Question

If the KE of a particle becomes four times its initial value, then the new momentum will be more than
its initial momentum by percent
(A) 50%
(C) 125%
(D) 150%
BY100%
A
: 10
Ashi:​

Answer 1

Answer:

Given:

New KE is 4 times the initial value

To find:

% increase in momentum

Formulas used:

$ke = \dfrac{ {p}^{2} }{2m}$

$= > p = \sqrt{2m(ke)}$

Calculation:

Initial Momentum:

$= > p1 = \sqrt{2m(ke)}$

Final Momentum:

$= > p2 = \sqrt{2m(4ke)}$

$= > p2 = 2 \times \sqrt{2m(ke)}$

$= > p2 = 2 \times (p1)$

So change in Momentum:

$\Delta p = p2 - p1$

$= > \Delta p =2( p1) - p1$

$= > \Delta p = \: p1$

So, % Change in Momentum :

$= \frac{ \Delta p}{p1} \times 100\%$

$= \frac{p1}{p1} \times 100\%$

$= 100\%$

So final answer is 100%

Answer 2

$\huge{\underline{\underline{\red{\mathfrak{AnSwEr :}}}}}$

• Kinetic Energy is increased by 4 times

$\small{\underline{\blue{\sf{Given :}}}}$

$\rule{200}{1}$

$\small{\underline{\green{\sf{Solution :}}}}$

We have formula :

$\large \star {\boxed{\sf{K.E \: = \: \dfrac{p^2}{2m}}}} \\ \\ \implies {\sf{p^2 \: = \: K.E \: \times \: 2m}} \\ \\ \implies {\sf{p \: = \: \sqrt{K.E \: (2m)}}}$

$\rule{200}{1}$

• Initial Momentum

$\implies {\sf{p \: = \: \sqrt{K.E(2m)}}}$

• Final Momentum

$\implies {\sf{p' \: = \: \sqrt{4( K.E) \: (2m)}}} \\ \\ \implies {\sf{p' \: = \: 2 \sqrt{K.E(2m)}}} \\ \\ \implies {\sf{p' \: = \: 2p}}$

Now, take change

$\large \star{\boxed{\sf{\Delta p \: = \: p' \: - \: p}}} \\ \\ \implies {\sf{\Delta p \: = \: 2p \: - \: p}} \\ \\ \implies {\sf{\Delta p \: = \: p}}$

$\rule{200}{1}$

As we know that Formula for change is :

$\large \star {\boxed{\sf{Change \: = \: \dfrac{\Delta p}{p} \: \times \: 100 \%}}} \\ \\ \implies {\sf{Change \: = \: \dfrac{\cancel{p}}{\cancel{p}} \: \times \: 100 \%}} \\ \\ \implies {\sf{Change \: = \: 100 \: \%}}$

∴ Change is 100 %

Answer is (E)