Question

A 3-kilogram block slides along on a frictionless surface at 4 m/s north, and strikes an obstacle that exerts an impulse on the block of 6 N-s to the south. What is the speed of the block after the collision?

Answer 1

2 m/s

Step-by-step explanation:

The impulse exerted by the obstacle on the block is equal to the change in momentum of the block:

`I=\Delta p = -6 Ns`

where we put a negative sign because the direction of the impulse is south, which is opposite to the initial direction of travelling of the block (north).

The change in momentum is also equal to

`\Delta p = m (v-u)`

where

m = 3 kg is the mass of the block

v is the final velocity

u = +4 m/s is the initial velocity

Solving the equation for v, we find the final velocity of the block:

`v=u+(\Delta p)/(m)=+4 m/s+(-6 Ns)/(3 kg)=+2 m/s`

and the positive sign means the direction is still north.