A 4 kg bowling ball rolls at 8 m/s to the right and strikes a 6 kg bowling ball that is rolling at 2 m/s to the left. If the 4 kg ball bounces back with a velocity of 4 m/s to the left, what is the velocity of the 6 kg ball after the collision?

Answer & Explanation:

use the principle of conservation of momentum

take right as positive and left as negative

momentum is the product of mass and velocity

(4 x 8) + (6 x -2) = (4 x -4) + (6 x Y)

32 - 12 = -16 + 6Y

20 + 16 = 6Y

Y = 6m/s

Final answer the velocity of the 6 Kg ball is 6m/s

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