(10%) Problem 8: A cue ball of mass m1-0.365 kg is shot at another billiard ball
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Question
(10%) Problem 8: A cue ball of mass m1-0.365 kg is shot at another billiard ball, with mass m2 OS6 kg, which is at rest. The cue ball has an initial speed of v 6.5 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, vr after the collision, in meters per second? Grade Summary Potential1 Submissions (2% per attempt) sinO cos0 tan7 89 cotan0 asinO acoso detailed view atan0 acotan0 sinh 0 tanhO cotanh Radians I give up Hint Hints: 2 deduction per hint. Hints remaining: - -& 50% Part(b) what is the horizontal component of the cue ball's final velocity,vir in meters per second?Explanation / Answer
m1 = 0.365 kg m2 = 0.56 kg
speeds before collision
v1i = 6.5 m/s v2i = 0 m/s
speeds after collision
v1f = ? v2f = ?
initial momentum before collision
Pi = m1*v1i + m2*v2i
after collision final momentum
Pf = m1*v1f + m2*v2f
from momentum conservation
total momentum is conserved
Pf = Pi
m1*v1i + m2*v2i = m1*v1f + m2*v2f .....(1)
from energy conservation
total kinetic energy before collision = total kinetic energy after collision
KEi = 0.5*m1*v1i^2 + 0.5*m2*v2i^2
KEf = 0.5*m1*v1f^2 + 0.5*m2*v2f^2
KEi = KEf
0.5*m1*v1i^2 + 0.5*m2*v2i^2 = 0.5*m1*v1f^2 + 0.5*m2*v2f^2 .....(2)
solving 1&2
we get
v1f = ((m1-m2)*v1i + (2*m2*v2i))/(m1+m2)
v2f = ((m2-m1)*v2i + (2*m1*v1i))/(m1+m2)
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part(a)
final speed of goalie
v1f = ((m1-m2)*v1i + (2*m2*v2i))/(m1+m2)
v1f = ((0.365-0.56)*6.5 + (2*0.56*0))/(0.356 + 0.56)
v1f = -1.4 m/s
=====================
part(b)
v2f = ((m2-m1)*v2i + (2*m1*v1i))/(m1+m2)
v2f = ((0.56-0.365)*0 + (2*0.56*0.365))/(0.365 + 0.56)
v2f = 0.44 m/s
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