# 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?

Dec 9, 2015

∆p = 11 Ns
$v = 4.7 m . {s}^{- 1}$

#### Explanation:

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

Impulse = change in momentum, so change in momentum = $11 k g . m . {s}^{- 1}$

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.