Ball 1, with a mass of 130 g and traveling at 12.0 m/s, collides head on with ba

Question

Ball 1, with a mass of 130 g and traveling at 12.0 m/s, collides head on with ball 2, which has a mass of 300 g and is initially at rest. What are the final velocities of each ball if the collision is perfectly elastic?

Explanation / Answer

Q What are the final velocities of each ball if the collision is perfectly elastic? Ball 1, with a mass of 140g and traveling at 12m/s , collides head on with ball 2, which has a mass of 330g and is initially at rest. What are the final velocities of each ball if the collision is perfectly elastic? What are the final velocities of each ball if the collision is perfectly inelastic? ans 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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