Question

How is the kinetic energy of a moving cart affected if (1) its mass is doubled , (2) its velocity is reduced to one-third of the initial velocity

Answer 1

we know that
K.E = 1/2•mV^2

a) if m = 2m
then

K.E = mV^2
i.e.
K.E gets double

b) V = 1/3v
then

K.E = 1/18(mv^2)

i.e.
K.E gets reduced by 1/9 times to it's initial value

Answer 2

Answer:

(1)Its mass is double-The kinetic energy becomes two times.

(2)Its velocity is reduced to one-third of the initial velocity-The kinetic energy reduces to one-ninth of its initial value.

Explanation:

The kinetic energy of a moving body is given as,

$K=\frac{1}{2}mv^2$        (1)

Where,

K=kinetic energy of the moving body

m=mass of the moving body

v=velocity with which the body is moving

               

Now according to the question,

Case I: mass is doubled

$m_2=2m$

So, equation (1) becomes,

$K_2=\frac{1}{2}\times 2m\times v^2$       (2)

By comparing equation (2) with equation (1) we get;

$K_2=2K$

The kinetic energy becomes two times its initial value.

Case II: velocity is reduced to one-third

$v'=\frac{v}{3}$

So, equation (1) becomes,

$K_2=\frac{1}{2}\times m\times( \frac{v}{3})^2=\frac{1}{2}\times m\times \frac{v^2}{9}$          (3)

By comparing equation (3) with equation (1) we get;

$K'=\frac{K}{9}$

The kinetic energy reduces to one-ninth of its initial value.

Hence, the kinetic energy of the moving cart is two times its initial value in case of mass is doubled and the kinetic energy reduces to one-ninth of its initial value in case the velocity is reduced to one-third of the initial velocity.