Question

A 0.20 kg cricket ball is traveling with a horizontal speed of 3 m/s and then is hit with a speed of 15 m/s in the opposite direction. What is the change in the ball’s momentum? What is the impulse?

Answer 1

Answer:

$\Delt p= J=-3.6\ kg-m/s$

Explanation:

Given that,

Mass of the cricket ball, m = 0.2 kg

Initial speed of the ball, u = 3 m/s

Final speed of the ball, v = -15 m/s (opposite direction)

To find,

The change in the ball's momentum and the impulse.

Solution,

The change in momentum of the ball is given by :

$\Delta p=m(v-u)$

$\Delta p=0.2\ kg\times (-15-3)\ m/s$

$\Delt p =-3.6\ kg-m/s$

Negative sign shows that the momentum is acting in opposite direction. We know that the change in momentum of an object is called impulse. So, the impulse of the ball is also 3.6 kg-m/s.

Answer 2

Answer:

-3.6 kg· m/s

Explanation:

Momentum is the product of velocity and mass.

p = m v

where m is mass and v is the velocity

Given:

m = 0.20 kg

u = 3 m/s

v = -15 m/s

Change in momentum:

Δp = m (v-u)

Δp = 0.20 (-15-3) = -3.6 kg· m/s

Impulse F.Δt = Δp

Impulse = Δp = -3.6 kg· m/s