A particle is projected at an angle of $30^{\circ}$ with the horizontal with a momentum $p$. At the highest point its momentum is

• Correct answer
  A

  $\frac{\sqrt{3}}{4} p$
• B

  $\frac{2}{\sqrt{3}} p$
• C

  $p$
• D

  $\frac{1}{2} p$

Solution

Let $m$ be the mass of particle and $u$ be its initial velocity, then Initial momentum, $p=m u \quad \text{...(i)}$ At highest point,

$\begin{aligned} v_{x} &=u \cos \theta=u \cos 30^{\circ} \\ &=\frac{\sqrt{3}}{2} u \\ v_{y} &=0 \end{aligned}$ $\therefore$ Momentum at highest point is $\begin{array}{rlr}p^{\prime}=m v_{x} & =m u\left(\frac{\sqrt{3}}{2}\right) & \\ & =\frac{\sqrt{3}}{2} p \quad \text { [using Eq. (i)] }\end{array}$

Hence, the correct answer is option: 'A'.