If the momentum of a body increases by 50 %, its kinetic energy will increase by

Question

If the momentum of a body increases by 50 %, its kinetic energy will increase by (A) 50% (B) 100% (C) 125% (D) 150%

Tardigrade Admin · Accepted Answer

Relation between kinetic energy (KE) and momentum (p) is,

K E = \frac{p ^{2}}{2 m}

Initial kinetic energy

K E_{1} = \frac{p _{2}^{2}}{2 m}

,Similarly, final kinetic energy

K E_{2} = \frac{p _{1}^{2}}{2 m}

∴ \frac{E _{2} - E _{1}}{E _{1}} = \frac{p _{2}^{2} /2 m - p _{1}^{2} /2 m}{p _{1}^{2} /2 m}

ie. Percentage increase in

K E = \frac{p _{2}^{2} - p _{1}^{2}}{p _{1}^{2}} \times 100

Now, let,

p_{1} = 100

then

p_{2} = 150

So,

%

increase in KE

= \frac{( 50 ) ^{2} - ( 100 ) ^{2}}{( 100 ) ^{2}} \times 100

or

%

increase in

E = 125

Q. If the momentum of a body increases by 50%, its kinetic energy will increase by