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

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The discussion centers on calculating the impulse exerted on a 2.4-kg ball that hits the floor and rebounds. The initial answer key indicates an impulse of 9.6 N/s, while one participant initially calculated it as 2.4 N/s. The correct approach involves recognizing a sign error in the change of velocity (Δv), which should account for the rebound direction. By correctly applying the impulse formula, the final calculation shows that the impulse is indeed 9.6 N/s. This highlights the importance of sign conventions in physics calculations.
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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?
 
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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

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!
 
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