15)A small ball of mass m is aligned above a larger ball of mass M = 1.2 kg (with a slight separation, as with the baseball and basketball of Figure (a)), and the two are dropped simultaneously from height h = 2.4 m. (Assume the radius of each ball is negligible compared to h.) (a) If the larger ball rebounds elastically from the floor and then the small ball rebounds elastically from the larger ball, what value of m results in the larger ball stopping when it collides with the small ball? (b) What height does the small ball then reach (see Figure (b))?
To answer part (a) of the question, let's consider the conservation of momentum.
When the two balls are dropped simultaneously, their initial momentum is zero since they are at rest. When they collide and the larger ball rebounds elastically from the floor, the momentum of the system after the collision is still zero. Therefore, the momentum of the smaller ball after the collision must also be zero for the larger ball to stop.
Let's denote the velocity of the smaller ball before the collision as v, and after the collision as u. The velocity of the larger ball after the collision will also be u.
Using conservation of momentum:
M * u = m * u
M = m
Therefore, the value of m that results in the larger ball stopping when it collides with the small ball is equal to the mass of the larger ball, which is 1.2 kg.
Now, let's move on to part (b) of the question and determine the height the small ball reaches after the collision.
Since the collision between the small and larger ball is elastic, we can apply the conservation of mechanical energy. The mechanical energy of the system is conserved throughout the collision and subsequent motion.
The initial mechanical energy of the system is given by:
E_initial = m * g * h
where g is the acceleration due to gravity.
After the collision, the small ball will reach its maximum height. At this point, its kinetic energy will be zero, and its potential energy will be at its maximum.
The final mechanical energy of the system is given by:
E_final = m * g * h_max
By equating the initial and final mechanical energies, we can solve for h_max:
m * g * h = m * g * h_max
h_max = h
Therefore, the height the small ball reaches after the collision is equal to the initial height h, which is 2.4 m.
In summary:
(a) The value of m that results in the larger ball stopping when it collides with the small ball is equal to the mass of the larger ball, which is 1.2 kg.
(b) The height the small ball reaches after the collision is equal to the initial height h, which is 2.4 m.
To solve this problem, we can use the principle of conservation of mechanical energy.
(a) First, let's calculate the speed of the larger ball just before colliding with the small ball.
The initial potential energy of the system is given by the formula: PE = mgh, where m is the mass of the small ball, g is the acceleration due to gravity, and h is the initial height.
The initial potential energy is then converted into kinetic energy just before the collision. So we have: KE = 1/2 * M * V1^2, where V1 is the velocity of the larger ball just before the collision.
Since the collision is elastic, the total mechanical energy is conserved. So we have the equation: PE = KE.
mgh = 1/2 * M * V1^2.
Now, we can solve for V1: V1 = sqrt((2 * mgh) / M).
During the collision, the larger ball will transfer its momentum to the smaller ball. Since the larger ball stops after the collision, the momentum transfer is equal to the initial momentum of the larger ball.
The initial momentum of the larger ball is given by: P = M * V1.
The momentum transfer is then equal to: P_transfer = m * V2, where V2 is the velocity of the smaller ball just after the collision.
Since momentum is conserved, we have the equation: P = P_transfer.
M * V1 = m * V2.
Substituting the value of V1, we get: M * sqrt((2 * mgh) / M) = m * V2.
Simplifying, we get: sqrt(2 * mgh * M) = V2.
Now, for the larger ball to stop, the velocity of the smaller ball just after the collision (V2) should be zero.
Therefore, we can set up the equation: sqrt(2 * mgh * M) = 0.
Solving for m, we find: m = 0.
This means that the larger ball will never stop if the smaller ball has zero mass.
(b) Since the larger ball never stops, the smaller ball's velocity and height will depend on the values of m and M.
If we consider a case where the smaller ball has non-zero mass, we can calculate its final velocity and height using the principle of conservation of mechanical energy.
The kinetic energy just after the collision is given by: KE = 1/2 * m * V2^2.
The final potential energy is given by: PE = mgh_final, where h_final is the final height reached by the smaller ball.
Again, the total mechanical energy is conserved, so we have: KE = PE.
1/2 * m * V2^2 = mgh_final.
Simplifying, we get: V2^2 = 2gh_final.
Solving for h_final, we find: h_final = V2^2 / (2g).
Therefore, the height reached by the small ball after the collision is h_final = V2^2 / (2g), where V2 is the final velocity of the smaller ball just after the collision, as calculated in part (a).
Please note that the calculation in part (b) depends on the values of m and M, for a specific mass of the smaller ball.
Assume on rebound they have the following velocities:
large ball: V, in upward direction
small ball: V, in downward direction
you can calculate V from the height of the drop.
Assume conservation of momentum, and if needed, conservation of energy.
Momentum
MV-mV=M*0+mV'
V'=(M/m -1) V
Conservation of energy
MV^2 + mV^2=mV'^2
MV^2+ mV^2=m(M/m-1)^2*V^2
notice V^2 divides out.
a bit of algebra to solve for m ensues, but it is tolerable.