Question

Part A

Answer 1

Answer:

v' = -18 m/s

Explanation:

• Assuming no external forces acting during the collision, total momentum must be conserved, as follows:

       $p_{o} = p_{f} (1)$

• The initial momentum can be expressed as follows (taking as positive the initial direction of the ball):

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

• The final momentum can be expressed as follows (since we know that v'b is opposite to the initial vb):

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

• If we assume that Mc >> mb, we can assume that the car doesn't change its speed at all as a result of the collision, so we can replace V'c by Vc in (3).
• So, we can write again (3) as follows:

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

• Replacing (2) and (4) in (1), we get:

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

• Simplifying, and rearranging, we can solve for v'b, as follows:
• $v'_{b} = -18 m/s (6)$, which is reasonable, because everything happens as if the ball had hit a wall, and the ball simply had  inverted its speed after the collision.