Question

A particle slides on a frictionless horizontal surface with some initial speed $V_0$. In time $t_0$ it loses $3/4$th of the kinetic energy, then in next time interval of $t_0$, it will lose _______ of kinetic energy.

• A. $1/4$th
• B. $1/6$th
• C. $2/9$th
• D. $9/16$th

Answer 1

The correct option is A $1/4$th

Let the initial kinetic energy be $K$.
So after $3K/4$ loss of energy, the particle will have $K/4$ energy.
As the initial velocity was $V_0$,
$K/4=\frac{1}{4}\times \frac{1}{2}m{V}_0^2=\frac{1}{2}m{V}^2$
$\Rightarrow V=V_0/2$

The speed becomes half. Assuming the deceleration to be $a$.
We can write,
$V=V_0-a{t}_0$
$\frac{V_0}{2}=V_0-a{t}_0$
$\Rightarrow t_0=\frac{V_0}{2a}$

In next time interval of $t_0$,
$\to V^{\prime }=V-a{t}_0=\frac{V_0}{2}-\frac{V_0}{2}=0$
$\to V^{\prime }=0$
Hence the particle comes to rest. So it loses the remaining kinetic energy i.e. $K/4$.