Q 1 :

A bullet of mass 50g is fired with a speed 100 m/s on a plywood and emerges with 40 m/s. The percentage loss of kinetic energy is             [2024]

  • 44%

     

  • 32%

     

  • 84%

     

  • 16%

     

(3)    

          Ki=12m(100)2 and Kf=12m(40)2

          %loss=|Kf-Ki|Ki×100=|12m(40)2-12m(100)2|12m(100)2×100

           =|1600-100×100|100=84%

 



Q 2 :

Four particles A, B, C, D of mass m2,m,2m,4m have same momentum, respectively. The particle with maximum kinetic energy is:                       [2024]

  • A

     

  • D

     

  • B

     

  • C

     

(1)    

         Kinetic energy = P22m

         P momentum (same for all)

         KE1mass

         Particle A is of least mass, has maximum kinetic energy

 



Q 3 :

When kinetic energy of a body becomes 36 times of its original value, the percentage increase in the momentum of the body will be                   [2024]

  • 600%

     

  • 500%

     

  • 60%

     

  • 6%

     

(2)     

          Kinetic energy (K)=P22mP=2mK

          If Kf=36 Ki. So Pf=6 Pi

         % increase in momentum

         =Pf-PiPi×100%=6Pi-PiPi×100%=500%P%=500%

 



Q 4 :

Three bodies A, B and C have equal kinetic energies and their masses are 400 g, 1.2 kg and 1.6 kg respectively. The ratio of their linear momenta is                  [2024]

  • 1:3:2

     

  • 1:3:2

     

  • 3:2:1

     

  • 2:3:1

     

(1)    

          KA=KB=KC

          Momentum P=2mK

          Pm

          PA:PB:PC=mA:mB:mC

         =400:1200:1600

          =20:203:40

          =1:3:2

 



Q 5 :

A particle of mass m moves on a straight line with its velocity increasing with distance according to the equation v=αx, where α is a constant. The total work done by all the forces applied on the particle during its displacement from x=0 to x=d, will be                    [2024]

  • m2α2d

     

  • md2α2

     

  • mα2d2

     

  • 2mα2d

     

(3)   

         v=αx

          at x=0:v=0 and at x=d;v=αd

          W.D. =Kf-Ki

          W.D.=12m(αd)2-12m(0)2

           W.D =mα2d2

 



Q 6 :

Two bodies of mass 4g and 25g are moving with equal kinetic energies. The ratio of magnitude of their linear momentum is                             [2024]

  • 3 : 5

     

  • 5 : 4

     

  • 2 : 5

     

  • 4 : 5

     

(3)     

           P122m1=P222m2

            P1P2=m1m2=25

 



Q 7 :

An artillery piece of mass M1 fires a shell of mass M2 horizontally. Instantaneously after the firing, the ratio of kinetic energy of the artillery and that of the shell is                  [2024]

  • M1M1+M2

     

  • M2M1+M2

     

  • M1M2

     

  • M2M1

     

(4)   

           |p1|=|p2|

           KE=p22M KE1m

          KE1KE2=p2/2M1p2/2M2=M2M1

 



Q 8 :

A particle is released from height S above the surface of the earth. At certain height its kinetic energy is three times its potential energy. The height from the surface of the earth and the speed of the particle at that instant are respectively          [2025]

  • S2, 3gS2

     

  • S2, 3gS2

     

  • S4, 3gS2

     

  • S4, 3gS2

     

(4)

Let height above the surface of earth be x,

v2=0+2g(Sx)=2g(Sx)          ... (i)

The potential energy at a height x from the surface of earth,

U = mgx          ... (ii)

3mgx=12mv2  3gx=12×2g(Sx)

4x=S  x=S4

 V=2g×3S4=3gS2



Q 9 :

A stone is projected at angle 30° to the horizontal. The ratio of kinetic energy of the stone at point of projection to its kinetic energy at the highest point of flight will be        [2023]

  • 1 : 2

     

  • 1 : 4

     

  • 4 : 1

     

  • 4 : 3

     

(4)

KEpopKEtop=12M(u)212M(ucos30°)2=43



Q 10 :

Given below are two statements:                                      [2023]

Statement I: A truck and a car moving with same kinetic energy are brought to rest by applying brakes which provide equal retarding forces. Both come to rest in equal distance.

Statement II: A car moving towards east takes a turn and moves towards north, the speed remains unchanged. The acceleration of the car is zero.

