(a) Let m1 be the mass (0.1 kg) of the ball that is moving with velocity u1 before collision. Let its final velocity be v1. Let m2 be the unkown mass of the ball that is initially at rest. After collsion it has velocity v2.
Because of momentum conservation,
m1*u1 + m2u2 = m1v1 + m2v2
0.1 u1 + 0 = 0.1*v1 + m2*u1/2
u1 = v1 + 5m2*u1
Since the collision is elastic, the relative velocity of the two balls, before and after collsion, does not change magnitude, but changes direction.
v1 - v2 = u2 - u1
(You may need to derive this separately, or consider what happens in a coordinate system traveling with the center of mass. The relative velocity, which is independent of coordinate system, changes sign but keeps the same magnitude)
v1 - u1/2 = 0 - u1
v1 = -u1/2
Combining two previous equations,
u1 = -v1/2 + 5m2*u1
Therefore
5m*u1 = 3u1/2
m = 3/10 = 0.3 kg
(b) For the kinetic energy,
KE(m2),final = 1/2 * 0.3 * (u1/2)^2 = 0.0375 u1^2
and KE(m1),initial = 1/2 x 0.1 x u1^2 = 0.05u1^2
The ratio of KE gain of m2 to intial KE is
= ÄKE/KE = 0.0375u1^2/0.05u1^2 = 0.75
A 0.100 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?
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