IIT JEEPhysics Paper I

PAPER –ITopic: Work power energy, Center of mass & Collisions + Kinematics and Newton’s

Laws of Motion SECITON –I (Single Correct Answer)1. The system shown in figure is in equilibrium where '' µ is the coefficient of friction between M

and table then maximum value of m is ______ (M > m).

a) [ ] ( )mMm 23max −= µ

b) [ ] ( )mMm −=3max

µ

c) [ ] ( )mMm −= 23max

µd) [ ] ( )mMm 2

8max −= µ

2. A particle moving along a straight line with uniform acceleration has velocities 7m/s at P and 17m/s at Q. R is the mid-point of PQ. Then

a) The average velocity between R and Q is 15 m/sb) The average velocity between P and R is 15 m/sc) The average velocity between P and Q is 25 m/sd) The average velocity between P and R is 24 m/s

3. A small particle of mass m is released form rest from point A inside a smooth hemispherical bowl as shown in figure. The ratio of (x) of magnitude of centripetal force and normal reaction on the particle at any point B varies with θ as:

a) b) c) d)

www.asifiitphysics.vriti.com 1

4. The potential energy of a particle of mass 5kg moving in the x-y plane is given by U = -7x+24y joule. x and y being in m. Initially at t = 0, the particle is at the origin, moving with a velocity

of

+

^^

7.04.26 ji m/s. The magnitude of force on the particle is:

a) 25 units b) 24 units c) 7 units d) None of these

5. As the work done by constant force in a closed path is always zero, a constant force is always conservative in nature. Suppose a constant force along x is given by Fx = -6N. Find the potential energy function associated with this force if U = 0 at x = 5.

a) -6x – 30 b) 6x – 30 c) 6x + 30 d) -6x + 30

6. A ball falls to the ground from height h and bounces to height h’. Momentum is conserved in the ball-earth system:

a) No matter what height h’ it reaches b) Only if h’ < hc) Only if h’ = h d) Only if h’ > h.

7. A block of mass M is tied to one end of a mass less rope. The other end of the rope is in the hands of a man of 2M as shown in the figure. The block and the man are resting on a rough wedge of mass M as shown in the figure. The whole system is resting on a smooth horizontal surface. The man pulls the rope. Pulley is mass less and frictionless. What is the displacement of the wedge when the block meets the pulley. (Man does not leave his position during the pull)

a) 0.5 m b) 1 m c) zero d) m3

2

8. Two light vertical springs with equal natural lengths and spring constants k1 and k2 are separated by a distance l. Their upper ends are fixed to the ceiling and their lower ends to the ends A and B of a light horizontal rod AB. A vertical downwards force F is applied at point C the rod. AB will remain horizontal in equilibrium if the distance AC

a) 2

lb)

12 kk

l

+ c) 12

1

kk

kl

− d) 12

2

kk

kl

+

SECITON –II (Multiple Correct Choice )


9. In the figure shown all the surface are smooth. All the blocks A, B and C are movable, x-axis is horizontal and y-axis vertical as shown. Just after the system is released form the position as shown.

a) Acceleration of ‘A’ relative to ground is in negative y-directionb) Acceleration of ‘A’ relative to B is in positive x-direction.c) The horizontal acceleration of ‘B’ relative to ground is in negative x-direction.d) The acceleration of ‘C’ relative to ‘B’ is directed along the inclined surface.

10. Starting from rest on her swing at initial height h0 above the ground, Saina swing forward. At the lowest point of her motion, she grabs her bag that lies on the ground. Saina continues swining forward to reach maximum height h1. She then swings backward and when reaching the lowest point of motion again, she simply lets go of the bag, which drops to the ground. Saina’s backward swing then reaches maximum height h2. Neglecting air resistance, how are the three heights related?

a) 10 hh > b) 21 hh > c) 21 hh = d) 21 hh >

11. Two blocks A and B, having masses m and 5m, are connected by a spring of constant k. The blocks are released from rest on a horizontal surface S (coefficient of friction 3.0=µ ) with

the spring stretched by k

mg from its natural length

a) The instantaneous acceleration of the block B with respect to A is -0.7 g^

ib) The force of friction exerted on B by the surface S is 1.5mg

^

i

c) The force of friction exerted on the surface S by block A is 0.3mg ^

id) The total momentum of the system, at any instant during the subsequent motion, is zero.


12. A body of mass m moving with velocity v has a one-dimensional collision with another body and comes to rest in time t∆ and in distance x∆ . If F is the instaneous force acting on the body during the process,

let ∫∫∆∆

==xt

dxFIandFdtI00

21

Then

a) vm

I =1 b) m

I1 need not be equal to v

c) vm

I =22d)

m

I22need not be equal to v

SECITON –III (Questions with Comprehension Type) PASSAG -I (13 to 15):

Two strings I and II of lengths l and 2l, respectively are arranged as shown in the figure. The ends of string I are connected to two point masses of the mass m and 2m while the ends of string II are connected to mass 2m and m as shown in the figure. The entire system is placed on a smooth horizontal surface. Initially the string I is just on the verge of getting taut and II is slacked in the position shown. The mass m attached to string II is given a horizontal velocity v in a direction perpendicular to line joining 2m and m.

