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A 250 g ball moving horizontally with ve...

A 250 g ball moving horizontally with velocity 10.0m/s strikes inclined surface of a 720g smooth wedge as shown in the figure. The wedge is placed at rest on a frictionless horizontal ground. If the coefficient of restitution is 0.8, calculate the velocity of the wedge after the impact.

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Let us consider the ball as the body A and the wedge as the body B. Before collision let velocity of ball is u^a after the impact, the ball bounces with velocity `v_(A)` and the wedge advances in horizontal direction with velocity `v_(B)`. These velocities and their t and n-components are immediately before and after the impact are shown in the following figures.

Component along t-axis The ball is free to move before and after the impact, therefore its t-component of momentum conserved. hence, t-component of velocities of the ball remains unchanged.
`v_(At)=U_(At)=10cos37.=8m//s` ...(i)
Momentum Conservation In the horizontal direction, there is no extrnal force on both the bodies. Therefore horizontal component of total momentum of both the bodies remain conserved. `m_(A) u_(A) = m_(A)(v_(At)cos37. -v_(An) cos53.) +m_Bv_(B)`
`0.25xx10=0.25(8xx4/5-(3v_(An))/5)+0.72v_B`
Coefficient of restitution Concept of coefficient of restitution e is applicable only for the n-component velocities.
`v_(Bn)-v_(An)=e(u_An-U_Bn)rarr 3v_B/5-v_An=0.8(6-0)` ...(iii)
From equations (i), (ii) and (iii), we obtain `v_(B) = 2.0 m//s`
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