Calculate Impulse of 0.4 kg Ball Kicked at 5m/s

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SUMMARY

The impulse imparted by a 0.4 kg ball kicked at a speed of 5.0 m/s is calculated to be 2 Ns. This conclusion is derived from the formula J = mvf - mvo, where vf is the final velocity and v0 is the initial velocity. The angle of the kick, 60° above the horizontal, does not affect the magnitude of the impulse, as it is aligned with the direction of the velocity. Therefore, the correct answer to the homework problem is ||J|| = 2 Ns.

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Homework Statement


A ball of mass 0.4 kg is initially at rest on the
ground. It is kicked and leaves the kicker’s
foot with a speed of 5.0 m/s in a direction 60◦
above the horizontal.
The magnitude of the impulse k~Ik imparted
by the ball to the foot is most nearly:

1. ||J|| = 1Ns.
2. ||J|| =(2/√3) Ns.
3. ||J|| = 2Ns.
4. ||J|| = 4Ns.
5. ||J|| = √3 Ns.


Homework Equations


J = Ft = change in p


The Attempt at a Solution


I figured the angle didn't matter in this situation since the impulse would be in the same direction as the velocity...

So using
J = mvf - mvo
J = 0.4 kg (5.0 m/s) - 0.4 kg(0 m/s)
J = 2 Ns

Did I do this correctly?
 
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Yes, from Newton's second law the impulse equals the change in momentum as you write which gives the answer you provide.

Your thoughts on the indifference of the angle it is correct, you can think of it as that the impulse has "nowhere else to go" but the foot.
 

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