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

more than, less

less than, more

d. platinum would have the greatest temperature increase because it has the lowest specific heat.