A 3.5 kg block moving with a velocity of +4.2 m/s makes an elastic collision wit

Question

A 3.5 kg block moving with a velocity of +4.2 m/s makes an elastic collision with a stationary block of mass 2.5 kg. a) Use conservation of momentum and the fact that the relative speed of recession equals the relative speed of approach to find the velocity of each block after the collision b) Check your answer by calculating the initial and final kinetic energies of each block. I get: V1= .95m/s V2= 5.15m/s Initial KE for m1: 30.87J Initial KE for m2: 0J Final KE for m1=1.58J Final KE ffor m2=33.15 But all my answers are wrong but the initial KEs

Explanation / Answer

3.5*4.5 = 3.5*v + 2.5*u

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