A 4.53 kg bowling ball travelling down the lane with a velocity of magnitude Vo = 7.15 m/s collides with two stationary 1.53 kg pins?

Question

After the collision, pin 1 moves away at an angle of 30° clockwise relative to Vo(vector), pin 2 moves away at an angle of 30° counter-clockwise relative to Vo(vector), and direction of the bowling ball’s motion does not change. What are the speeds of the pins? (both pins have same final speed)

Answer 1

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Ellastic colision. Both kinetic energy and momentum are preserved.

v1 is speed od ball after collision

v2 is speed od pins after collision

Mvo^2/2 = Mv1^2/2 + mv2^2.  (energy)

Mvo = Mv1 + 2mv2 cos 30°   (momentum)

divide first eq. with M/2, second with M

vo^2 = v1^2 + 2v2^2 (m/M)

vo = v1 + 2v2 cos 30° (m/M)

For simplicity, let ratio k=m/M

Isolate terms containing v1 to the left

v1^2 = vo^2 - 2 k v2^2

v1 = vo - 2 k v2 cos 30°

square second eq. and equalize right sides

vo^2 - 2kv2^2 = vo^2 - 4k vov2 cos 30° + 4k^2v2^2 cos^2 30°

cancel vo^2

 - 2kv2^2 = -4kvov2cos 30°+ 4k^2v2^2cos^2 30°

Divide by 2k and rearrange

v2^2 * (2k cos^2 30° + 1) - 2vov2 cos 30° = 0

We discard trivial solution  v2 = 0 because that would mean the ball missed the pins. We can therefore divide quadratic by v2 to get linear

v2 * (2kcos^2 30° + 1) - 2vo cos 30° = 0

v2 = 2vo cos 30° /(2kcos^2 30° + 1)

substitute back k=m/M

v2 = 2vo cos 30° /(2(m/M)cos^2 30° + 1)

v2 = 2*7.15 cos 30° / (2(1.53/4.53)cos^2 30°+ 1)

v2 = 8.22 m/s (Pins after colision)

v1 = 7.15 - 2*8.22 cos 30° (1.53/4.53)

v1 = 2.34 m/s (ball after collision)

Answer 2

conserve momentum horizontally:

4.53kg * 7.15m/s = 4.53kg * V + 1.53kg * 2 * v * cos30º

where v is the speed of the pins

and V is the velocity of the bowling ball (post-collision)

This rearranges to

V = 32.4kg·m/s - 2.65kg*v / 4.53kg = 7.15m/s - 0.585v

conserve energy (elastic collision assumed, or there is no solution)

4.53kg*(7.15m/s)² = 4.53kg*(7.15m/s - 0.585v)² + 2*1.53kg*v²

which has a non-trivial solution at

v = 8.22 m/s