Respuesta :
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!!!
Answer:
The soft sticky ball
Explanation:
The collision produced by the soft ball is much more inelastic that the other one (with the hard baseball), hence, the soft ball will give almost all its energy to the door, and in fact, almost all its momentum. This will produce that the door gains more energy and it will shut faster. The collision with the hard baseball is more an elasctic collision and only for a moment the hard baseball will give part of its momentum to the door, ant it is not enough.
