Question

While playing pool, Larris observes a 165 g cue ball moving at 3.5 m/s striking another (stationary) ball of equal mass head-on. The cue ball is brought to rest by the collision, while the second ball moves at 3.5 m/s following the collision. What was the impulse of the collision?

a) 0.58 Ns
b) 0.67 Ns
c) 0.78 Ns
d) 0.91 Ns

Answer 1

The impulse of the collision is 0.58 Ns.

Step-by-step explanation:

The impulse of a collision is given by the change in momentum of an object. In this case, the cue ball goes from a velocity of 3.5 m/s to 0 m/s, so its momentum changes by -3.5 m/s. The second ball goes from 0 m/s to 3.5 m/s, so its momentum changes by +3.5 m/s. Since the masses of the two balls are equal, the impulse of the collision is equal to the change in momentum of either ball.

The impulse is given by the formula Impulse = Mass x Change in Velocity. Substituting the values, the impulse of the collision is equal to 165 g x 3.5 m/s = 0.57825 Ns ≈ 0.58 Ns.

Therefore, the correct answer is (a) 0.58 Ns.