Answer:
v' = -18 m/s
Explanation:
- Assuming no external forces acting during the collision, total momentum must be conserved, as follows:
[tex]p_{o} = p_{f} (1)[/tex]
- The initial momentum can be expressed as follows (taking as positive the initial direction of the ball):
[tex]m_{b} * v_{b} -M_{c}*V_{c} = m_{b} * 18 m/s + (-M_{c}* 20 m/s) (2)[/tex]
- The final momentum can be expressed as follows (since we know that v'b is opposite to the initial vb):
[tex]-(m_{b} * v'_{b}) + M_{c}*V'_{c} (3)[/tex]
- 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:
[tex]-(m_{b} * v'_{b}) +(- M_{c}*V_{c}) = -(m_{b} * v'_{b}) + (-M_{c} * 20 m/s) (4)[/tex]
- Replacing (2) and (4) in (1), we get:
[tex]m_{b} * 18 m/s + (-M_{c}* 20 m/s) = -(m_{b} * v'_{b}) + (-M_{c} * 20 m/s) (5)[/tex]
- Simplifying, and rearranging, we can solve for v'b, as follows:
- [tex]v'_{b} = -18 m/s (6)[/tex], 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.