A 2.0-kg ball moving at 10 m/s makes an off-center collision with a3.0-kg ball t

Question

A 2.0-kg ball moving at 10 m/s makes an off-center collision with a3.0-kg ball that is initially at rest. After the collision, the2.0-kg ball is deflected at an angle of 30° from its originaldirection of motion and the 3.0-kg ball is moving at 4.0 m/s. Findthe speed of the 2.0-kg ball and the direction of the 3.0-kg ballafter the collision.
Hint: sin2 +cos2 = 1. speed of 2.0-kg ball 1 m/s direction of 3.0-kg ball 2° from original direction ofmotion speed of 2.0-kg ball 1 m/s direction of 3.0-kg ball 2° from original direction ofmotion

Explanation / Answer

m = 2.0 kg ball: initial velocity = u = 10 m/s, final speed v = ?,angle = 30o m' = 3.0 kg ball: initial velocity = 0, final speed v' = 4.0 m/s,angle ' = ? momentum conservation: mu = mvcos + m'v'cos', so m'v'cos' = mu - mvcos 0 = mvsin - m'v'sin', so m'v'sin' = mvsin (m'v')2 = (mu - mvcos)2 +(mvsin)2 (m'v')2 = (mu)2 - 2m2uvcos +(mv)2 (mv)2 - 2m2uvcos + (mu)2 -(m'v')2 = 0 4v2 - 69.28v + 256 = 0 v = 5.34 m/s ' = arcsin[mvsin/(m'v')] = 26.4o

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