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pushpa_bhagat · Oct 11, 2010, 08:27 AM

FadedMaster · Oct 11, 2010, 08:51 AM

Please restate your question so that it can be understood.

Unknown008 · Oct 11, 2010, 10:04 AM

I assume what I'll be posting is what you are looking for...

Let the initial kinetic energy be E.

$E = \frac12 mv_1\ ^2$

It increases by 100%, or doubles, to become:

$E' = \frac12 mv_2\ ^2$

Where E' = 2E

Now, comparing the two:

$\frac{E'}{E} = \frac{\frac12 mv_2\ ^2}{\frac12 mv_1\ ^2} = \frac{v_2\ ^2}{v_1\ ^2}$

Now, momentum is given by:

$p = mv_1$

$p' = mv_2$

$\frac{p'}{p} = \frac{mv_2}{mv_1} = \frac{v_2}{v_1}$

Substituting what we found earlier;

$\frac{E'}{E} = \frac{p'^2}{p^2}$

Since this equals 2;

$\frac{p'^2}{p^2} = 2$

$\frac{p'}{p} = \sqrt2$

Can you finish it now?

Post what you get!

pushpa_bhagat · Oct 12, 2010, 02:11 AM

I want ans in % in this Q. If increase in K.E is 300% , increase in momentum in %
A.29% B .100% C .none
Thank for 1st Q . Tell me this Q.

Unknown008 · Oct 12, 2010, 02:23 AM

Just change it into %...

$\sqrt2 = 100 \times 1.41 = 141%$

When this equals to 3, we get:

$\frac{p'}{p} = \sqrt{3}$

$\sqrt3 = 100 \times 1.73 = 173%$

pushpa_bhagat · Oct 14, 2010, 03:50 AM

Thant u Igot it