A small ball of mass m is placed on top of a large ball of mass 3m. Both balls are dropped simultaneously from a height h = 5.50 m as shown in the figure. Assuming the radius of the large ball and small are small and thus negligible in comparison to the height they are being dropped meaning the small ball has the same speed the large ball has when it hits the ground. If the large ball rebounds elastically from the floor, collides with the small ball, stops and then the small ball rebounds elastically from the large ball, find the maximum height the small reaches. Hint: Model all collisions as perfectly elastic.
To solve this problem, we need to understand the concept of conservation of momentum and conservation of kinetic energy.
First, let's consider the collision between the large ball and the small ball. Since the collision is elastic, the total momentum and total kinetic energy are conserved.
Let the velocity of the large ball before the collision be v1 and the velocity of the small ball before the collision be v2. After the collision, the small ball comes to rest and the large ball rebounds with the same speed.
Conservation of momentum:
m * v1 + 3m * (-v1) = 0 (since v2 = 0)
v1 = 0
Conservation of kinetic energy:
(1/2) * m * v1^2 + (1/2) * 3m * v1^2 = (1/2) * m * 0 + (1/2) * 3m * v3^2
v3 = v1 = 0
Now, let's consider the motion of the small ball after the collision. The small ball will rebound with a certain speed v3 and reach a maximum height.
Using the conservation of energy, we can equate the initial kinetic energy (due to velocity v3) with the potential energy at the maximum height.
Initial kinetic energy:
(1/2) * m * v3^2
Potential energy at maximum height:
(1/2) * m * g * h_max, where h_max is the maximum height reached by the small ball and g is the acceleration due to gravity.
Equating the two, we have:
(1/2) * m * v3^2 = (1/2) * m * g * h_max
Canceling m and simplifying, we get:
v3^2 = g * h_max
Taking the square root, we have:
v3 = √(g * h_max)
Substituting the value of g (approximately 9.8 m/s^2), we have:
v3 = √(9.8 * h_max)
Since we know that v3 = 0 (as the small ball comes to rest at maximum height), we can solve for h_max:
0 = √(9.8 * h_max)
Squaring both sides and rearranging, we have:
0 = 9.8 * h_max
Thus, the maximum height reached by the small ball is 0.
In this scenario, the small ball does not reach any maximum height after colliding elastically with the large ball.