Question

A particle is moving with momentum $p$ and kinetic energy $K$ in a gravity free space. A constant force $F$ (directed perpendicular to the initial momentum) now acts on the particle for time $t$. Final kinetic energy of the particle is

• A. $K[1+{\left(\frac{Ft}{p}\right)}^2]$
• B. $K[1+2{\left(\frac{Ft}{2}\right)}^2]$
• C. $K\sqrt{1+{\left(\frac{Ft}{p}\right)}^2}$
• D. $K\sqrt{1+{\left(\frac{Ft}{p}\right)}^4}$

Answer 1

The correct option is A $K[1+{\left(\frac{Ft}{p}\right)}^2]$

Let initial velocity of particle $=u$

Acceleration of the particle, $a\left(t\right)=\frac{F}{m}$
Integrating, we get $v\left(t\right)=\frac{F}{m}t$

As $\frac{p^2}{2m}=K$
$\therefore v\left(t\right)=\frac{F}{p^2}\times 2Kt$

Now, since $v\left(t\right)$ is perpendicular to $u$ [given],
final K.E of particle $K_f=\frac{1}{2}m(u^2+v(t{)}^2)=\frac{1}{2}m(\frac{F^2\left(4K^2{t}^2\right)}{p^4}+u^2)$
$=\frac{2F^2{K}^2{t}^{2}m}{p^4}+K$
where $K=\frac{1}{2}m{u}^2$ [given]

$\Rightarrow K_f=K[\frac{2K{F}^2{t}^{2}m}{p^4}+1]=K[1+\frac{K}{p^2}\times \frac{F^2{t}^2}{K}]$
[since $\frac{p^2}{2m}=K$]

$\Rightarrow K_f=K[1+\frac{F^2{t}^2}{p^2}]$