What is the impulse on the ball from the wall?

AI Thread Summary
The discussion revolves around calculating the impulse on a 250 g ball that strikes a wall at a speed of 6.6 m/s and rebounds with the same speed and angle after 6 ms of contact. The impulse is determined using the change in momentum formula, where the initial and final velocity vectors are broken into components. The initial momentum is calculated as 0.825i + 1.428j, while the final momentum is 0.825i - 1.428j. The net impulse, calculated as the difference between final and initial momentum, results in a value of -2.856 in the j direction. Participants suggest re-evaluating the initial velocity components to resolve discrepancies in the calculations.
A_lilah
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Homework Statement



a 250 g ball with a speed v of 6.6 m/s strikes a wall at an angle θ (the angle between the wall and the path of motion) of 30° and then rebounds with the same speed and angle. It is in contact with the wall for 6 ms. What is the impulse on the ball from the wall?


Homework Equations



J is the impulse
J = mvfinal - mvinitial
the change in momentum (P) = Jnet
P = mv


The Attempt at a Solution



First I broke the velocity vectors into components:

6.6m/s * cos(30) = 5.716m/s
6.6m/s * sin(30) = 3.3 m/s, so

Vo = 3.3i + 5.716j (m/s)
and
Vfinal = 3.3i - 5.716j (m/s)

and the momentum of each is the mass of the object * the velocity:

(.25kg)Vo = Po = .825i + 1.428j
(.25kg)vfinal = Pfinal = .825i - 1.428j

so Jnet = Pfinal - Pinitial = (.825-.825) i + (-1.428 -1.428)j
so Jnet = -2.856 in the j direction
so the answer should be -2.856, but it is not (and it is not 2.856 either)...

any help is appreciated.
Thanks!
 
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Have a re-look at your v(initial).
 
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