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Billiard ball A (0.35 kg) is struck such that it moves at 10 m/s toward a second identical ball (Ball B) initially at rest. After the collision, Ball A continues to move in the same direction at 2 m/s. What is the magnitude of the velocity for Ball B after the collision?

Respuesta :

Answer:

Explanation:

We shall apply law of conservation of momentum during the collision of ball A and B .

Total momentum before collision of A and B = .35 x 10 = 3.5 kg m/s

Let the velocity of B after collision be v .

Total momentum after collision  = .35 x 2 + .35v

According to law of conservation of momentum

.35 x 2 + .35v  = 3.5

.35 v = 2.8

v = 8 m /s .

The direction of B will be same as direction of A .

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