Question

A 5.75-kg bowling ball moving at 9.25 m/s collides with a 0.925-kg bowling pin, which is scattered at an angle of θ = 23° from the initial direction of the bowling ball, with a speed of 11.1 m/s. (a) Calculate the direction, in degrees, of the final velocity of the bowling ball. This angle should be measured in the same way that θ is. (b) Calculate the magnitude of the final velocity, in meters per second, of the bowling ball.

Answer 1

Solution mys .25 m/s Pinyl o l. 23. 0975kg Qu ball 5.75kg Applying Camservation of mamentum m x-deveckan. (5-75)x(9.25 ) = (5.75) (COSC) + (0-925) x 111 COS 23 Ves = (5-75 X 9-25) -(6-925x1101 x cos(23")) 5-75 U copo = 7.606 - 0 Applying can senverdian g momondom m y-divechten. 0 = 0.925 X 11-1 Sim 23° - 5.75 v sino V Sino = 0.698 - from Ey 010 stano = '0.698 U Sino 7.606 -698 = tomto 7.606) O = 5.2433 close direction, o ·-5.24 6 Puting value of om €9. O U COM (5.24 ) = 7.606 6 = 2-606. = 7.638 m/s - Cor 5:24 L = 7638 m/s

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