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A 2g particle moving at 8 m/s makes a perfectly elastic head-on collision with a 1g particle at rest. Find the speed of the 2 g particle after collision.

Physics
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The smaller body will be sent off at 10.67 m/s and the larger body will reduce to 2.67 m/s
Assume A to be larger body and B to be smaller body. Let u be velocities before impact and v be velocities after impact. Now, by principle of conservation of momentum \[m_{a} * u_{a} +m_{b} * u_{b} = m_{a} * v_{a} +m_{b} * v_{b}\] and for perfectly elastic collision \[u_{a} + v_{a} = u_{b} + v_{b}\] Here, \[u_{b} = 0\] and \[u_{a} = 8 m/s\] Solving, you'll get two equations which give \[v_{a} = 2.667 m/s\] and \[v_{b} = 10.667 m/s\]
U can use this equation for the 2g particle: V(final)=(M'-M)V(initial)/M'+M V(final)=(2-1)*8/2+1=8/3m/s for the other u can use V(final)=2M'V/M'+M V(final)=2*2*8/2+1=32/3m/s

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