contestada

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

Respuesta :

pstnonsonjoku

Answer:

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

asuwafoh

Answer:

-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

Otras preguntas