Calculate the final speed of the pin

[JoeCool9]

Assume that a bowling ball, initially traveling at 12.0m/s, has 5 times the mass of a pin and that the pin goes off at 80* from the original direction of the ball. A) Calculate the final speed of the pin, B) the final speed of the ball, C) and the angle through which the ball was deflected. Assume the collision is elastic.

 

[mrjeffy321]

This relies on the law of conservation of momentum, initial momentum = final momentum.

In the initial conditions, the only thing that is moving is the bowling ball, moving "up" the lane at 12 m/s. We know that the bowling ball is 5 times the mass of the pin (m_p). So we can use this to calculate the initial momentum as,
(12 m/s)*(5 m_p) = 60 m_p up.

So we know that the final momentum must equal this.

Say you put a coridinate grid out on the bowling lane, with up (the original direction of travel by the bowling ball) in the +y, and right as +x.
The final conditions saw the pin flies off at 80* from the original bowling ball direction (we will say right, in the +x direction), meaning the pin has some momentum in the +x direction. This was not the case in the original conditions, all the momentum was stright up, so in order to cancel this out, the bowling ball must be traveling at some angle to the left in the -x direction.
Also we know the pin [and bolwing ball] have some final momentum in the stright up (+y) direction. We know that the ball's and the pin's momentums must add together to give the final upward momentum, which is equal to the initial upward momentum.

I will leave all the math up to you.

 

[quicknote]

A) To calculate the final speed of the pin, we can use the conservation of momentum and energy equations. Since the collision is assumed to be elastic, we can equate the initial and final momentum and kinetic energy of the system. Let m1 be the mass of the bowling ball and m2 be the mass of the pin.

Initial momentum of the system = m1v1 + m2v2 = (5m)(12.0m/s) + m(0) = 60m/s

Final momentum of the system = m1v1' + m2v2' = (5m)v1' + mv2'

Since the pin goes off at an angle of 80*, we can break down the final velocities into their x and y components:

v1' = v1cos(80*) = (12.0m/s)(0.1736) = 2.08m/s
v2' = v2sin(80*) = (v2)(0.9848)

Now, equating the initial and final momentum, we get:

60m/s = (5m)(2.08m/s) + (mv2)(0.9848)
Solving for v2, we get v2 = 57.7m/s

B) To calculate the final speed of the ball, we can use the same equations as above:

v1' = v1cos(80*) = (12.0m/s)(0.1736) = 2.08m/s
v2' = v2sin(80*) = (v2)(0.9848) = (57.7m/s)(0.9848) = 56.8m/s

C) To calculate the angle through which the ball was deflected, we can use trigonometric properties:

tan(angle) = v2y/v2x
tan(angle) = (v2sin(80*))/(v2cos(80*))
tan(angle) = (57.7m/s)(0.9848)/(57.7m/s)(0.1736)
tan(angle) = 5.663
angle = tan^-1(5.663) = 80*

Therefore, the angle through which the ball was deflected is also 80*.

In conclusion, the final speed of the pin is 57.7m/s, the final speed of the ball is