A 2.00-kg block of ice is moving on a frictionless horizontal surface. At t = 0 the block is moving to the right with a velocity of magnitude 3.00 m/s. (a) Calculate the velocity of the block (magnitude and direction) after a force of 5.00 N directed to the right has been applied for 4.00 s; (b) if instead a force of 7.00 N directed to the left is applied from t=0 to t = 4.00 s, what is the final velocity of the block?

a)

We shall apply the concept of impulse.

Impulse = force x time = change in momentum

= 5 x 4 = 2 (V - 3), where V is final velocity of the object

20 = 2V - 6

V = 13 m / s

b)

Impulse applied = - 7 x 4 = - 28 kg m/s (negative as direction of force is opposite motion)

If v be the final velocity

2 x 3 - 28 = 2 v (initial momentum - change in momentum = final momentum)

- 22 = 2v

v = - 11 m / s

object will move with 11 m / s in opposite direction.