Question

A 0.5 kg ball is dropped from such a height that it takes 4s to reach the ground. Calculate the change in momentum of the ball.

Answer 1

Answer:

change in momentum = 19.6 kgm/s

Explanation:

a = v - u/t

v= u + at

v = 0 + 9.8×4

v= 19.6

∆p = m(v - u)

∆p = 0.5(39.2 - 0)

∆p = 19.6

Answer 2

The change in momentum of the ball was 19.6 kg.m/s

Given: A 0.5 kg ball is dropped from such a height that it takes 4s to reach the ground.

To Find: The change in momentum of the ball.

Solution:

• We know that the momentum of a body can be described as the product of the mass and velocity of a body. Momentum depends upon the mass and the velocity of the body.

• The momentum of a body can be calculated by the formula,

                   P = m × v                                                           ...(1)

Where P = momentum, m = mass of the body, v = velocity of the body.

• Since momentum depends on velocity ( which is a vector quantity ), momentum is also considered to be a vector quantity.

• We shall find the velocity of the ball after it reaches the ground ( starting from rest ) by the formula;

                  v = u + at                                                          ....(2)

Where v = final velocity, u = initial velocity, a = acceleration, t = time.

Coming to the numerical, we are given;

The mass of the ball was = 0.5 kg

The time is taken for the ball to reach the ground = 4 sec

The initial velocity (u) of the ball = 0

The acceleration (a) = 9.8 m/s²

Putting respective values in (2), we get;

               v = u + at

           ⇒ v = 0 + ( 9.8 × 4 )

           ⇒ v = 39.2 m/s

Now, the change in momentum can be found by putting the respective values in (1), we get;

               P = m × V  

           ⇒ Δ P = m × Δ V

           ⇒ Δ P = m × ( v - u )

           ⇒ Δ P = 0.5 × ( 39.2 - 0 )

           ⇒ Δ P =  19.6 kg.m/s

Hence, the change in momentum of the ball was 19.6 kg.m/s

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