3 3 7. You\'re playing pool and you line up a shot on the 3 ball. You want the 3
ID: 1580474 • Letter: 3
Question
3 3 7. You're playing pool and you line up a shot on the 3 ball. You want the 3 ball to be deflected by an angle of 0s 25o at a velocity v 1.3 m/s so that it will make it into a side pocket. What initial velocity must you give the cue ball in order to make this shot and what will be the velocity and angle of the cue ball following the collision? Recall that the mass of a cue ball is 0.17 kg, while the mass of a 3 ball is 0.16 kg. Hint: it might be useful to express the mass of the 3 ball as a fraction of the mass of the cue ball.Explanation / Answer
initial momentum Pi = mc*vc0 i
final momentum Pf = mc*vc1*costhetai + mc*vc1*sinthetaj + m3*v3*costheta3i - m3*v3*sinthetaj
from momentum conservation
Pf = Pi
mc*vc1*costhetai + mc*vc1*sinthetaj + m3*v3*costheta3i - m3*v3*sinthetaj = mc*vc0 i
0.17*vc1*costheta i + 0.17*vc1*sintheta j + 0.16*1.3*cos25i - 0.16*1.3*sin25 j = 0.17*vc0i
0.17*vc1*sintheta j - 0.16*1.3*sin25 j = 0
0.17*vc1*sintheta - 0.16*1.3*sin25 = 0
0.17*vc1*sintheta = 0.16*1.3*sin25
vc1*sintheta = 0.517 .........(1)
vc1*costheta = sqrt(1-(vc1*sintheta)^2)
vc1*costheta = sqrt(1-0.517^2)
vc1*costheta = 0.856 .............(2)
0.17*vc1*costheta i + 0.16*1.3*cos25i = 0.17*vc0i
0.17*vc1*costheta + 0.16*1.3*cos25 = 0.17*vc0 .........(3)
using 2 in 3
0.17*0.856 + 0.16*1.3*cos25 = 0.17*vc0
speed of cue ball Vc0 = 1.96 m/s <<<<<-----------ANSWER
after collison
from 1 & 2
tantheta = 0.517/0.856
direction of cue ball after colliosn theta = 31.1 degrees <<<---------ANSWER
speed of cue ball after collison vc1 = 0.517/sin31.1 = 1 m/s <<<---------ANSWER
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