Bow & Arrow - Impulse-Momentum Theorem

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SUMMARY

The discussion centers on the application of the impulse-momentum theorem to a scenario where a 64.2 kg man fires a 0.3 kg arrow at a velocity of 63 m/s while sliding backward at -0.26 m/s. The conservation of momentum equation, represented as m1v1 + m2v2 = m1v1' + m2v2', is utilized to determine the archer's final velocity after firing the arrow. The correct value for the arrow's velocity post-firing is confirmed to be 63 m/s, clarifying the confusion regarding the arrow's motion after release.

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Homework Statement


A 64.2kg man fires a .3kg arrow. The arrow is measured to have a velocity of 63m/s by a stationary bystander. All these happen while the archer sliding backwards at -.26m/s. What will be his velocity (in the x-direction) after firing this arrow?

Homework Equations


pbefore = pafter


The Attempt at a Solution


I used the eq'n m1v1 + m2v2 = m1v1' + m2v2'
I was able to solve the left side to be 2.208.
But I'm unsure of what to use for v2' on the right side... The arrow surely doesn't still fly at 63m/s right? Nor does it come to rest?

Thank you!
 
Physics news on Phys.org
v2' will be 63m/s.why do you have a doubt here. what have you taken as v2
 

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