A ball moving with velocity 2 m/s collides h | vedclass

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(A) Given: $m_{1} = m$,$m_{2} = 2m$,$u_{1} = 2 \ m/s$,$u_{2} = 0$,and coefficient of restitution $e = 0.5$.Let $v_{1}$ and $v_{2}$ be the velocities a

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(A) Given: $m_{1} = m$,$m_{2} = 2m$,$u_{1} = 2 \ m/s$,$u_{2} = 0$,and coefficient of restitution $e = 0.5$.Let $v_{1}$ and $v_{2}$ be the velocities after collision.Applying the law of conservation of linear momentum:$m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}$$m(2) + 2m(0) = mv_{1} + 2mv_{2}$$2 = v_{1} + 2v_{2} \quad ...(i)$By the definition of the coefficient of restitution:$e = \frac{v_{2} - v_{1}}{u_{1} - u_{2}}$$0.5 = \frac{v_{2} - v_{1}}{2 - 0}$$1 = v_{2} - v_{1} \quad ...(ii)$Adding equations $(i)$ and $(ii)$:$(v_{1} + 2v_{2}) + (v_{2} - v_{1}) = 2 + 1$$3v_{2} = 3 \implies v_{2} = 1 \ m/s$Substituting $v_{2} = 1$ in equation $(ii)$:$1 = 1 - v_{1} \implies v_{1} = 0 \ m/s$Thus,the velocities after collision are $0 \ m/s$ and $1 \ m/s$.

$A$ ball moving with velocity $2 \ m/s$ collides head-on with another stationary ball of double the mass. If the coefficient of restitution is $0.5$,then their velocities (in $m/s$) after collision will be

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