Answer:
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!!!
