(6%) Problem 14: A 6.75-kg bowling ball moving at 9.25 m/s collides with a 0.825
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(6%) Problem 14: A 6.75-kg bowling ball moving at 9.25 m/s collides with a 0.825-kg bowling pin, which is scattered at an angle of ? =27" from the initial direction of the bowling ball, with a speed of 11 m/s ? 1S 50% Part (a) Calculate the direction in degrees, of the final velocity of the bowling ball. This angle should be measured in the same way that ? Grade Summary Deductions 0% Potential 100% cosO cotanO asin acos0 Submissions Attempts remaining: S (0% per attempt) detailed view sin acotan sinh0 cosh0 tanhO cotanh0 0 END Degrees O Radians BACKSPACE | DELI cleAR Submit Hint I give up! Hints: 0% deduction per hint. Hints remaining: 2 Feedback: 0% deduction per feedback. -? 50% Part (b) Calculate the magnitude of the final velocity, in meters per second, of the bowling ballExplanation / Answer
a.
Tan x1 = - [ m2*v2*sin x2 ] / [ (m1v1) - (m2v2*cos x2) ]
= - [0.825*11*sin 27] / [ (6.75*9.25) - (0.825*11*cos 27) ]
= - 0.0758
x1 = arctan(-0.1755)
= -4.335 deg
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(b)
v = -m2*v2*sin x2 / m1sin x1
= -[0.825*11*sin 27] / [6.75*sin(-4.334)]
= 8.076 m/s
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