A 2.1-kg ball traveling at 5 m/s directly hits a concrete wall. The ball bounces off the wall and then travels in the opposite direction (perpendicularly) at 5 m/s. The impulse, in the direction of the ball's final velocity, that the concrete wall exerts on the ball is
A) 0 N.s.
B) 21 N.s.
C) 5 N.s.
D) 11 N.s.

Question

Quizplus · Accepted Answer

Impulse is given by the change in momentum, which is $ \Delta p = m(v_f - v_i) $. Here, $ m = 2.1 $ kg, $ v_i = 5 $ m/s (initial velocity), and $ v_f = -5 $ m/s (final velocity, since it's in the opposite direction). Thus, $ \Delta p = 2.1 \times (-5 - 5) = 2.1 \times -10 = -21 $ N.s. The direction of the impulse is in the direction of the final velocity, so the magnitude of the impulse is 21 N.s.

A 21-Kg Ball Traveling at 5 M/s Directly Hits a Concrete