Two students are relaxing in their dorm room and are too lazy to get up and shut the door. One of the students suggests throwing something at the door but there are only two things close at hand:
1) A hard baseball of mass m
2) A soft sticky ball of clay, also of mass m
If the baseball bounces off the door and the clay ball sticks to the door, which collision will cause the door to shut fastest?

The collision of the hard baseball that bounces back will shut the door fastest.

Explanation:

Let the mass of the door be M

The mass of the hard baseball and soft clay Is m

Let the initial velocity of both the hard baseball and the soft clay be v₀

Let the Final velocity of the hard baseball be v₁

Then let the final velocity of the door be v

Using the law of conservation of Momentum,

Momentum before collision = Momentum after collision

For the hard baseball system

Momentum before collision = (m)(v₀) + (M)(0) = mv₀ (the initial velocity of the door is 0 m/s since it is at rest)

Momentum after collision = (m)(-v₁) + (M)(v)

= (-mv₁ + Mv) (the velocity of the hard baseball after collision has a minus sign because it is in the opposite direction to the initial velocity)

mv₀ = (-mv₁ + Mv)

Mv = mv₀ + mv₁ = m(v₀ + v₁)

v = m(v₀ + v₁)/M

For the soft clay,

Momentum before collision = (m)(v₀) + (M)(0) = mv₀

Momentum after collision = (m + M)(v)

mv₀ = (m + M) v

v = mv₀/(m + M)

Comparing these two answers for the velocity of the door after collision

With hard baseball that bounces back

v = m(v₀ + v₁)/M

With soft clay that sticks to the door

v = mv₀/(m + M)

v for hard baseball has a bigger numerator and a smaller denominator, hence, it has the higher value and this means the door will close faster with a numerically higher final velocity from the elastic collision of the hard baseball with the door.

Hope this Helps!!!

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