A 1.2lb ball A is moving with a velocity vA when it is struck by a 2.4lb ball B

Question

A 1.2lb ball A is moving with a velocity vA when it is struck by a 2.4lb ball B which has a velocity of vB of magnitude vB = 18 ft/s. Knowing that the velocity of ball B is zero after impact and that the coefficient of restitution is 0.8, determine the velocity of the ball A (a) before impact and (b) after impact.

Explanation / Answer

mass of A = 1.2lb mass of B = 2.4lb velocity of B before collision= 18ft/s velocity of B after collision = 0 let the velocity of A before collision = V1 and after collision = V2 then conservation of momentum V1*1.2 - 2.4*18 = V2*1.2 + 0 =>V2 - V1 = -36 ....eqn1. coefficient of restitution = -velocity of separation/velocity of approach 0.8 = -(V2 -0) /(V1-18) =>0.8 = V2/(18-V1) =>14.4 - 0.8V1 = 18V2 = 18(V1-36) (from eqn.1) =>14.4 -0.8V1 = -648 + 18V1 =>-662.4 = -18.8V1 =>V1 = 35.234 => V2 = -36+35.234 = -0.766 direction of particle velocity after collision reversed(indicated by -ve sign of V2) plz rate! let me know if u have any doubt...

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