Finding the mass of a hockey puck - inertia

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SUMMARY

The mass of a hockey puck can be determined using the impulse-momentum theorem. Given an impulse of 4.5 kg*m/s and a final velocity of 12 m/s in the opposite direction, the mass can be calculated using the equation jx = m(vf)x - m(vi)x. By rearranging this equation, the mass can be isolated and solved, confirming that friction and drag are negligible in this scenario.

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Finding the mass of a hockey puck - inertia!

An ice hockey puck slides along the ice at 12m/s. A hockey stick delivers an impulse of 4.5 kg * m/s, causing the puck to move off in the opposite direction with the same speed. What is the mass of the puck. Friction and drag are negligible.

Any Ideas? I am completely lost.

Thanks!
 
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Units...

Thanks!

Yes my units are correct. I was skimming over my notes from class and I believe we have to use the ∆px=jx=area under the curve equation

to find the ball's momentum:

m(vf)x-m(vi)x but then I don't have "m" this whole thing is equal to jx

and then the area under the curve

jx=fmax • ∆time

but I have jx and the velocities, but not the mass or the fMax
 
crewrules101 said:
Thanks!

Yes my units are correct. I was skimming over my notes from class and I believe we have to use the ∆px=jx=area under the curve equation

to find the ball's momentum:

m(vf)x-m(vi)x but then I don't have "m" this whole thing is equal to jx

and then the area under the curve

jx=fmax • ∆time

but I have jx and the velocities, but not the mass or the fMax
oh sorry, you're actually ahead of me :-p hopefully someone will be able to help you tonight.
 
Last edited:

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