Question

A thin block of soft wood with a mass of 0.078 kg rests on a horizontal frictionless surface. A bullet with a mass of 4.67 g is fired with a speed of 613 m/s at a block of wood and passes completely through it. The speed of the block is 23 m/s immediately after the bullet exits the block.
Determine the speed of the bullet as it exits the block. (m/s)

Answer 1

Answer:

Final speed of the bullet is 228.3 m/s

Explanation:

As we know that there is no external force on the system of wooden block and the bullet

so we can say momentum of the system is conserved here

so here we can say

$P_i = P_f$

$m_1v_1 = m_1v_{1f} + m_2v_{2f}$

$4.67\times 10^{-3}(613) = 4.67 \times 10^{-3}v_{1f} + (0.078)(23)$

so we will have

$2.86 = 4.67\times 10^{-3} v_{1f} + 1.794$

$v_{1f} = 228.3 m/s$