[php snippet=1]

A 1 kg gold ball is thrown straight down from a height of 1m onto a horizontal floor and bounces up. It’s speed at release is 4 m/s. The speed before just after impact is 10 m/s. Determine the magnitude of the impulse delivered to the floor by the gold ball.

Loading ...

Correct Answer: **A.** *16 N-s*

This question combines two principles, translational motion and impulse. The magnitude of impulse is given by the equation

J = Ft = m(vf-vi)

In this case, we are not given time (t) so to find the magnitude of impulse, we must focus on the change in momentum. The change in momentum is given with the equation

m(vf – vi) = 1 kg * (10 m/s – vi)

vf is the velocity directly after impact with the floor. vi is velocity directly before impact. It is important to note that vi is also the final velocity of the ball in the reference frame of throwing the ball to the ground. Thus,

vf^{2} = vi^{2} +2a?x = 4^{2} + 2(10)(1) = 36

Our velocity right before impact is thus 6 m/s. Because velocity is a vector function, we must remember that this velocity is in the **opposite **direction of vf. Finally, we return to our previous impulse equation substituting

-6 m/s for vi and we see that the magnitude of impulse (J) is 16 N-s.

**Subscribe below** to receive the MCAT Question of the Day delivered straight to your inbox every morning.