Ball 1, with a mass of 130g and traveling at 11m/s , collides head on with ball

Question

Ball 1, with a mass of 130g and traveling at 11m/s , collides head on with ball 2, which has a mass of 300g and is initially at rest. Part A What are the final velocities of each ball if the collision is perfectly elastic? Part B What are the final velocities of each ball if the collision is perfectly inelastic?

Explanation / Answer

Let m1 and m2 be the masses of the two balls and u and 0 their initial velocities. Let v1 and v2 be their velocities after the collision in the direction of u. Since the collision is head on, the motion of both balls is along a line. By law of conservation of momentum, m1*u + 0 = m1*v1 + m2*v2 ... ( 1 ) As the collision is elastic, kinetic energy is conserved => (1/2)m1*u^2 = (1/2)m1*v1^2 + (1/2)m2*v2^2 => m1u^2 = m1v1^2 + m2v2^2 ... ( 2 ) Plugging v2 = (m1/m2)(u - v1) from eqn. ( 1 ) into eqn. ( 2 ) m1u^2 = m1v1^2 + m2*[(m1/m2)(u - v1)]^2 => u^2 = v1^2 + (m1/m2)(u - v1)^2 => u^2 - v1^2 = (m1/m2)(u - v1)^2 => u + v1 = (m1/m2)(u - v1) => (u + v1)/(u - v1) = m1/m2 => v1/u = (m1-m2)/(m1+m2) => v1 = u * (m1-m2)/(m1+m2) Plugging u = 12, m1 = 140, m2 = 330 v1 = 12 * (140-330)/(140+330) = - 4.851 m/s Plugging v1 = - 4.851 in eqn. v2 = (m1/m2)(u - v1), v2 = (140/330) * (12 + 4.851) = 7.149 m/s. For inelastic collision, v1 = v2 = m1u/(m1+m2) = 140*12/(140+330) = 4.541 m/s.

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