Question

The kinetic Energy of an object of mass m
moving with the velocity of anal 4 ms–¹ is 16 J.
what will be the kinetic Energy when
a) its velocity is doubled.
b) reduced to half...​

Answer 1

Mass is constant in the system for both cases.

v = 4 m/s²

m = m

K.E = 1/2 mv²

16 = 1/2 m(4)²

32 = 16m

m = 2 Kg

Case I:

Normal K.E of the object = 1/2 mv²

mass, m = m

velocity, v = 2v

We know that,

$\qquad \quad \; \boxed{E_{k.e} = \frac{1}{2}mv^2}$

RATIO:

$\Rightarrow \qquad \frac{\cancel{\frac{1}{2}m}v^2}{\cancel{\frac{1}{2}m}(2v)^2} \\ \Rightarrow \qquad \frac{\cancel{v^2}}{4\cancel{v^2}} \\ \Rightarrow \qquad \frac{1}{4} \; or \; 1 : 4$

Hence, Kinetic Energy will increase by four times.

Case II:

Normal K.E of the object = 1/2 mv²

mass, m = m

velocity, v = v/2

We know that,

$\qquad \quad \; \boxed{E_{k.e} = \frac{1}{2}mv^2}$

RATIO:

$\Rightarrow \qquad \frac{\cancel{\frac{1}{2}m}v^2}{\cancel{\frac{1}{2}m}\left( \frac{v}{2} \right)^2} \\ \Rightarrow \qquad \frac{\cancel{v^2}}{\frac{v^2}{4}} \\ \Rightarrow \qquad \frac{4\cancel{v^2}}{\cancel{v^2}} \\ \qquad \quad \; \frac{4}{1} \; or \; 4 : 1$

Hence, Kinetic Energy will reduce by four times.

Hope this helps you, bro!