posted by Anonymous .
A 4.0-kg particle is moving horizontally with a speed of 5.0 m/s when it strikes a vertical wall. The particle rebounds with a speed of 3.0 m/s. What is the magnitude of the impulse delivered to the particle?
Physics (4) -
I will help with this, then you do the rest.
impulse = change in momentum = force * time
initial momentum = 4*5 = 20
final momentum = 4*-3 = -12
change = -12 - 20 = -32
magnitude of change of momentum = impulse = 32