Question

If kinetic energy of a body is increased by 100% then by how much percentage it's momentum will increase? ​

Answer 1

Given:

• Kinetic energy is increased by 100%

━━━━━━━━━━━━━━━━

Need to find:

• By how much percentage the momentum will increase ?

━━━━━━━━━━━━━━━━

Solution:

let Initial momentum = P

→ P = $\sqrt{2m \times KE}$

• here m= mass
• KE = Initial Kinetic Energy

Final momentum = P'

→P' = $\sqrt{2m \times KE'}$

When kinetic energy increases by 100%, new KE will be :

KE' = KE + 100% of KE

→ KE' = KE + KE

$\purple{\bold{\boxed{\rightarrow KE' = 2KE----i}}}$

━━━━━━━━━━━━━━━━

We know,

$KE = \dfrac{ {P}^{2} }{2m}$

Here,

• P= momentum
• KE'= Final Kinetic Energy

KE' = 2 KE (from i)

→ $\dfrac{ {P'}^{2} }{2 m} = 2 \times \dfrac{ {P}^{2} }{2m}$

→${P'}^{2} = 2 {P}^{2}$

→$P' = \sqrt{2} P$

$\purple{\bold{\boxed{\rightarrow P' = 1.41 P ----ii}}}$

━━━━━━━━━━━━━━━━

% Change in P:

=$\dfrac{P'-P}{P} \times 100\%$

=$\dfrac{1.41P- P}{P} \times 100\%$

$= 0.41 \times 100\%$

$\red{\large{\bold{\boxed{\boxed{= 41\%}}}}}$

Answer 2

let Initial momentum = P

→ P =

   here m= mass

   KE = Initial Kinetic Energy