Question

1) What will be the final KE if the velocity of a body is doubled keeping mass constant
2)What will be the final KE If the mass of a body is doubled keeping velocity constant.
3)What will be the final KE If the mass of a body is halved keeping velocity constant.
4)What will be the final KE If the velocity of a body is halved keeping mass constant.
with full process plz............

Answer 1

Question 1:

What will be the final KE if the velocity of a body is doubled keeping mass constant?

Solution:

Initial Kinetic Energy be KE ,

$\sf{ \therefore \: KE = \dfrac{1}{2} m {v}^{2} }$

Final Kinetic Energy be KE_(2);

$\sf{ \therefore \: KE_{2} = \dfrac{1}{2} m {(2v)}^{2} }$

$\sf{ = > \: KE_{2} = 4 \times \dfrac{1}{2} m {v}^{2} }$

$\sf{ = > \: KE_{2} = 4 \times KE }$

So, kinetic energy becomes four times as that of the initial kinetic energy.

Question 2:

What will be the final KE If the mass of a body is doubled keeping velocity constant?

Solution:

Let initial kinetic energy be KE:

$\sf{ \therefore \: KE = \dfrac{1}{2} m {v}^{2} }$

Let final kinetic energy be KE_(2)

$\sf{ \therefore \: KE_{2} = \dfrac{1}{2} (2m) {v}^{2} }$

$\sf{ = > \: KE_{2} = 2 \times \dfrac{1}{2} m {v}^{2} }$

$\sf{ = > \: KE_{2} = 2 \times KE }$

So, kinetic energy becomes twice as that of the initial kinetic energy.

Question 3:

What will be the final KE If the mass of a body is halved keeping velocity constant?

Solution:

$\sf{ \therefore \: KE = \dfrac{1}{2} m {v}^{2} }$

Let final kinetic energy be KE_(2);

$\sf{ \therefore \: KE_{2} = \dfrac{1}{2} ( \dfrac{1}{2} m) {v}^{2} }$

$\sf{ = > \: KE_{2} = \dfrac{1}{2} \times \dfrac{1}{2} m {v}^{2} }$

$\sf{ = > \: KE_{2} = \dfrac{1}{2} \times KE }$

So, kinetic energy becomes halved.

Question 4:

What will be the final KE If the velocity of a body is halved keeping mass constant?

Solution:

$\sf{ \therefore \: KE = \dfrac{1}{2} m {v}^{2} }$

Let final kinetic energy be KE_(2);

$\sf{ \therefore \: KE_{2} = \dfrac{1}{2} m {( \dfrac{1}{2} v)}^{2} }$

$\sf{ = > \: KE_{2} = \dfrac{1}{4} \times \dfrac{1}{2} m { v}^{2} }$

$\sf{ = > \: KE_{2} = \dfrac{1}{4} \times KE }$

So, kinetic energy will be ¼ times the initial kinetic energy.

HOPE IT HELPS.