Ball A of mass 6 kg moving with a velocity of 20 $m s^{- 1}$ collides with another ball B of mass 2 kg moving in the opposite direction with a velocity of 10 $m s^{- 1}$ . After the collision, ball A comes to rest. Find the speed of ball B after collision.

m s^{- 1}

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m s^{- 1}

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m s^{- 1}

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m s^{- 1}

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Solution

The correct option is D 50 $m s^{- 1}$
By the law conservation of momentum:
$m_{A}$ $u_{A}$ + $m_{B}$ $u_{B}$ = $m_{A}$ $v_{A}$ + $m_{B}$ $v_{B}$
6 × 20 + 2 × (-10) = 6 × 0 + 2 × $v_{B}$
(Velocity is a vector quantity. Since the balls are approaching from opposite directions, $u_{B}$ is negative.)
$v_{B}$ = $\frac{6 \times 20 - 20}{2}$
$v_{B}$ = 50 $m s^{- 1}$

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Ball A of mass 6 kg moving with a velocity of 20 ms−1 collides with another ball B of mass 2 kg moving in the opposite direction with a velocity of 10 ms−1. After the collision, ball A comes to rest. Find the speed of ball B after collision.

The correct option is D 50 ms−1By the law conservation of momentum: mAuA + mBuB = mAvA + mBvB 6 × 20 + 2 × (-10) = 6 × 0 + 2 × vB (Velocity is a vector quantity. Since the balls are approaching from opposite directions, uB is negative.) vB = 6×20−202 vB = 50 ms−1

Ball A of mass 6 kg moving with a velocity of 20 $m s^{- 1}$ collides with another ball B of mass 2 kg moving in the opposite direction with a velocity of 10 $m s^{- 1}$ . After the collision, ball A comes to rest. Find the speed of ball B after collision.