Impulse that redirects a hockey puck 90°

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An ice hockey puck initially moves at 18 m/s before being redirected to 33 m/s after a slap shot. The impulse required to change the puck's direction and speed is calculated using momentum principles, specifically the change in momentum. Initial calculations suggested an impulse of 64 Ns, but further analysis using the Pythagorean theorem indicated a more accurate value of approximately 6.39 Ns. The direction of the impulse was determined to be around 28.6°. The discussion highlights the importance of vector analysis in impulse calculations.
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Homework Statement



An ice hockey player receives a pass so that the puck, with a velocity of 18 m/s, glides across the ice, perpendicularly to the side axis (translation?) of the rink. After a slap shot, the puck moves exactly parallel to side axis of the rink, towards the goal with the velocity 33 m/s. Determine the impulse (magnitude and direction) of the hitting force the hockeystick affected the puck with. The mass of the puck is 170 g.
Answer: Imp=64 Ns , direction 29°.

Sorry for the awkward translation, I hope it is comprehensible!

Homework Equations



Imp=Δp

The Attempt at a Solution



So I take it you have to use the momentum, not Imp=FΔt, since you have all the needed quantities for it. I get the direction right, both the velocity and the momentum vectors are obviously perpendicular so I just use the tangent and get the angle. I have trouble with the magnitude of the impulse however. I understand you can't directly use the equation because of the different directions but I can't fit the impulse magnitude in with the above reasoning and the directions of the puck's momentums.

Any help is appreciated. :)
 
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Hello, Daltohn, welcome to PF!

Can you show more detail of your calculation and state the answer that you got?

I believe that 64 Ns is not the right answer.
 
Hi! My initial thought was to just use the Pythagorian theorem and calculate the magnitude of the vector, whose direction I had calculated to 28.6...°. I'm thinking that the impulse has to "cancel out" p1=0.17kg*18m/s, and also move give the puck p2=0.17kg*33m/s, in p2's direction. So Imp^2=3.06^2+5.61^2=40.83...→Imp=6.39...Ns.
 
And that is very similar indeed to the "correct answer", which means they made a typo. :) Didn't realize!
 
Daltohn said:
Hi! My initial thought was to just use the Pythagorian theorem and calculate the magnitude of the vector, whose direction I had calculated to 28.6...°. I'm thinking that the impulse has to "cancel out" p1=0.17kg*18m/s, and also move give the puck p2=0.17kg*33m/s, in p2's direction. So Imp^2=3.06^2+5.61^2=40.83...→Imp=6.39...Ns.

I think that's right.
 
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