(17%) Problem 9: A cue ball of mass m1-0.35 kg is shot at another billiard ball,
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(17%) Problem 9: A cue ball of mass m1-0.35 kg is shot at another billiard ball, with mass m2-0.51 kg, which is at rest. The cue ball has an intial speed of v 5.5 m/s in the positive direction. Assume that the collision is elastic and exactly head-on. 50% Part (a) What is the horizontal component of the cue ball's velocity, vzr, after the collision, in meters per second? Grade Summary Potential Attempts remaini 4% per attempt detalled view 4 5 6 cotano asin acosO atan acotan sinh cosh tanh0 cotanh Degrees Radians Submit Hint I give upt Hints: .4%, deduction per hint. Hints remaining: 2 Feedback:% deduction per feedback. Submission History Answe Feedback Totals 0% 0% ??50% Part (b) What is the horizontal component of the cue ball's finai velocity,v16 in meters per second?Explanation / Answer
Two balls m1= 0.35 and m2= 0.51.
Before: v1 = v = 5.5 and v2 = 0
After v1f and v2f
Elastic collision=Total momentum is constant (1) and
Total kinetic energy is constant (2)
m1* v + m2* 0 = m1* v1f + m2* v2f (1)
m1* v²/2 + 0 = m1* (v1f)²/2 + m2* (v2f)²/2 (2)
Let m2=k*m1 and simplify
v = v1f + k* v2f (1)
= v1f = v - k* v2f plug in (2)
v² = (v1f)² + k* (v2f)² ....(2)
v² = (v - k* v2f)² + k* (v2f)²
v² = v² - 2k* v*v2f + (k*v2f)² + k* (v2f)² , but v2f ? 0 divide by v2f
0 = -2k* v + k² * v2f + k* v2f
2k*v = v2f* (k² + k) , divide by k
v2f = 2v / (k+1) , now we plug in v = 5.5 and k = m2/m1 = 0.51/0.35 = 1.457
v2f = 2*5.5 / (1.457+1)
= 4.477m/s
From above we have v1f = v - k* v2f
= 5.5 - 1.457* 4.477
= - 1.0229m/s
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