A bullet of mass 0.1 kg moving horizontally with speed 400 m s^-1 hits a wooden block of mass 3. 9 kg kept on a horizontal rough surface. The bullet gets embedded into the block and moves 20 m before…

Question

A bullet of mass 0.1 kg moving horizontally with speed 400 m s^-1 hits a wooden block of mass 3. 9 kg kept on a horizontal rough surface. The bullet gets embedded into the block and moves 20 m before coming to rest. The coefficient of friction between the block and the surface is_______.

Accepted Answer

Correct answer: D. (d) Let the mass of the bullet be m and mass of the block be M. It is given that V = 400 m s^-1. By conservation of momentum, mV = (m + M)v ⇒0.1 x 400 = (0.1 + 3.9)v ⇒ v = 10 ms^-1 The force acting on the block f=μ m'g ⇒a=μ g Using v'^2 - v^2 = -2aS ⇒0^2−1 0^2=2_(−μ g_)(S_) ⇒μ= 1 0 02×1 0×2 0=0.2 5

$A bullet of mass 0.1 kg moving horizontally with speed 400 m s^{- 1} hits a wooden block of mass 3. 9 kg kept on a horizontal rough surface. The bullet gets embedded into the block and moves 20 m before coming to rest. The coefficient of friction between the block and the surface is_______.$

Solution

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