16 m above ground,before reaching the ground at the point C , as shown above.
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(a) the time of the flight from A to C , (5)
Vertical component of velocity is vsinα = 19.2ms-1
Therefore t = 4.77sec
Calculate
21s ut at
2
Using With s = -20, u = 19.2, a = -9.8
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(b ) the distance OC , (3)
Horizontal component of velocity = vcosα =25.6ms-1
Distance = speed ×time = 25.6 × 4.77
= 122m
Calculate
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(c ) the speed of P at B , (4)
Using
Vertical component of velocity = 21.14ms-1
Horizontal component remains constant = 25.6ms-1
Calculate
2 2v u 2as Where s = -4, a = -9.8 and u = 19.2
Therefore by Pythagoras speed = 33.2ms-1
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(d ) the angle that the velocity of P at B makes with the horizontal. (3)
Calculate
21.14Tan
25.6
39.6