A 0.421-kg ball has an elastic, head-on collision with a second ball that is initially at rest. The second ball moves off with half the original speed of the first ball.
(a) What is the mass of the second ball?
(b) What fraction of the original kinetic energy (ΔK/K) is transferred to the second ball?
Why did the balls go to therapy? They were having a collision of emotions!
(a) Let's call the mass of the second ball "m2". According to the law of conservation of momentum, the momentum before the collision equals the momentum after the collision. Since the first ball is at rest after the collision, the momentum of the system is just the momentum of the second ball, which is mass times velocity. So, we have:
0 = (0.421 kg) * v1f + m2 * v2f
But we know that the second ball moves off with half the original speed of the first ball, so v2f = (1/2) * v1f. Substituting this into the equation, we get:
0 = (0.421 kg) * v1f + m2 * (1/2) * v1f
Simplifying:
0 = (0.421 kg + m2 * (1/2)) * v1f
Since v1f cannot be zero, we can say that (0.421 kg + m2 * (1/2)) must equal zero. Solving for m2:
m2 = -(0.421 kg)/(1/2)
m2 = -0.842 kg
Oh no! It seems I made a mistake with the calculation, and I got a negative mass for the second ball. That's quite an anomaly! So let's review the problem and try to figure out where I went wrong.
To solve this problem, we can use the principle of conservation of momentum and kinetic energy.
(a) Let's denote the mass of the second ball as m2.
According to the principle of conservation of momentum, the momentum before the collision is equal to the momentum after the collision. The momentum of an object is given by the product of its mass and velocity.
Before the collision:
The first ball has a mass of 0.421 kg and an initial velocity, v1.
The second ball is at rest, so its initial velocity, v2, is 0.
After the collision:
The first ball continues to move with a final velocity, v1f, which is not given.
The second ball moves off with half the original speed of the first ball, so its final velocity, v2f, is (1/2)v1.
Using the conservation of momentum, we can write:
m1 * v1 + m2 * v2 = m1 * v1f + m2 * v2f
Plugging in the values we know:
(0.421 kg) * v1 + m2 * (0 m/s) = (0.421 kg) * v1f + m2 * ((1/2)v1)
We don't know the final velocity of the first ball (v1f), so we can't solve for m2 directly. However, we can use the principle of conservation of kinetic energy to find the ratio ΔK/K.
(b) The initial kinetic energy (K) of the system is equal to the sum of the kinetic energies of the two balls before the collision.
K = (1/2) * m1 * v1^2 + (1/2) * m2 * v2^2
The final kinetic energy (K') of the system is equal to the sum of the kinetic energies of the two balls after the collision.
K' = (1/2) * m1 * v1f^2 + (1/2) * m2 * v2f^2
The ratio of the change in kinetic energy (ΔK) to the initial kinetic energy (K) can be calculated as follows:
ΔK/K = (K' - K)/K
Plugging in the values we know:
ΔK/K = [(1/2) * m1 * v1f^2 + (1/2) * m2 * v2f^2 - ((1/2) * m1 * v1^2 + (1/2) * m2 * v2^2)] / [(1/2) * m1 * v1^2 + (1/2) * m2 * v2^2]
This equation does not depend on the specific value of the masses. However, substituting our earlier equation for m2 in terms of m1 and simplifying, we get:
ΔK/K = (m1/m2) * [(v1f^2 - v1^2) / (v1^2)]
Now, we can substitute the known values:
ΔK/K = (m1/m2) * [(v1f^2 - v1^2) / (v1^2)]
= (0.421 kg / m2) * [(v1f^2 - v1^2) / (v1^2)]
Unfortunately, we still can't solve for m2 without knowing v1 and v1f, but we can at least find the fraction ΔK/K if we know the ratio (v1f^2 - v1^2) / (v1^2).
To solve this problem, we can use the principle of conservation of momentum and conservation of kinetic energy.
(a) Mass of the second ball:
Using the principle of conservation of momentum, we know that the total momentum before the collision is equal to the total momentum after the collision. In this case, the initial momentum of the first ball is zero since it is at rest, and the final momentum is half the momentum of the first ball.
The equation for conservation of momentum is:
m1 * v1i + m2 * v2i = m1 * v1f + m2 * v2f
Where:
m1 and m2 are the masses of the first and second balls, respectively.
v1i and v2i are the initial velocities of the first and second balls.
v1f and v2f are the final velocities of the first and second balls.
Given:
m1 = 0.421 kg (mass of first ball)
v1i = 0 m/s (initial velocity of first ball)
v2i = 0 m/s (initial velocity of second ball)
v1f = ? (final velocity of first ball)
v2f = 0.5 * v1f (final velocity of second ball)
Since the first ball is at rest, we have v1f = 0. Substituting these values into the conservation of momentum equation, we get:
0.421 kg * 0 m/s + m2 * 0 m/s = 0.421 kg * 0 m/s + m2 * 0.5 * v1f
Simplifying, we have:
0 = 0 + m2 * 0.5 * v1f
Since the equation holds true for any value of v1f, we can conclude that the mass of the second ball (m2) is zero. Therefore, the second ball has zero mass.
(b) Fraction of the original kinetic energy transferred to the second ball:
Using the principle of conservation of kinetic energy, we know that the total kinetic energy before the collision is equal to the total kinetic energy after the collision.
The equation for conservation of kinetic energy is:
K1i + K2i = K1f + K2f
Where:
K1i and K2i are the initial kinetic energies of the first and second balls, respectively.
K1f and K2f are the final kinetic energies of the first and second balls.
Given:
K1i = 0.5 * m1 * v1i^2 (initial kinetic energy of first ball)
K2i = 0 (initial kinetic energy of second ball)
K1f = 0.5 * m1 * v1f^2 (final kinetic energy of first ball)
K2f = 0.5 * m2 * v2f^2 (final kinetic energy of second ball)
Since the second ball has zero mass (as calculated in part (a)), we have m2 = 0 in the equation for K2f. Hence, K2f = 0.
Substituting these values into the conservation of kinetic energy equation, we get:
0.5 * m1 * v1i^2 + 0 = 0.5 * m1 * v1f^2 + 0
Simplifying, we have:
0.5 * m1 * v1i^2 = 0.5 * m1 * v1f^2
Cancelling the factors of 0.5 and m1, we have:
v1i^2 = v1f^2
Taking the square root of both sides, we get:
v1i = v1f
This means that in this elastic head-on collision, the final velocity (v1f) of the first ball is equal to its initial velocity (v1i). Therefore, the fraction of the original kinetic energy transferred to the second ball is zero. No energy is transferred to the second ball.