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!

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%

Thant u Igot it