GREPhysics.NET: GR0177 #54

GR0177 #54

Problem

GREPhysics.NET Official Solution

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Mechanics $\Rightarrow$ }Impulse

Impulse is defined as $J=\int F dt$ . But, for this problem, one doesn't have to evaluate a messy integral. Instead, the area under the curve is just the sum of two triangles. $J=2\frac{1}{2}(2)(1)=2kgm/s$ , as in choice (C).

Alternate Solutions

ramparts
2009-10-08 10:53:17

Better yet, it's one triangle :P $\frac{1}{2}*2*2 = 2$ kg m/s.

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Comments

kevintah
2015-10-02 10:29:02

I mean . . . . when you look at it, it just screams area under the curve scenario. . . so you just do area of triangle.

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Almno10
2010-11-09 12:57:21

better yet, you can write the equations of the lines F = 2t on [0,1] and F = -t+4 on [1,2] and simply sum the integrals! j/k

Dr. D.R. Dopetec
2011-09-28 19:38:06

ha yes

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fiber
2009-11-05 22:09:52

better yet, put the triangles together and it's one rectange with $A=x\cdot y = 1 \cdot 2$

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ramparts
2009-10-08 10:53:17

Better yet, it's one triangle :P $\frac{1}{2}*2*2 = 2$ kg m/s.

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Better yet, it's one triangle :P $\frac{1}{2}*2*2 = 2$ kg m/s.

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