A 0.45 kg ball is moving horizontally with a speed of 5.3 m/s when it strikes a vertical wall. The ball rebounds with a speed of 2.4 m/s. What is the magnitude of the change in linear momentum of the ball

1.31Kgms-1

Explanation:

∆p = m∆v

Where:

∆p = change in momentum

m = mass of the ball

∆v = change in velocity of the ball = (5.3-2.4) = 2.9ms-1

Therefore, substituting appropriately with the values above:

∆p = 0.45*2.9 = 1.31Kgms-1

-3.465 kgm/s

Explanation:

Momentum; This can be defined as the product of the mass of a body and its change in velocity. The S. I unit of momentum is kgm/s

From the question,

ΔM = m (v-u) ... Equation 1

Where ΔM = change in momentum, m = mass of the ball, v = final velocity, u = initial velocity

Note: Let the direction of the initial velocity be the positive direction

Given: m = 0.45 kg, v = - 2.4 m/s (Rebound), u = 5.3 m/s

Substitute into equation 1

ΔM = 0.45 (-2.4-5.3)

ΔM = 0.45 (-7.7)

ΔM = - 3.465 kgm/s

The negative sign means that the change in momentum is along the direction of the final velocity