Question

If the linear momentum is increased by 50% , then kinetic energy will be increased by... *​

Answer 1

Answer:

K.E.=P²/2M,P being increased 50% =>it becomes 1.5P=>new K.E.=(1.5P)²/2M=2.25P²/2M=>2.25K.E.

=>the KE is changed by (2.25–1)×100%=125%.Ans…

Answer 2

Given:-

• Linear momentum increased = 50%

To find:-

• Kinetic energy will be increased by?

Solution:-

If the initial momentum is $\sf p_0$

then the initial kinetic energy is $\sf k_0$ = $\sf{ \dfrac{ \sf p^{2}_0 }{2m}}$

When momentum is increased by 50%,

then the momentum become p = $\sf{ \sf p_0 + \bigg( \sf p_0 \times \dfrac{50}{100} \bigg) = \dfrac{3}{2} \sf p_0 }$

Now kinetic energy,

$\sf{ K = \dfrac{p^2}{2m} = \bigg( \dfrac{9}{4} \bigg) \times \bigg( \dfrac{ \sf p^{2}_0 }{2m} \bigg) = \dfrac{9}{4} \sf K_0 }$

Thus the percentage increased in kinetic energy = $\sf{ \dfrac{K - \sf K_0 }{ \sf K_0 } \times 100 = \bigg[ \dfrac{9}{4} - 1 \bigg]100 = 125 \%}$

Hence, Kinetic energy will be increased by 125%

$\underline{\rule{261}{2}}$