Question

A body starting from rest is moving at constant acceleration. After travelling a certain distance ${}^{\prime }{x}^{\prime }$, its momentum is ${}^{\prime }{k}^{\prime }$ units. What would be the momentum of the body at double the initial distance?

• A. $\sqrt{12}$ times the initial momentum
• B. $\sqrt{3}$ times the initial momentum
• C. $\sqrt{2}$ times the initial momentum
• D. $\sqrt{4}$ times the initial momentum

Answer 1

The correct option is C $\sqrt{2}$ times the initial momentum

Initial velocity $\left(u\right)$= 0

we know that $v^2-u^2=2as$

$v^2=2as$

it is clear that $v\propto \sqrt{s}$

$⟹$ $mv\propto \sqrt{s}$

$⟹$ $p\propto \sqrt{s}$

Now let momentum be $k_1$

so, $\frac{k_1}{k}=\frac{\sqrt{2s}}{\sqrt{s}}$

$⟹k_1=k\sqrt{2}$