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mihalych1998 [28]
3 years ago
6

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 e

merges 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?
Physics
1 answer:
Zanzabum3 years ago
3 0

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

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Explanation:

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Since we know the net force, we can use this to calculate q₃. As q₁ is at the right side of q₃ and q₁ and q₃ have opposite signs, the force F₁₃ points to the right. In a similar way, as q₂ is at the left side of q₃, and q₂ and q₃ have equal signs, the force F₂₃ points to the right. That means that the resultant net force is the sum of these two forces:

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