Question

A breaking force of 10 N applied for 5 seconds is applied to stop a 2.5 kg cart traveling at 20 m/s. What is the impulse exerted on the cart?

Answer 1

`50N.s`

Explanation:

Impulse (`I`) is defined as change in momentum `m(v-u)` (where, `m` is the mass of the object,`u` is the initial velocity and `v` is the final velocity)

And ,impulsive force (`F`) is defined as `(m(v-u))/t` (where,`t` is the time of impact or contact of the object with the force `F`)

So, `F×t = m(v-u)=I=10×5N` i.e `50 N.s`