In the light of given statements, choose the most appropriate answer from the options given below.

  • Statement I is correct but Statement II is incorrect

     

  • Statement I is incorrect but Statement II is correct

     

  • Both Statement I is correct but Statement II are incorrect

     

  • Both Statement I is incorrect but Statement II are correct

     

(1)

Work done=ΔKE

Work done=-FS=0-K

S=KF,  Hence statement 1 is correct.

ΔV=Vf-Vi

Velocity is changing  a0



Q 11 :

Two bodies are having kinetic energies in the ratio 16 : 9. If they have same linear momentum, the ratio of their masses respectively is         [2023]

  • 4 : 3

     

  • 3 : 4

     

  • 16 : 9

     

  • 9 : 16

     

(4)

K1K2=p122m1×2m2p22=m2m1=169

m1m2=916



Q 12 :

An object of mass 'm' initially at rest on a smooth horizontal plane starts moving under the action of a force F=2 N. In the process of its linear motion, the angle θ (as shown in the figure) between the direction of force and the horizontal varies as θ=kx, where 'k' is a constant and 'x' is the distance covered by the object from its initial position. The expression of kinetic energy of the object will be E=nksinθ. The value of 'n' is ________ .             [2023]



(2)

θ=kx

W=KEf-KEi

dW=Fdx

W=F1dx=Fcoskxdx

                        =2k(sinkx)

  E=W=2ksinkx=2ksinθ



Q 13 :

A lift of mass M = 500 kg is descending with speed of 2 m s-1. Its supporting cable begins to slip thus allowing it to fall with a constant acceleration of 2 m s-2. The kinetic energy of the lift at the end of fall through a distance of 6 m will be ________ kJ.                 [2023]



(7)

v2=u2+2as=22+2(2)(6)

      =4+24=28

KE=12mv2=12(500)(28)=7000 J=7 kJ



Q 14 :

A particle of mass 10 g moves in a straight line with retardation 2x, where x is the displacement in SI units. Its loss of kinetic energy for above displacement is (10x)-nJ. The value of n will be _________ .             [2023]



(2)

Loss of K.E.=work done against retarding force

                        =0xmadx=0xm2xdx=mx2

                        =(10-2kg)x2J=(10x)-2J

So n=2



Q 15 :

A body is dropped on ground from a height 'h1' and after hitting the ground, it rebounds to a height 'h2'. If the ratio of velocities of the body just before and after hitting ground is 4, then percentage loss in kinetic energy of the body is x4. The value of x is _________.        [2023]



(375)

Let V1 and V2 be velocities just before and just after hitting the floor.

V1V2=4V1=4V2

KEbefore=12mV12

KEafter=12mV22=12mV1216

ΔKE=12mV12=(116-1)=-1532mV12

% change=ΔKEKEbefore×100%=-1516×100%=-3754%



Q 16 :

The momentum of a body is increased by 50%. The percentage increase in the kinetic energy of the body is ________ %.            [2023]



(125)

Kinetic energy of body=p22m

Initial kinetic energy=pi22m

Final kinetic energy=pf22m=(1.5pi)22m=2.25pi22m

% increase in KE=2.25pi22m-pi22mpi22m×100=125%



Q 17 :

A closed circular tube of average radius 15 cm, whose inner walls are rough, is kept in a vertical plane. A block of mass 1 kg just fits inside the tube. The speed of the block is 22 m/s when it is introduced at the top of the tube. After completing five oscillations, the block stops at the bottom region of the tube. The work done by the tube on the block is ______ J.  [Given g=10 m/s2]              [2023]



(245)

ravg=15 cm

Wf+Wg=ΔKE

Wf+10×0.3=-12×484

Wf=-245 J



Q 18 :

To maintain a speed of 80 km/h by a bus of mass 500 kg on a plane rough road for 4 km distance, the work done by the engine of the bus will be ______ kJ.  (The coefficient of friction between tyre of bus and road is 0.04.)            [2023]



(784)

For constant speed, WD by engine+WD by friction=0 [by WET]

WDengine=-WDfriction=-[-μmgx]

                  =0.04×500×9.8×4×103=784 kJ



Q 19 :

A car accelerates from rest to u m/s. The energy spent in this process is E J. The energy required to accelerate the car from u m/s to 2u m/s is nE J. The value of n is ______.              [2023]



(3)

E1=12mu2-0=12mu2=E

E2=12m(2u)2-12mu2=32mu2=3E