Two blocks move along a linear path on a nearly frictionless air track. One bloc

Question

Two blocks move along a linear path on a nearly frictionless air track. One block, of mass 0.103 kg, initially moves to the right at a speed of 5.40 m/s, while the second block, of mass 0.206 kg, is initially to the left of the first block and moving to the right at 6.80 m/s. Find the final velocities of the blocks, assuming the collision is elastic.
velocity of the 0.103 kg block =

velocity of the 0.206 kg block =

Explanation / Answer

Initial momentum = final momentum 0.103*5.40-0.206*6.80 = 0.103*(-V1) + 0.206*V2 =>-0.8446 = 0.103*(-V1) + 0.206*V2 Also initial energy = final energy => 0.5*0.103*5.4^2 + 0.5*0.206*6.80^2= 0.5*0.103*V1^2 + 0.5*0.206*V2^2 =>12.5282 = 0.103 V1 ^2 +0.206 V2 ^2 => putting V2 from 1 we get V2 = (0.103 V1-0.8446)/0.206 => 12.5282 = 0.103 V1^2 + (0.0106 V1^2 +0.7133 - 0.1744 V1)/0.042436 => 0.5316 = 0.0149 V^2 +0.7133-0.1744 V1 =>0.0149 V^2-0.1744 V1+0.1817 =0 V1 = 10.54 m/sec [ of 0.103 kg block towards left] V2 = 1.16 m/sec[ of 0.206 kg block towards right]

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