Question

A 1.2-kg object moving with a speed of 8.0 m/s collides perpendicularly with a wall and emerges with a speed of 7.2 m/s in the opposite direction. If the object is in contact with the wall for 2.0 ms, what is the magnitude of the average force on the object by the wall

Answer 1

480N

Step-by-step explanation:

Given data

M1= 1.2kg

U1= 8m/s

V1=7.2 m/s

T= 2ms= 0.002

We know that the impulse expression is

P=FT

M(U1-V1)= FT

substitute

1.2(8-7.2)= F*0.002

1.2*0.8= F*0.002

0.96= F*0.002

F= 0.96/0.002

F=480N

Hence the force is 480N