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Question

A carnival game requires you to knock over a wood post by throwing a
ball at it. You're offered a very bouncy rubber ball and a very sticky
clay ball of equal mass. Assume that you can throw them with equal
speed
and equal accuracy. You only get one throw.

Which ball will you choose? Why?

Answer 1

Choose the bouncy ball, because of Conservation of Momentum.

Let's say before the collision, the ball has mass Mb and is traveling in the +x direction with a speed of Sb. The wooden post has a mass of Mp and an initial speed of 0. So the total momentum before the collision is 0 + Mb Sb in the +x dir.

After the collision, the speed of the ball is Sb' in the NEGATIVE x direction, which would give it a momentum of -Mb Sb'. The post has some speed Sp' and therefore a momentum in the positive x direction of Mp Sp'.

Since the momentum before equals momentum after...
0 + Mb Sb = Mp Sp' - Mb Sb'
so... Sp' = Mb(Sb + Sb')/Mp

Contrast this with the case where the ball sticks to the post, and thus has a velocity in the POSITIVE x direction after the collision. Indeed, Sb' = Sp'
0 + Mb Sb = Mp Sp' + Mb Sp'
Sp' = Mb Sb/(Mp + Mb)

Whatever positive values you chooses for Mb, Mp and Sb (and possibly Sb'), the value for Sp' is greater (in the +x direction) in the 'bouncy ball' case than in the 'sticky ball' case.

Let's try some numbers:
Bouncy:
Mb = 50g
Mp = 100g
Sb = 100 m/s
Sb' = -50 m/s
Sp = 0
Sp' = ???
Mb*Sb + Mp*Sp = Mb*Sb' + Mp*Sp'
5000 g m/s = -2500 g m/s + 100g*Sp'
75 m/s = Sp' in the bouncy ball case

Sticky:
Mb = 50g
Mp = 100g
Sb = 100 m/s
Sb' = Sp'
Sp = 0
Sp' = ???

Mb*Sb + Mp*Sp = Mb*Sb' + Mp*Sp'
5000 g m/s = (Mb + Mp) * Sp'
5000 g m/s = 150g * Sp'
33.(3) m/s = Sp' in the sticky ball case

With these made-up numbers, the final speed of the post is more than twice as large in the bouncy ball case. If anyone thinks the sticky ball case is more likely to knock the post over, I'd love to see ANY numbers to back up that assertion.

Answer 2

I'd take the clay ball since the total momentum has to be conserved and, in an inelastic collision, all of the original momentum is retained in the resulting system instead of being 'shared' between the collision by-broducts (as is the case in an elastic collision) as they travel in different directions.

Doug