Solving Physics Question: Momentum Change of 0.42 kg Ball

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SUMMARY

The discussion centers on calculating the change in linear momentum of a 0.42 kg ball that strikes a wall while moving at 5.5 m/s and rebounds at 2.0 m/s. The correct formula for change in momentum is Δp = mv2 - mv1, where m is the mass, v2 is the rebound speed, and v1 is the initial speed. The magnitude of the change in momentum is calculated as Δp = (0.42 kg * 2.0 m/s) - (0.42 kg * 5.5 m/s), resulting in a definitive value of -1.26 kg·m/s, indicating a reversal in direction.

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A 0.42 kg ball is moving horizontally with a speed of 5.5 m/s when it strikes a vertical wall. The ball rebounds with a speed of 2.0 m/s. What is the magnitude of the change in linear momentum of the ball?

Might seem like a simple problem, but everytime i input an answer into it, it says I am wrong, can neone help?
 
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draconicspectre said:
A 0.42 kg ball is moving horizontally with a speed of 5.5 m/s when it strikes a vertical wall. The ball rebounds with a speed of 2.0 m/s. What is the magnitude of the change in linear momentum of the ball?

Might seem like a simple problem, but everytime i input an answer into it, it says I am wrong, can neone help?

Since the change in linear momentum is a vector given by (mv2-mv1)i, its magnitude is the square root of (mv2-mv1)^2, which equals mv2-mv1. This should work.
 

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