13. The impulse experienced by point mass of 2m due to jerking of string II, is

a) ( )0202

02

7cos27sin363

7cos

3 +++ mumu

b) 0202

0

7cos27sin36

7cos3

++mu

c) 00 7cos7sin4

3mud) Zero

14. The component of velocity of 2m perpendicular to the length of string II just after the strings get taut, is

a) 4

7cos7sin3 00ub) [ ]0202

00

7cos27sin362

7sin7cos3

++u

c) [ ]0202

00

7cos27sin362

7sin7cos3

+++ u

u d) Zero

15. The velocity of m (connected to string I) just after the string becomes taut, is


a) 4

7cos7sin3 00ub)

0202

0

7cos27sin36

7cos3

++u

c) [ ]0202

00

7cos27sin362

7sin7cos3

++u

d) Zero

PASSAG -II (16 to 18):In the case of a moving object making a head-on collision with another object of equal mass at

rest we have momentum conservationMV0 = MV1+MV2 ……….(1)

And for elastic collision, energy conservation2

22

12

0 2

1

2

1

2

1MVMVMV += ………..(2)

In equation (1) and (2), V0 is the incoming speed, V1 is the speed of the incident object after the collision, and V2 is the final speed of the initially stationary object, Eliminating common factors form both sides of equation (1) and (2) yields.

V1 + V2 = V0 ………..(3)And 2

02

22

1 VVV =+ ………..(4)

The V1 – V2 graph for a two-body collision with M1 = M2

Equations (3) and (4) are two simultaneous equations in two unknowns, which are readily solved. Rather than solve them algebraically it is very helpful to plot these two equations in a two-dimensional V1-V2 graph. This plot is shown in figure. the circle of radius V0 arising from energy conservation and the straight line from momentum conservation. The two points of intersection provide two mathematical solutions to equation (3) and (4). One of these solutions (V1 = V0, V2 = 0) is prohibited on physical grounds since it would require object 1 to pass through object 2 in the collision. The other intersection point (V1=0, V2 =V0) provides the well-known and unique physical solution.

16. For an inelastic collision in a similar situation, momentum is still conserved, but the kinetic energy of system reduces. In this case:

a) The momentum conservation equation will yield a different straight line, but the energy equation will yield a circle of smaller radius

b) The momentum conservation equation will yield the same straight line, but the energy equation will yield a circle of larger radius

c) The momentum conservation equation will yield a different straight line, but the energy equation will yield a circle of larger radius

d) The momentum conservation equation will yield the same straight line, but the energy equation will yield a circle of smaller radius


17. If the collision is completely inelastic, how can this be represented on V1-V2 graph.

a) b) c) d)

18. In a case where both objects are of mass 1kg, and 1st object is traveling with velocity of 2m/s. The V2-V1 graph is shown here. What is the velocity V2 after the collision:

a) 1.5 m/s b) 0.5 m/s c) 2 m/s d) ZeroSECITON –IV (Match following Type)19. A block of mass m is released form rest on a smooth inclined plane (wedge), the wedge itself is

resting on smooth horizontal surface as shown in the figure. Assume that the block undergoes a vertical displacement of h. For this situation, match the entries of Column-I with the entries of Column-II.

Column-I Column-IIA) Work done by normal reaction force on the block, is P) PositiveB) Work done by normal reaction force exerted by block on the wedge is Q) NegativeC) Total work done by normal reaction force (acting between block and wedge) on block and wedge is R) ZeroD) Total work done by all forces on block is S) Less than mgh in magnitude

20. Column-I Column-II

A) Kinetic energy of the system is minimum p) 0=→

CMa

B) Momentum of the system is not zero q) 0=→

CML

C) Net external force acting on system is zero r) 0≠→

CMV

s) ∑ =→

0p


PAPER –I KEY1) c 2) a 3) a 4) b 5) a6) b 7) a 8) d 9) a,b,cd 10

)a,c

11)

a,c 12)

a,d 13)

a 14)

b 15)

b

16)

d 17)

a 18)

a 19)

(A-P,S)(B-P,S)(C-Q,S)(D-Q,S)

20)

(A-s)(B-r)(C-p)

SOLUTIONS

1. From Basics 2. Let a be the acceleration and l be the distance between P and Q, then using asuv 222 += , we

get.

