Q 1 :

A particle moves in x-y plane under the influence of a force F→ such that its linear momentum is p→(t)=i^cos(kt)-j^sin(kt). If k is constant, the angle between F→ and p→ will be                    [2024]

  • π2

     

  • π4

     

  • π3

     

  • π6

     

(1)   

        π2

 



Q 2 :

A player caught a cricket ball of mass 150 g moving at a speed of 20 m/s. If the catching process is completed in 0.1s, the magnitude of force exerted by the ball on the hand of the player is                    [2024]

  • 150 N

     

  • 3 N

     

  • 30 N

     

  • 300 N

     

(3)   30 N

 



Q 3 :

A cricket player catches a ball of mass 120 g moving with 25 m/s speed. If the catching process is completed in 0.1 s then the magnitude of force exerted by the ball on the hand of player will be (in SI unit)              [2024]

  • 24

     

  • 12

     

  • 25

     

  • 30

     

(4)   30

 



Q 4 :

Figures a, b, c and d show variation of force with time.

The impulse is highest in figure.                [2023]

  • Fig (c)

     

  • Fig (b)

     

  • Fig (a)

     

  • Fig (d)

     

(2)

Impulse=Area under F=t curve

(a)  12×1×0.5=14N·s

(b)  0.5×2=1 N·s (maximum)

(c)  12×1×0.75=38 N·s

(d)  12×2×0.5=12 N·s



Q 5 :

An average force of 125 N is applied on a machine gun firing bullets each of mass 10 g at the speed of 250 m/s to keep it in position. The number of bullets fired per second by the machine gun is                  [2023]

  • 5

     

  • 50

     

  • 100

     

  • 25

     

(2)

F=nmv

where n=number of bullets fired per second

      n=Fmv=12510×10-3×250=50



Q 6 :

A body of mass 500 g moves along the x-axis such that its velocity varies with displacement x according to the relation v=10x m/s. The force acting on the body is    [2023]

  • 166 N

     

  • 25 N

     

  • 125 N

     

  • 5 N

     

(2)

v=10x⇒v2=100x

2v dvdx=100⇒a=50 m/s2

Hence, F=25 N



Q 7 :

Three forces F1=10 N, F2=8 N, F3=6 N are acting on a particle of mass 5 kg. The forces F2 and F3 are applied perpendicular so that the particle remains at rest. If the force F1 is removed, then the acceleration of the particle is                  [2023]
 

  • 2 m s-2

     

  • 0.5 m s-2

     

  • 4.8 m s-2

     

  • 7 m s-2

     

(1)

Resultant of F→2 and F→3 should be opposite to F→1

        a=105=2 m/s2



Q 8 :

A bullet of 10 g leaves the barrel of a gun with a velocity of 600 m/s. If the barrel of the gun is 50 cm long and the mass of the gun is 3 kg, then the value of impulse supplied to the gun will be                [2023]

  • 12 Ns

     

  • 6 Ns

     

  • 36 Ns

     

  • 3 Ns

     

(2)

By momentum conservation

0=3(-v)+0.01(600-v)

v=2 m/s

Impulse on gun = 3×2=6 Ns



Q 9 :

A particle of mass m = 2 kg is moving at 5 m/s in a given direction and is struck by an impulsive force F that deflects its direction of motion through 60° and doubles the magnitude of its velocity. The same impulsive force F is applied to a mass of 5m at rest. What is the resulting kinetic energy of mass 5m (in J)?



(15)

FΔt=|Δp→|=mv3

        =103 kg m/s

Now FΔt=5mV0

K=12 5mV02=15J



Q 10 :

The particles M1 & M2 and the pulley P are lying on smooth horizontal surface. Initially the string is loose.

Assertion: The particle M1 is projected with a speed V0. Then the particle M2 moves with a velocity (M1V0M1+M2) just after the collision.

Reason: The momentum of system (M1+M2) is conserved along x-direction.

  • Assertion is True, Reason is True; Reason is a Correct explanation for Assertion

     

  • Assertion is True, Reason is True; Reason is NOT a Correct explanation for Assertion

     

  • Assertion is True, Reason is False

     

  • Assertion is False, Reason is True

     

(3)

Assertion is True, Reason is False as impulse will be imparted by string