Question

In the non-relativistic regime, if the momentum, is increased by 100%, the percentage increase in kinetic energy is :

• A. 100
• B. 200
• C. 300
• D. 400

Answer 1

Answer: (C)

300

Answer 2

kinetic energy,

Since m remains constant

$\therefore \mathrm { K } \propto \mathrm { p } ^ { 2 }$

$\frac { \mathrm { k } _ { 1 } } { \mathrm { k } _ { 2 } } = \left( \frac { \mathrm { p } _ { 1 } } { \mathrm { p } _ { 2 } } \right) ^ { 2 } = \left( \frac { \mathrm { p } _ { 1 } } { 2 \mathrm { p } _ { 1 } } \right) ^ { 2 } = \frac { 1 } { 4 }$ or $\mathrm { k } _ { 2 } = 4 \mathrm { k } _ { 1 }$

Percentage increase in kinetic energy

$= \frac { \mathrm { k } _ { 2 } - \mathrm { k } _ { 1 } } { \mathrm { k } _ { 1 } } \times 100 = \frac { 4 \mathrm { k } _ { 1 } - \mathrm { k } _ { 1 } } { \mathrm { k } _ { 1 } } \times 100 = 300 \%$