A bullet of mass 0.02 kg travelling horizontally with velocity 250 m s − 1 strikes a block of wood of mass 0.23 kg which rests on a rough 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 of the rough surface is ( g = 9 .8 m s − 2 )

Question

A bullet of mass 0.02 kg travelling horizontally with velocity 250 m s − 1 strikes a block of wood of mass 0.23 kg which rests on a rough 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 of the rough surface is ( g = 9 .8 m s − 2 ), 0.75, 0.61, 0.51, 0.30

MARKS App · Accepted Answer

After impact the bullet and block move together and come's to rest after covering a distance of 40 m . By conservation of momentum, m 1 u 1 + m 2 u 2 = m 1 v 1 + m 2 v 2 or 0 .02 × 250 + 0 .23 × 0 = 0 .0 2 v + 0 .23 v 5 + 0 = v 0 .25 v = 5 0 0 2 5 = 2 0 m s - 1 Now, by conservation of energy 1 2 M ⁡ v ⁡ 2 = μ N.d or 1 2 × 0 .25 × 400 = μ × 0 .25 × 9 .8 × 40 ⇒ μ = 200 9 .8 × 40 = 0 .51

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