A 5.85-kg bowling ball moving at 10.0 m/s collides with a 1.60-kg bowling pin, s

Question

A 5.85-kg bowling ball moving at 10.0 m/s collides with a 1.60-kg bowling pin, scattering it with a speed of 8.00 m/s and at an angle of 36.5° with respect to the initial direction of the bowling ball.

(a) Calculate the final velocity (magnitude and direction) of the bowling ball. magnitude m/s

and direction ° counterclockwise from the original direction of the bowling ball

(b) Ignoring rotation, what was the original kinetic energy of the bowling ball before the collision? J

(c) Ignoring rotation, what is the final kinetic energy of the system of the bowling ball and pin after the collision? J

(d) Is the collision elastic or inelastic? elastic inelastic

Explanation / Answer

(a) Conserve momentum vertically and horizontally (alley runs along x-axis).

Vertically:
0 m/s = 1.60kg * 8.00m/s * sin36.5º - 5.85kg * v * sin
where is measured clockwise w/r/t the initial direction of the bowling ball.
0 = 7.61kg·m/s - 5.85kg * v * sin
v*sin = 1.30 m/s #1

Horizontally:

5.85kg * 10.0m/s = 1.60kg * 8.00m/s * cos36.5º + 5.85kg * v * cos
48.21 kg·m/s / 5.85kg = v*cos
v*cos = 8.24 m/s #2

Divide #1 by #2:
v*sin / v*cos = tan = 1.30 / 8.24 = 0.157
= arctan0.157 = 8.92º direction
v = 8.24m/s / cos8.92 = 8.34 m/s magnitude

Check: v = 1.30m/s / sin8.92º = 8.38 m/s

(b) initial KE = ½mv² = ½ * 5.85kg * (10.0m/s)² = 292.5 J

(c) final KE = ½(5.85kg * (8.34m/s)² + 1.60kg * (8.00m/s)²) = 305.85 J

collision elastic   

Elastic collision: KE and momentum are both conserved. No energy is lost to things like deformation. Collisions between hard objects like billiard balls tend to be elastic.

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