Question

What impulse occurs when an average force of 9 N is exerted on a 2.3 kg cart, initially at rest, for 1.2 s? What change in momentum does the cart undergo? What is the cart's final speed?

Answer 1

`∆p = 11 Ns`
`v = 4.7 m.s^(-1)`

Explanation:

Impulse (∆p)
`∆ p = Ft = 9 × 1.2 = 10.8 Ns`
Or `11 Ns` (2 s.f.)

Impulse = change in momentum, so change in momentum = `11 kg.m.s^(-1)`

Final velocity
m = 2.3 kg, u = 0, v = ?

`∆p = mv - mu = mv - 0 ⇒ v = (∆p)/m = 10.8/2.3 = 4.7 m.s^(-1)`

The direction of the velocity is in the same direction as the force.