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Throwing and Catching at Same Height

A ball P movi...

Question

A ball P moving with a speed of $v m s^{- 1}$ collides directly with another identical ball Q moving with a speed $10 m s^{- 1}$ in the opposite direction. P comes to rest after the collision. If the coefficient of restitution is 0.6, the value of $v$ is:

30 m s^{- 1}

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40 m s^{- 1}

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50 m s^{- 1}

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60 m s^{- 1}

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Solution

The correct option is C $40 m s^{- 1}$

As momentum is conserved, we can say,
$m (v - 10) = m v_{2}$

$v_{2} = (v - 10)$

$e = \frac{v_{2} - v_{1}}{u_{1} + u_{2}} = \frac{(v - 10) - 0}{(v + 10)}$

$0.6 = \frac{v - 10}{v + 10}$

$0.6 v + 6 = v - 10$

$0.4 v = 16$

$v = 40 m s^{- 1}$

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A ball P moving with a speed of v ms−1 collides directly with another identical ball Q moving with a speed 10 ms−1 in the opposite direction. P comes to rest after the collision. If the coefficient of restitution is 0.6, the value of v is:

The correct option is C 40 ms−1 As momentum is conserved, we can say,m(v−10)=mv2 v2=(v−10) e=v2−v1u1+u2=(v−10)−0(v+10) 0.6=v−10v+10 0.6v+6=v−10 0.4v=16 v=40 ms−1

A ball P moving with a speed of $v m s^{- 1}$ collides directly with another identical ball Q moving with a speed $10 m s^{- 1}$ in the opposite direction. P comes to rest after the collision. If the coefficient of restitution is 0.6, the value of $v$ is:

The correct option is C $40 m s^{- 1}$

As momentum is conserved, we can say,
$m (v - 10) = m v_{2}$

$v_{2} = (v - 10)$

$e = \frac{v_{2} - v_{1}}{u_{1} + u_{2}} = \frac{(v - 10) - 0}{(v + 10)}$

$0.6 = \frac{v - 10}{v + 10}$

$0.6 v + 6 = v - 10$

$0.4 v = 16$

$v = 40 m s^{- 1}$