# Impulse - I am suspecting a wrong answer on the key.

• ajdrausal
In summary, the impulse exerted on the ball by the floor is 9.6N/s. This is calculated by finding the change in momentum, which is equal to the mass of the ball multiplied by the change in velocity. The correct way of doing this is by making sure the signs of the velocities are correct in the calculation.

#### ajdrausal

Impulse -- I am suspecting a wrong answer on the key.

A 2.4-kg ball falling vertically hits the floor with a speed of 2.5 m/s and rebounds with a speed of 1.5 m/s. What is the magnitude of the impulse exerted on the ball by the floor?

* 9.6N/s
* 2.4N/s
* 6.4N/s
* 1.6N/s
* 1.0N/s

My calculations say 2.4N/s

Impulse = ∆Momentum

∆Momentum = mv(Initial) - mv(final)

∆Momentum = (2.4kg)(2.5m/s) - (2.4kg)(1.5m/s)

∆Momentum = 6 - 3.6 = 2.4 N/s

What is the right way of doing this?

That is a cute question.

I=Δp=mΔv

You have Δv wrong (sign error).

Δv=(-2.5-1.5)m/s

Do this for practice
A 40 000 Kg sled is sliding towards you at 50 m/s what impulse do you need to apply to change its velocity to
a)49.999 m/s towards you
b)49.999 m/s away from you

Last edited:

lurflurf said:
That is a cute question.

I=Δp=mΔv

You have Δv wrong (sign error).

Δv=(-2.5-1.5)m/s

So then

Impulse = ∆Momentum

∆Momentum = mv(Initial) - mv(final)

So the v(final) should be negative (my coordinate plane being down as the positive direction)
then

∆Momentum = (2.4kg)(2.5m/s) - (2.4kg)( - 1.5m/s)

which gives me
∆Momentum = 6 + 3.6 = 9.6

Thanks!