Question

A 3.9 kg ball traveling towards a soccer player at a velocity of -3.5 m/s rebounds off the soccer player's foot at a velocity of +15.9 m/s. If the Force of contact between the ball and the player's foot was 25.9 N. For how long was the force applied?

Answer 1

Answer: 2.92 s

Step-by-step explanation:

Given

Mass of ball is
`m=3.9\ kg`

The initial velocity of the ball is
`u=-3.5\ m/s`

Velocity after the rebound is
`v=15.9\ m/s`

Force during the contact is
`F=25.9\ N`

We know, change in momentum is Impulse

`\Rightarrow F\cdot \Delta t=m(\Delta v)`

`\Rightarrow 25.9\cdot \Delta t=3.9(15.9-(-3.5))\\\\\Rightarrow \Delta t=(3.9* 19.4)/(25.9)=2.92\ s`

Thus, the force is applied for 2.92 s