Question

The momentum of a body is double .what is the % eg measure in kenitic energy.

Answer 1

Question :

If the momentum of a body is double .what is the % increase in kinetic energy.

Theory :

We know that

$\rm{K\propto\:P^2}$

$\implies\rm{\dfrac{K_{2}}{K_{1}}=(\dfrac{P_2}{P_1})^2}$

Solution :

Let the initial Momentum of a body be P and Kinteic energy is K.

i.e $\rm{P_1=P\:and\:K_1=K}$

Now Momentum is doubled

$\rm{P_2=2P}$

Therefore,

$\implies\rm{\dfrac{K_{2}}{K_{1}}=(\dfrac{P_2}{P_1})^2}$

$\implies\rm{\dfrac{K_{2}}{K}=\dfrac{(2P)^2}{P^2}}$

$\implies\rm{K_{2}=4K}$

Change in Kinteic energy

$\sf{=K_{2}-k_{1}}$

$\sf{=4k-k}$

$\sf{=3k}$

Therefore, Kinetic energy is increased by 300%

Answer 2

Answer:

Question :

If the momentum of a body is double .what is the % increase in kinetic energy.

Theory :

We know that

\rm{K\propto\:P^2}K∝P

2

\implies\rm{\dfrac{K_{2}}{K_{1}}=(\dfrac{P_2}{P_1})^2}⟹

K

1

K

2

=(

P

1

P

2

)

2

Solution :

Let the initial Momentum of a body be P and Kinteic energy is K.

i.e \rm{P_1=P\:and\:K_1=K}P

1

=PandK

1

=K

Now Momentum is doubled

\rm{P_2=2P}P

2

=2P

Therefore,

\implies\rm{\dfrac{K_{2}}{K_{1}}=(\dfrac{P_2}{P_1})^2}⟹

K

1

K

2

=(

P

1

P

2

)

2

\implies\rm{\dfrac{K_{2}}{K}=\dfrac{(2P)^2}{P^2}}⟹

K

K

2

=

P

2

(2P)

2

\implies\rm{K_{2}=4K}⟹K

2

=4K

Change in Kinteic energy

\sf{=K_{2}-k_{1}}=K

2

−k

1

\sf{=4k-k}=4k−k

\sf{=3k}=3k

Therefore, Kinetic energy is increased by 300%