A package of mass m is released from rest at a warehouse loading dock and slides down the h = 3.4 m - high, frictionless chute to a waiting truck. Unfortunately, the truck driver went on a break without having removed the previous package, of mass 2m, from the bottom of the chute. (Figure 1)
Suppose the collision between the packages is perfectly elastic. To what height does the package of mass m rebound?
To determine the height to which the package of mass m rebounds after the collision, we can apply the principle of conservation of mechanical energy.
First, let's analyze the situation before the collision. The package of mass m is released from rest at a height h, so it has potential energy equal to mgh, where g is the acceleration due to gravity.
Next, let's consider the situation after the collision. The initial package of mass m collides elastically with the stationary package of mass 2m at the bottom of the chute. In an elastic collision, both momentum and kinetic energy are conserved.
The momentum of the system before the collision is zero since the first package is at rest. After the collision, the packages move together as a single system with a common final velocity. Therefore, the momentum of the system after the collision is also zero.
Now let's look at the kinetic energy of the system before and after the collision. Before the collision, the package of mass m has no initial kinetic energy since it is released from rest. The package of mass 2m is stationary, so it also has zero kinetic energy.
After the collision, both packages move together with a common final velocity. Since the collision is elastic, we know that kinetic energy is conserved.
The initial potential energy of the m package is converted to the final kinetic energy of the combined system. Therefore, we have:
mgh = (3m)v²/2,
where v is the common final velocity of the packages after the collision.
Now let's determine v by using the principle of conservation of momentum. The initial momentum of the m package is m * 0 = 0. The initial momentum of the 2m package is 2m * 0 = 0. The final momentum of the combined system is (m + 2m) * v = 3mv. Applying conservation of momentum, we have:
0 + 0 = 3mv,
which gives us v = 0.
Since the final velocity is zero, the final kinetic energy is also zero. Therefore, all the initial potential energy is converted back into potential energy after the collision.
To determine the height to which the m package rebounds, we can equate the final potential energy to the initial potential energy:
mgh = mgh',
where h' is the height to which the package of mass m rebounds.
We can rearrange the equation to solve for h':
h' = h.
Therefore, the package of mass m rebounds to the same height h from which it was released.
In summary, the package of mass m rebounds to the same height h = 3.4 m from which it was released.