Question

In a ballistics test, a 24 g bullet traveling horizontally at 1200 m/s goes through a 31-cm-thick 320 kg stationary target and emerges with a speed of 910 m/s. The target is free to slide on a smooth horizontal surface. What is the targetâs speed just after the bullet emerges?

Answer 1

The velocity is
`v_t = 0.02175 \ m/s`

Step-by-step explanation:

From the question we are told that

The mass of the bullet is
`m_b = 0.024 \ kg`

The initial speed of the bullet is
`u_b = 1200 \ m/s`

The mass of the target is
`m_t = 320 \ kg`

The initial velocity of target is
`u_t = 0 \ m/s`

The final velocity of the bullet is is
`v_b = 910 \ m/s`

Generally according to the law of momentum conservation we have that

`m_b * u_b + m_t * u_t = m_b * v_b + m_t * v_t`

=>
`0.024 * 1200 + 320 * 0 = 0.024 * 910 + 320 * v_t`

=>
`v_t = 0.02175 \ m/s`