A 0.280 kg croquet ball makes an elastic head-on collision with a second ball 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 (ΔKE/KE) gets transferred to the second ball?
In order to solve these two questions, we can use the principle of conservation of momentum and conservation of kinetic energy.
Conservation of momentum states that in a closed system, the total momentum before a collision is equal to the total momentum after the collision.
(a) To find the mass of the second ball, let's assign variables to the given values. Let the mass of the first ball be m1 = 0.280 kg and the mass of the second ball be m2 (which we need to find). Also, let the initial velocity of the first ball be v1 and the final velocity of the second ball be v2.
Using the conservation of momentum, we have:
(m1 * v1) + (m2 * 0) = m1 * (v1/2) + m2 * v2
0.280 kg * v1 = 0.280 kg * (v1/2) + m2 * v2
Simplifying this equation, we get:
v1 = (v1/2) + (m2/m1) * v2
Multiplying through by 2:
2 * v1 = v1 + (2m2/m1) * v2
Subtracting v1 from both sides:
v1 = (2m2/m1) * v2
Now, we know that the second ball moves off with half the original speed of the first ball. Therefore, we can write this equation as:
v1 = (2m2/m1) * (v1/2)
Cancelling out v1 on both sides:
1 = (2m2/m1) * (1/2)
Simplifying this equation, we obtain:
1 = m2/m1
Therefore, the mass of the second ball is equal to the mass of the first ball. Hence, the mass of the second ball is 0.280 kg.
(b) To find the fraction of the original kinetic energy transferred to the second ball, we can use the conservation of kinetic energy.
The kinetic energy before the collision is given by:
KE1 = (1/2) * m1 * v1^2
The kinetic energy after the collision is given by:
KE2 = (1/2) * m1 * (v1/2)^2 + (1/2) * m2 * v2^2
Simplifying this equation:
KE2 = (1/4) * m1 * v1^2 + (1/2) * m2 * v2^2
Dividing KE2 by KE1, we get:
(KE2/KE1) = [(1/4) * m1 * v1^2 + (1/2) * m2 * v2^2] / [(1/2) * m1 * v1^2]
Simplifying further:
(KE2/KE1) = [(1/4) * m1 * v1^2] / [(1/2) * m1 * v1^2] + [(1/2) * m2 * v2^2] / [(1/2) * m1 * v1^2]
Canceling out common factors:
(KE2/KE1) = (1/2) + (m2/m1) * (v2/v1)^2
Since we found in part (a) that m2 = m1 and v2 = (v1/2), we can substitute these values into the equation:
(KE2/KE1) = (1/2) + (m1/m1) * [(v1/2)/v1]^2
Simplifying this equation:
(KE2/KE1) = (1/2) + 1/4
(KE2/KE1) = 3/4
Therefore, the fraction of the original kinetic energy transferred to the second ball is 3/4 or 0.75.
Does the first ball stop? If so the second has twice the mass.
KE = 1/2mv^2