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

  • Thread starter ajdrausal
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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

Answer key says 9.6N/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?
 

Answers and Replies

  • #2
lurflurf
Homework Helper
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138


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:
  • #3
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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!
 

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