Question

A bullet with a mass of 0.00658 kg is fired through a 1.95 kg block of wood on a frictionless surface. The initial speed of the bullet is 822 m/s, and the speed of the bullet after it exits the block is 439 m/s. At what speed does the block move after the bullet passes through it?

Answer 1

The final velocity of the block is 1.29 m/s

Explanation:

We can solve this problem by using the principle of conservation of momentum: in fact, the total momentum of the system must be conserved before and after the collision. Therefore, we can write:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

where:

$m_1 = 0.00658 kg$ is the mass of the bullet

$u_1 = 822 m/s$ is the initial velocity of the bullet

$v_1 = 439 m/s$ is the final velocity of the bullet

$m_2 = 1.95 kg$ is the mass of the block

$u_2 = 0$ is the initial velocity of the block

$v_2$ is the final velocity of the block

Re-arranging the equation and solving for v2, we find:

$v_2 = \frac{m_1 u_1 - m_1 v_1}{m_2}=\frac{(0.00658)(822)-(0.00658)(439)}{1.95}=1.29 m/s$

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