113. A wooden block of mass M rests on a horizontal surface. A bullet of mass m moving in the horizontal direction strikes and gets embedded in it. The combined system covers a distance x on the surface. If the coefficient of friction between wood and the surface is μ, the speed of the bullet at the time of striking the block is (where m is mass of the bullet)
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Thus the speed of the bullet is u1 = m + M / m √ 2μgx
Explanation:
As the system consists of block + bullet
By conservation of momentum:
mv = ( m + M) V system
v^2 - u^2 = 2ax
0 - v^2 = - 2 μgx
F (ext) = 0 = dp / dt
=> Δp = 0
mu1 = ( m + M ) √ 2μgx
u1 = m + M / m √ 2μgx
Thus the speed of the bullet is u1 = m + M / m √ 2μgx
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