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# Momentum Help (page 2)

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By McGraw-Hill Professional
Updated on Sep 4, 2011

## Momentum Practice Problems

#### Problem 1

Suppose that an object of mass 2.0 kg moves at a constant speed of 50 m/s in a northerly direction. An impulse, acting in a southerly direction, slows this mass down to 25 m/s, but it still moves in a northerly direction. What is the impulse responsible for this change in momentum?

#### Solution 1

The original momentum p 1 is the product of the mass and the initial velocity:

P 1 = 2.0 kg × 50 m/s = 100 kg · m/s

in a northerly direction. The final momentum p 2 is the product of the mass and the final velocity:

p 2 = 2.0 kg × 25 m/s = 50 kg · m/s

in a northerly direction. Thus, the change in momentum is p 2p 1 :

p 2p 1 = 50 kg · m/s − 100 kg · m/s = −50 kg · m/s

in a northerly direction. This is the same as 50 kg · m/s in a southerly direction. Because impulse is the same thing as the change in momentum, the impulse is 50 kg · m/s in a southerly direction.

Don’t let this result confuse you. A vector with a magnitude − x in a certain direction is the same as a vector with magnitude x in the exact opposite direction. Problems sometimes will work out to yield vectors with negative magnitude. When this happens, just reverse the direction and then take the absolute value of the magnitude.

Practice problems of these concepts can be found at: Momentum, Work, Energy, And Power Practice Test

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