A 19g ball of clay is shot to the right (in the positive -direction) at 21m/s to

Question

A 19g ball of clay is shot to the right (in the positive -direction) at 21m/s toward a 38g ball of clay at rest. The two balls of clay collide and stick together. Call this reference frame S. 1. What is the total momentum in frame S? 2. What is the velocity V of a reference frame S' in which the total momentum is zero? 3. After the collision, what is the velocity in frame S' of the resulting 57g ball of clay? Answering this question requires only thought, no calculations. 4. Use your answer to part C and the Galilean transformation of velocity to find the post-collision velocity of the 57g ball of clay in reference frame S.

Explanation / Answer

Momentum p = mv Conservation of Momentum: Pi = Pf = m1v1 + m2v2 = m3v3 [particularly for inelastic collisions] (19kg)(21m/s) + (38kg)(0m/s) = (57kg)v3   [Eqn.1] 1) Pf = Pi = (19kg)(21m/s) + 0 = 399kg*m/s From [Eqn. 1]: 2) v3 = V = VS' = (19kg)(21m/s)/(57kg)= 7.00 m/s Since VS' (velocity of Referrence Frame S') = 7 m/s, velocityof clay in S' frame is zero 3) v'' = 0 m/s Gallilean Transformation, or classical velocity-additionformula: v' = v - V v' = velocity in referrence frame S' = 0 m/s v = velocity in referrence frame S V = relative velocity of referrence frames = VS' = 7.00m/s solving for v, v = v' + V = 0m/s + 7.00m/s 4) v = 7.00 m/s that is, velocity of 57 kg clay ball in original Referrenceframe S is 7.00 m/s, which is what we expected because we alreadyknew v3 = 7.00m/s ----- I hope this solution was a Lifesaver for you. Please remember to always conserve your energy andmomentum.

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