A 5.65 kg bowling ball moving at 9.25 m/s collides with a 0.855 kg bowling pin,

Question

A 5.65 kg bowling ball moving at 9.25 m/s collides with a 0.855 kg bowling pin, which is scattered at an angle of 85.0° to the initial direction of the bowling ball and with a speed of 12.5 m/s. (Assume the bowling ball is initially moving in the +x-direction.)

Calculate the final velocity of the bowling ball. (Find the magnitude in units of m/s and direction in degrees counterclockwise from the +x-axis).

magnitude m/s direction ° counterclockwise from the +x-axis

Is the collision elastic?

Linear kinetic energy is greater after the collision. Discuss how spin on the ball might be converted to linear kinetic energy in the collision.

Explanation / Answer

Let's assume the initial direction was along the x-axis, or "horizontal."

a) Conserve vertical momentum:
0 = 0.855kg * 12.5m/s * sin85º - 5.65kg * v * sin
where is measured clockwise from the +x axis.
Then v*sin = 1.8843 m/s #1

Conserve horizontal momentum:
5.65kg * 9.25m/s = 0.855kg * 12.5m/s * cos85º + 5.65kg * v * cos
v*cos = 9.0851 m/s #2

Divide #1 by #2:
v*sin / v*cos = tan = 1.8843 / 9.0851 = 0.2074
= 11.71º to the initial direction

By putting this value in equation-1 , we get

v * cos11.71º = 9.0851 m/s
v = 9.27 m/s velocity


b) The velocity of the bowling ball has slightly INCREASED. Considering only linear kinetic energy, this is not possible, since this total has also INCREASED. Since "elastic" means "energy is conserved" (IMO), I'd say this collision is NOT elastic.

c) If there is spin on the ball, then it initially has more KE than its speed alone suggests. The spin might "kick" the pin a little bit extra; it might also allow the ball to "grab" onto the pin somewhat in order to pick up speed. Of course, the rotational kinetic energy of the ball would necessarily decrease as a result; the overall (translational plus rotational) kinetic energy cannot increase after the collision.

I suspect that this surprising result occurs not because of spin on the ball, but overstatement of either the scatter angle or the pin's velocity, or both.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.