Question

A bullet of mass 0.4 kilograms is fired towards a stationary wood block. The wood block is kept on a rough horizontal surface. The mass of the block is 0.6 kilograms. After the collision the bullet gets embedded in the block and they move as a single object. The coefficient of friction between the block and the horizontal surface is 0.4. calculator the velocity of the block with the bullet just after the collision.

Answer 1

Explanations & Calculations

• The friction from the floor is neglected during the momentum transfer as the time is too small. Therefore, the momentum of the bullet-block system is conserved.

$\qquad\qquad \begin{aligned} \small (m+M)V&=\small mv\\ \small V&=\small \frac{mv}{(m+M)}\\ &=\small \frac{0.4\,kg\times9.8\,ms^{-2}}{(0.4\,kg+0.6\,kg)}\\ &=\small 3.92\,ms^{-1} \end{aligned}$