What is their velocity just after the tackle?

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SUMMARY

The discussion centers on calculating the final velocity of two players after a perfectly inelastic tackle. A 90-kg fullback moving east at 5 m/s collides with a 95-kg opponent running north at 3 m/s. The correct approach involves applying the conservation of momentum in both the x (east-west) and y (north-south) directions separately. The final velocity is determined to be 2.9 m/s, contrasting with the incorrect initial calculation of 3.9 m/s.

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okay, more questions for me!

this really, really easy question is not working out for me!

A 90-kg fullback moving east with a speed of 5 m/s is tackled by a 95-kg opponent rnning north at 3m/s. what is their velocity just after the tackle?

Perfectly inelastic, right?

so (90)(5) + (95)(3) = (m1 + m2) Vf
So, according to this, Vf= 3.9 m/s but the book answer is 2.9 m/s. What am I doing wrong?
 
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Remember that momentum is a VECTOR QUANTITY. You have to consider momentum conservation in both the East-West (x) and North-South (y)directions separately. That is,

p_{x, initial} = p_{x, final} \ \ \ \ \ p_{y, initial} = p_{y, final}

After you find what the final velocities in the x and y directions remember that the magnitude of the velocity is

v_{final} = \sqrt{v_{x, final}^2 + v_{y, final}^2}
 
Last edited:
oooooohhhhh yeah. thanks! let's see if i can get the answer now...
 

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