Question

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.

Answer 1

Answer:

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

Answer 2

Answer:

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