Question

A bullet of mass 0.02 kg travelling horizontally with velocity 250 m/s strikes a block of wood of mass 0.23 kg which rests on a tough horizontal surface. After the impact, the block and bullet move together and come to rest after travelling a distance of 40 m. The coefficient of sliding friction on the rough surface is (g = 9.8 m/s²). ​

Answer 1

Explanation:

According to the Law of Conservation of Momentum

M1 U1 + M2 U2 = ( M1 + M2 ) V

( 0.02 × 250 ) + ( 0.23 × 0 ) = ( 0.02 + 0.23 ) V

5 = 0.25 V

V = 20 m/s

Now ,

$\frac{ {v}^{2} }{2 \alpha g} = s$

where , v = 20 m/s

g = 9.8 m/s^2

S = 40 m

$\alpha = coefficient \: of \: friction$

$\frac{ {20}^{2} }{2 \times \alpha \times 9.8} = 40$

$\alpha = 1.96$

Therefore, the coefficient of friction is 1.96

I hope it helps you. If you have any doubts, then don't hesitate to ask.

Answer 2

According to the Law of Conservation of Momentum

M1 U1 + M2 U2 = ( M1 + M2 ) V

( 0.02 × 250 ) + ( 0.23 × 0 ) = ( 0.02 + 0.23 ) V

5 = 0.25 V

V = 20 m/s

Now ,

\frac{ {v}^{2} }{2 \alpha g} = s

2αg

v

2

=s

where , v = 20 m/s

g = 9.8 m/s^2

S = 40 m

\alpha = coefficient \: of \: frictionα=coefficientoffriction

\frac{ {20}^{2} }{2 \times \alpha \times 9.8} = 40

2×α×9.8

20

2

=40

\alpha = 1.96α=1.96

Therefore, the coefficient of friction is 1.96