Q 1 :

A body of mass 1000 kg is moving horizontally with a velocity 6 m/s. If 200 kg extra mass is added, the final velocity (in m/s) is      [2024]

  • 6

     

  • 2

     

  • 3

     

  • 5

     

(4)  

        Momentum will remain conserve

        1000×6=1200×v

          v=5m/s

 



Q 2 :

A stationary particle breaks into two parts of masses mA and mB, which move with velocities vA and vB respectively. The ratio of their kinetic energies (KB:KA) is             [2024]

  • mB:mA

     

  • 1:1

     

  • mBvB:mAvA

     

  • vB:vA

     

(4)

By conservation of linear momentum

0=mAVA→+mBVB→

⇒mAVA→=-mBVB→

⇒mAVA=mBVB

K=P22m⇒K∝1m

⇒KBKA=mAmB⇒KBKA=vBvA



Q 3 :

A nucleus at rest disintegrates into two smaller nuclei with their masses in the ratio of 2:1. After disintegration, they will move   [2024]

  • in opposite directions with speed in the ratio of 1:2 respectively

     

  • in opposite directions with speed in the ratio of 2:1 respectively

     

  • in the same direction with the same speed.

     

  • in opposite directions with the same speed.

     

(1)

From conservation of momentum

Pi=Pf

0=mv1-2mv2

v1v2=21, move in opposite direction with speed ratio 1 : 2.



Q 4 :

Consider two blocks A and B of masses m1 = 10 kg and m2 = 5 kg that are placed on a frictionless table. The block A moves with a constant speed v = 3 m/s towards the block B kept at rest. A spring with spring constant k = 3000 N/m is attached with the block B as shown in the figure. After the collision, suppose that the blocks A and B, along with the spring in constant compression state, move together, then the compression in the spring is, (Neglect the mass of the spring)         [2025]

  • 0.2 m

     

  • 0.4 m

     

  • 0.1 m

     

  • 0.3 m

     

(3)

Conservation of Linear momentum,

m1v1+m2v2=(m1+m2)vcm

vcm ⇒10×310+5 ⇒ 3015=2 m/s 

Conservation of mechanical energy,

12kx2=12(10)(3)2–[12(15)(2)2]

⇒ 3000x2=90–60=30

⇒ x2=303000=1100

⇒ x=110m



Q 5 :

A machine gun of mass 10 kg fires 20 g bullets at the rate of 180 bullets per minute with a speed of 100 ms-1 each. The recoil velocity of the gun is       [2023]

  • 0.02 m/s

     

  • 2.5 m/s

     

  • 1.5 m/s

     

  • 0.6 m/s

     

(4)

20×10-3×18060×100=10 V

⇒v=0.6 m/s



Q 6 :

100 balls each of mass m moving with speed v simultaneously strike a wall normally and reflected back with same speed, in time t s. The total force exerted by the balls on the wall is                [2023]

  • 100mvt

     

  • 200mvt

     

  • 200mvt

     

  • mv100t

     

(2)

P→i=Nmvi^,   P→f=-Nmvi^

N is the number of balls striking the wall

N=100

ΔP→=P→f-P→i=-2Nmvi^=-200Nmvi^

F→total=ΔP→Δt=-200mvtt

|F→|=200 mvt



Q 7 :

The figure represents the momentum--time (p-t) curve for a particle moving along an axis under the influence of a force. Identify the regions on the graph where the magnitude of the force is maximum and minimum respectively, if (t3-t2)<t1.                   [2023]

  • c and a

     

  • b and c

     

  • c and b

     

  • a and b

     

(3)

|dp→dt|=|F→|⇒ dp→dt=slope of curve

Maximum slope: (c)

Minimum slope: (b)



Q 8 :

A body of mass 5 kg is moving with a momentum of 10 kg ms-1. Now a force of 2 N acts on the body in the direction of its motion for 5 s. The increase in the kinetic energy of the body is ________ J.                  [2023]



(30)

Given: M=5 kg

Pi=10 kg m/s  (initial momentum)

Impulse=FΔt=ΔP=Pf-Pi

⇒2×5=Pf-10

Pf=20 kg m/s  (final momentum)

Increase in KE=KEf-KEi

=Pf22m-Pi22m=4002×5-1002×5=40-10=30 J



Q 9 :

A body of mass 14 kg initially at rest explodes and breaks into three fragments of masses in the ratio 2 : 2 : 3. The two pieces of equal masses fly off perpendicular to each other with a speed of 18 m/s each. The velocity of the heavier fragment is ____ m/s.   [2026]

  • 12

     

  • 102

     

  • 122

     

  • 242

     

(3)

Apply C.O.M

M1V1+M2V2+M3V3=0

2(-18i)+2(-18j)+3V3=0

V3=12i+12j

|V3|=122 m/s



Q 10 :

Given below are two statements :

Statement I : For a mechanical system of many particles total kinetic energy is the sum of kinetic energies of all the particles.

Statement II : The total kinetic energy can be the sum of kinetic energy of the center of mass w.r.t the origin and the kinetic energy of all the particles w.r.t. the center of mass as the reference.

In the light of the above statements, choose the correct answer from the options given below :        [2026]

  • Statement I is false but Statement II is true

     

  • Both Statement I and Statement II are false

     

  • Statement I is true but Statement II is false

     

  • Both Statement I and Statement II are true

     

(4)

KE=12m1v12+12m2v22

KE=12(m1+m2)vcm2+12m1m2m1+m2|v→1-v→2|2

So Ans. is (4)
 



Q 11 :

A wedge of mass M = 4m lies on a frictionless plane. A particle of mass m approaches the wedge with speed v towards inclined side. There is no friction between the particle and the plane or between the particle and the wedge. The maximum height climbed by the particle on the wedge is given by:

  • v2g

     

  • 2v27g

     

  • 2v25g

     

  • v22g

     

(3)

mv=(m+M)V'

Or v=mvm+M=mvm+4m=v5

Using conservation of ME, we have

12mv2=12(m+4m)(v5)2+mgh

or    h=2v25g



Q 12 :

Assertion A: In case of bullet fired from a free gun, the ratio of kinetic energy of gun and bullet is equal to ratio of mass of bullet and gun

Reason R: In the above firing, total momentum of gun+bullet system remains conserved.

  • A is false but R is true

     

  • A is true but R is false

     

  • Both A and R are true and R is the correct explanation of A

     

  • Both A and R are true but R is NOT the correct explanation of A

     

(3)

0=P→gun+P→bullet

or  Pgun=Pbullet

KgunKbullet=Pgun2/2mgunPbullet2/2mbullet=mbulletmgun