Billiard ball A of mass mA = 0.115 kg moving with speed vA = 2.60 m/s strikes ball B, initially at rest, of mass mB = 0.140 kg. As a result of the collision, ball A is deflected off at an angle of 32.0 ∘ with a speed vA1 = 2.35 m/s .
a) Solve for the final speed, vB, of ball B. Do not assume the collision is elastic.
b) Solve for the angle, θB, of ball B. Do not assume the collision is elastic.
Conservation of momentum:
ma*vai + mb*vbi = ma*vaf + mb*vbf
Initially mb is at rest, vbi = 0
ma*vai = ma*vaf + mb*vbf
In terms of vbf:
vbf = ma/mb*vai - ma/mb*vaf
vbf = ma/mb*(vai - vaf)
where
vai = 2.6i + 0j m/s
vaf = 2.35 m/s * (cos(32)i + sin(32)j) = 1.993i + 1.245j m/s
Now plug in the numbers
vbf = 0.115 kg / 0.140 kg * ((2.6i + 0j) m/s - (1.993i + 1.245j) m/s)
vbf = 0.115/0.140*((2.6-1.993)i + (0 - 1.245)j m/s)
vbf = 0.115/0.140*(0.607i - 1.245j m/s)
vbf = 0.499i – 1.02j m/s
(a) ||vb|| = sqrt(i^2 + j^2) = 1.14 m/s
(b) direction = atan(j/i) = -64 deg i.e. 64 degrees below +x axis
or = 296 deg ccw from +x axis
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