You are standing on a skateboard, initially at rest. A friend throws a very heavy ball towards you. You have two choices about what to do with the ball: either catch the ball or deflect it back toward your friend with the same speed as it was originally thrown. Which choice should you make in order to maximize your speed on the skateboard?

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moya1316 moya1316 · Answer 1 · 2019-11-13T13:54:53+01:00

Answer:

the skateboard speed is maximum for when the ball bounces

Explanation:

To determine which speed is maximum we use the moment, for each case

Case 1. Catch the ball

Before crash

p₀ = M v

After catching

[tex]p_{f}[/tex] = (m + M) [tex]v_{f}[/tex]

po = [tex]p_{f}[/tex]

M v = (m + M) [tex]v_{f}[/tex]

[tex]v_{f}[/tex] = M / (m + M) v

[tex]v_{f}[/tex] = v M / (m + M)

Case 2. The ball bounces.

In this case the final speed of the ball is [tex]v_{f}[/tex] = -v

[tex]p_{f}[/tex] = m [tex]v_{f2}[/tex] + M [tex]v_{f}[/tex]

[tex]p_{f}[/tex] = m [tex]v_{f2}[/tex] - M v

po = [tex]p_{f}[/tex]

M v = m [tex]v_{f2}[/tex] - M v

[tex]v_{f2}[/tex] = 2M v / m

Let's find the relationship between speeds

[tex]v_{f1}[/tex] / [tex]v_{f2}[/tex] = (v M / (m + M)) / (2M v / m)

[tex]v_{f1}[/tex] /[tex]v_{f2}[/tex] = m / (2 (m + M)

[tex]v_{f1}[/tex] / [tex]v_{f2}[/tex] = ½ [m / M (1 + m / M)]

As m <M we can approximate (1 + m /M) = 1

[tex]v_{f1}[/tex] / [tex]v_{f2}[/tex] = ½ m / M

[tex]v_{f2}[/tex] = (2 M / m) [tex]v_{f1}[/tex]

Consequently vf2 is greater than the speed vf1, so the skateboard speed is maximum for when the ball bounces