# Question 1c8be

Apr 28, 2015

Based on the Impulse-Momentum theorem $J = \Delta p = F \cdot \Delta t$
This tells us that force can be found if you know the change in momentum as well as the time of the collision.

EX)# A $1 k g$ ball going $5 \frac{m}{s}$ to the east collides with a wall and stops completely. The collision takes $0.1 s$. How much force did the ball apply on the wall?

$J = \Delta p = F \cdot \Delta t \Rightarrow \Delta p = {p}_{f} - {p}_{i}$
${p}_{i} = m \cdot {v}_{i} = 1 \cdot 5 = 5 k g \frac{m}{s}$
${p}_{f} = m \cdot {v}_{f} = 1 \cdot 0 = 0 k g \frac{m}{s}$
$0 - 5 = F \cdot 0.1 \Rightarrow F = \pm 50 N$

Depending on which object the question asks that is applying the force, the direction may change, and so the sign of the force will also change. In this case, the ball is applying the force, and since the velocity of the ball is to the east (positive direction) the force is $50 N$.