Anonymous
Anonymous asked in Science & MathematicsPhysics · 1 month ago

# 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?

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)

Relevance

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)

• 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