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A particle is projected at an angle of $30^{\circ}$ with the horizontal with a momentum $p$. At the highest point its momentum is
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Verified Answer
The correct answer is:
$\frac{\sqrt{3}}{4} p$
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
$\begin{array}{rlr}p^{\prime}=m v_{x} & =m u\left(\frac{\sqrt{3}}{2}\right) & \\ & =\frac{\sqrt{3}}{2} p \quad \text { [using Eq. (i)] }\end{array}$
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
$\begin{array}{rlr}p^{\prime}=m v_{x} & =m u\left(\frac{\sqrt{3}}{2}\right) & \\ & =\frac{\sqrt{3}}{2} p \quad \text { [using Eq. (i)] }\end{array}$
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