COMEDK · Physics · 4. Motion In Two Dimensions
A particle is projected at an angle of \(30^{\circ}\) with the horizontal with a momentum \(p\). At the highest point its momentum is
- A \(\frac{\sqrt{3}}{4} p\)
- B \(\frac{2}{\sqrt{3}} p\)
- C \(p\)
- D \(\frac{1}{2} p\)
Answer & Solution
Correct Answer
(A) \(\frac{\sqrt{3}}{4} p\)
Step-by-step Solution
Detailed explanation
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…
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