Question

A ball of mass 2 kg is dropped from a height of 5 m. On striking the ground it rebounds with same speed

with which it struck the ground. Find the magnitude of net change in its momentum just after striking the

ground





plz don't copy old question​

Answer 1

Given:-

• Mass of the ball = 2kg

• Height = 5m

• Initial Velocity = 0m/s ( As it was dropped)

• Acceleration due to gravity = +10m/s²

To Find:-

• The magnitude of net change in momentum just after Striking the ground.

Formulae used:-

• v² - u² = 2as

• Change in momentum = mv - mu

Now, we will find the Velocity with which it strikes the ground.

So,

→ v² - u² = 2as

→ v² - (0)² = 2 × 10 × 5

→ v² = 100

→ √v² = √100

→ v = 10

Hence, The Final Velocity of the ball is 10m/s.

Now,

Atq, the ball rebounds with the same speed with which it struck the ground.

• When the ball struck the ground take the Velocity as negative because it moves downard So, u = -10m/s

• When the ball rebounds after Striking we take Velocity as Positive Because it moves upward. So, v = + 10m/s

Now,

→ Change in momentum = mv - mu

→ 2 × 10 - {(2 × (-10)}

→ 20 + 20

→ 40kgm/s

Hence, The Change in momentum is 40kgm/s.