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

Answer:

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

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

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

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

=>    $v_t = 0.02175 \ m/s$