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 got: 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 for m2: 33.15J The initial KEs are right but the rest I got wrong

Explanation / Answer

a)m1 = 3.5, u1 = 4.2 m/s m2 = 2.5, u2 = 0 m/s We know that, v1-v2 = u2-u1 =>v1-v2 = -4.2 =>v2 = v1+4.2 Conservation of momentum: m1u1+m2u2 = m1v1 + m2v2 =>(3.5*4.2) = 3.5v1 + 2.5v1 + (2.5*4.2) =>6v1 = 4.2 =>v1 = 0.7 m/s =>v2 = 0.7+4.2 = 4.9 m/s b)Initial K.E. of both the masses = 0 + (0.5*3.5*(4.2^2)) = 30.87 J Final K.E. of both the masses = (0.5*3.5*(0.7^2))+(0.5*2.5*(4.9^2)) = 30.87 J So K.E. is conserved. So it is elastic collision. Hence checked.

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