Therefore, velocity at R is

2..22 laVV PR +=

laVP += 2

2

7177

222 −+=


132

177 22

=+=

Therefore, average velocity between P and R is

21RP VV

V+=>< [∴ a is constant]

sm /102

137 =+=

Therefore, average velocity between R and Q is

smVV

V QR /152

1713

21 =+=+

=><

3. Using conservation of Mechanical energy, we have,Los in P.E. = gain in K.E.

2

2

1mvmgy =⇒

2

2

1sin mvmgR =⇒ θ

θsin22 gRv =⇒ ………..(1)

Centripetal force, R

mvFC

2

=

Again, using Newton’s second law along the radially inward direction, we have,

R

mvmgN

2

sin =− θ

R

mvmgN

2

sin =−⇒ θ

2

2

2

sin1

1

/sin

//

v

RgRmvmg

RmvNFx C θθ +

=+

==∴

3

2

2

11

1 =+

=

4. 7=∂∂−=x

UFx

24−=∂∂=y

UFy

2522 =+= yx FFF


5. conxtaxta WUU −=− =5

∫=

−=x

x

x dxF5

.

xFx ∆− . [∴ Fx is constant]( ) ( )56 −−−= x

( ) 3060 −=−⇒ xxU

( ) 306 −=⇒ xxU

6. When we consider “ball + earth” as a single system, there is no external force is acting on the system during the collision. Hence, momentum of the system is conserved.

7. As the external horizontal force the system is zero and initially the system is at rest.0=∆ mcX [ Assuming X direction horizontally away form the pulley]

0321 =∆+∆+∆⇒ xmxmxm wedgemanblock

( ) 0..22 =++−+⇒ lMlMlM [Assuming lxx +=∆=∆ 32 ]

M

Ml

4

2=⇒ = 0.5m

8. Let x be the elongation in the spring and distance AC be y. As the rod is in equilibrium.

xkxkF 21 += ………….(1)

And ( )ylxkyxk −+ .. 21 ………… (2)Solving (1) and (2), we get

21

2

kk

kly

+=

11. First we check for static friction Block A

N1=mgF1=mgAs f1>0.3mg static friction not possible.

( )g

m

mgmga 07

3.01 =−=∴

Block B


N2 = 5mgF2 = mg< 0.3 (N2) ∴ static friction prevails = aB = 0

( )^^___

7.07.0 igigaaa ABBA ==−=∴mgf =2

→= mgf 3.01 (Note that this is the force on the surface = 0.3mg^

iSystem A+BAs 0111 ≠− ffThe total momentum is not conserved

12. mvpI =∆=1

KEI ∆=2 only if it is elastic collision. KEI ∆=2 if inelastic. For example, part of I2 may be lost as heat or in permanent deformation of the body (increases in strain energy called potential energy)

13, 14, 15.

Position just before the string gets taut


Situation just after the string becomes taut.The situations are shown just before the string gets taut and just after the string gets taut. The point masses are shown as A, B and C.Along the length of string, the two particle connected to two ends of string have same velocity just after the string becomes taut. Let T1 and T2 be impulsive tensions in strings I and II, respectively. Let v1 be the velocity of A along the length of string II which is same for B also, but due a component of T1 perpendicular to II, A also acquires a velocity v3 as shown.Velocity of C along the length of string I is v1.

For C, 1122 2

3

2

3mv

muummvJdtT −=

−−==∫ ……….(i)

For B, ∫ ∫ =− 10

12 27cos mvdtTdtT

10

12 27cos mvJJ =−⇒ ………..(ii)For B, along a direction perpendicular to length of string II,

∫ == 30

10

1 27sin7sin mvJdtT …………(iii)

For c, ∫ == 211 mvJdtT ………….(iv)

Velocity of B and C along the string I would be the same, therefore.0

30

12 7sin7cos vvv −= …………(v)Solving above equation (v), we get,

( )0202

0

1 7cos27sin363

7cos3

6

3

++−=

2uuv

0202

0

2 7cos27sin36

7cos3

++−=

2uv

[ ]0202

00

3 7cos27sin362

7cos7sin3

++−=

2uv

0202

0

1 7cos27sin36

7cos3

++= mu

J

( )0202

02

27cos27sin363

7cos

3 +++= mumu

J

In Question No. 13 we have to find J2.In Question No. 14 we have to find v3.In Question No. 15 we have to find v2.

16. V1+V2 = V0 (remain same st. line)


201

22

21 2

1

2

1

2

1mvCmvmv =++

C1 is loss in kinetic energyOn solving, 2

22

21

20 CVVV ++=

( )2

22

02

22

1 CVVV −=+

( )022

0 VCVR <−= therefore yield circle of smaller radius.

17. In inelastic collision radius will be less but straight line of momentum conservation will be same.

18. Velocity of incident I object will always be less than II object, therefore from above two solutions (1.5, 0.5) is correct.

19. In the diagram, the direction of normal contact force between block & wedge and velocity of block & wedge are shown.

Here, v1 = velocity of block wrt wedge

v2 = velocity of wedge wrt ground

For block angle between N and v (wrt ground) is greater than 2

π, so work done is negative.

From work energy theorem, dK = WN + Wmg

⇒ mghWN <For wedge WN > OIf we take the entire system, then

KEblock + KEwedge = Wgravity force

WN (block + wedge) together is zero.


IIT JEEPhysics Paper I

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