Question

19. A bullet having mass 20 gm is fired horizontally with a velocity 10 m/s into a stationary block of wood having thickness of 0.2 m. The bullet leaves the block with velocity 6 m/s. The resistive force acting on bullet is - *​

Answer 1

Answer:

Correct option is

A

0.16

Mass of the bullet =0.2kg=m

1

Mass of the block =10kg=m

2

Initial velocity of bullet=500m/s=v

1

Initial velocity of block=0m/s=v

2

Final velocity of bullet=v

1

′

=100m/s

Final velocity of block=v

2

′

From the conservation of momentum:

m

1

v

1

+m

2

v

2

=m

1

v

1

′

+m

2

v

2

′

0.02×500+10×0=0.02×100+10v

2

′

10=2+10v

2

′

v

2

′

=0.8m/s

After moving a distance of 0.2 the block stops.Therefore change in kinetic energy =work done by friction

0−

2

1

×10×(0.8)

2

=−μ×10×10×0.2

μ=0.16