: 11/1/2016 12:00:00 AM-Due I 16 11:59:00 PM 0%) Proble n 10: A 625 kg bowling b

Question

: 11/1/2016 12:00:00 AM-Due I 16 11:59:00 PM 0%) Proble n 10: A 625 kg bowling ball moving at 9.4 mis collides with a 0875-kg bowling pin, which is scatter d at an angle of 86-from the initial direction of the bowling ball, with a speed of 155 m/s. 50% Part (a) Calculate the direct in degrees, of the final velocity of the bowling ball. This angle be that is Grade Summary Potential 100% Submissions sinO 7 8 9 atano acotan0 sinh0 0 cos Hint 50% Part (b) Calculate the magnitude of the final velocity in meters per second, of the bowling ball.

Explanation / Answer

Applying momentum conservation for collision,

6.25 x 9.4i + 0.875 x 0 = 6.25 vf + 0.875 x 15.5(cos86i + sin86j)

58.75i = 0.946i + 13.53j + 6.25vf

vf = 9.25i - 2.16 j

(A) direction = tan^-1(2.16 / 9.25) = 13.14 deg to the initial direction of ball.

(B) |vf| = sqrt(9.25^2 + 2.16^2) = 9.50 m/s

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