A ball of mass 0.440 kg moving east (+X --direction) with a speed of 3.30 m/s co

Question

A ball of mass 0.440 kg moving east (+X --direction) with a speed of 3.30 m/s collides head-on with a 0.220 kg ball at rest. If the collision is perfectly elastic, what will be the speed and direction of each ball after the collision.

Explanation / Answer

m1u1 + m2u2 = m1v1 + m2 v2 .440 * 3.3 = .440 v1 + .220 v2 => 6.6 = 2v1+ v2 -----------eqn 1 => v2 = 6.6 -2v1 0.5m1u1^2 = .5m1v1^2+ .5m2v2^2 as m1 = 2m2 .5 * 2m2 * 3.3^2 = .5 * 2m2 *v1^2 + .5 m2* v2^2 cancelling out m2 10.89 = v1^2 + .5v2^2 2v1^2 + v2^2 = 21.78 2v1^2 + (6.6 - 2v1)^2 = 21.78 4v1^2 + 43.56 -26.4v1 = 21.78 on solving v1 = 5.633 (rejected since it will not satisfy eqn 1) , 0.966 v2 = 6.6 -2 * .966 = 4.668 m/sec both along positive diretion

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