A ball of mass 2.0 kg falls vertically and hits the ground with speed 7.0 ms-1 as shown below.

The ball leaves the ground with a vertical speed 3.0 m/s. The magnitude of the change in momentum of the ball is

A. zero.

B. 8.0 Ns.

C. 10 Ns.

D. 20 Ns.

Magnitude of the change in momentum of the ball is 8.0Ns

Explanation:

Momentum of a body is defined as the product of mass of the body and its change in velocity.

Momentum = mass * change in velocity

If change in velocity = final velocity - initial velocity

Momentum = velocity (final velocity-initial velocity)

Momentum = m (v-u)

Given;

Mass of the ball m = 2.0kg

If the ball leaves the ground with a vertical speed 3.0 m/s,

Initial velocity of the ball u = 3.0m/s

If the ball falls vertically and hits the ground with speed 7.0 m/s,

Final velocity of the ball v = 7.0m/s

Substituting the given datas into the momentum formula will be equivalent to;

Momentum = m (v-u)

Momentum = 2.0 (7-3)

Momentum = 2*4

Momentum = 8.0kgm/s or 8.0Ns

Magnitude of the change in momentum of the ball is 8.0Ns

8.0 Ns

Explanation:

Change in momentum is given as:

Final momentum - Initial momentum

= m*v - m*u

Where m = mass of ball

v = final velocity

u = initial velocity

Change in momentum = (2.0 * 3.0) - (2.0 * 7.0)

= 6.0 - 14.0 = - 8.0 Ns

The magnitude will be |-8.0| = 8.0 Ns