A billard ball moving at a speed of 2.2 metres per second
strikes an identical stationary ball a glancing blow. After
the collision one ball is found to be moving at a speed of
1.1 metres per second in a direction making an angle of 60.0
degrees with the original line of motion. Find the velocity
of the other ball. Can the collision be inelastic, given these
data? [25]
.
.
.
.
. 60 deg angle
.
.......................................o _ _ _ _
.
.
. ? deg angle
.
m1 = m2 The masses are the same
v1b = 2.2 m/s The speed of ball #1 before collision
v1a = 1.1 m/s The speed of ball #1 after collision
v2b = 0 m/s The speed of ball #2 before collision
v2a = ? m/s ? deg
Solution:
Since momentum is conserved, the total momentum along the y axis
is zero, while the total momentum along the x axis is m1 * v1b.
Therefore m1 * v1a * sin(60) = m2 * v2a * sin(?)
and since m1 = m2, v1a * sin(60) = v2a * sin(?).
Also, m1 * v1b = m1 * v1a * cos(60) + m2 * v2a * cos(?)
and since m1 = m2, v1b = v1a * cos(60) + v2a * cos(?).
Now we have two equations in two unknowns, v2a and angle ?.
v2a = v1a * sin(60) / sin(?) and
v2a = ( v1b - v1a * cos(60) ) / cos(?)
v1a * sin(60) = (v1b - v1a * cos(60)) * sin(?) / cos(?)
and so: tan(?) = v1a * sin(60) / (v1b - v1a * cos(60))
tan(?) = 1.1 * 0.866 / (2.2 - 1.1 * 0.500)
tan(?) = 0.5773
and ? = 30.0 degrees
AND THEREFORE we can show that kinetic energy is also conserved
and so the collision must be elastic (by pythagoreas' theorem).
Solving the second equation above for v2a we have:
v2a = ( v1b - v1a * cos(60) ) / cos(30)
v2a = 1.9 m/s
_
And so the answer is 1.9 m/s 90 degrees from the path of the
other ball.