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worty
3 months ago
11

Part A

Physics
1 answer:
kicyunya [3.2K]3 months ago
3 0

Answer:

v' = -18 m/s

Explanation:

  • Assuming no external forces act during the collision, the overall momentum must be preserved as follows:

       p_{o} = p_{f} (1)

  • The initial momentum can be defined this way (with the initial direction of the ball considered positive):

       m_{b} * v_{b} -M_{c}*V_{c} = m_{b} * 18 m/s + (-M_{c}* 20 m/s) (2)

  • The final momentum can be formulated as follows (given that v'b moves in the opposite direction to vb):

        -(m_{b} * v'_{b}) + M_{c}*V'_{c} (3)

  • If we consider that Mc >> mb, we assume the car's speed remains unchanged as a result of the collision, allowing us to substitute V'c for Vc in (3).
  • Consequently, we can rewrite (3) like this:

       -(m_{b} * v'_{b}) +(- M_{c}*V_{c}) = -(m_{b} * v'_{b}) + (-M_{c} * 20 m/s) (4)

  • By substituting (2) and (4) back into (1), we arrive at:

       m_{b} * 18 m/s + (-M_{c}* 20 m/s) = -(m_{b} * v'_{b}) + (-M_{c} * 20 m/s) (5)

  • By simplifying and rearranging, we can solve for v'b like this:
  • v'_{b} = -18 m/s (6), which makes sense because it's as though the ball collided with a wall and simply reversed its velocity post-collision.
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