Ask Me Help Desk - IF increase in K.E is of a body is 100% ,increase in momentum

IF increase in K.E is of a body is 100% ,increase in momentum

Please restate your question so that it can be understood.

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!

Comment on Unknown008's post

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.

Just change it into %...

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

When this equals to 3, we get:

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

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

Comment on Unknown008's post

Thant u Igot it