JEE Advanced · Physics · 4. Motion in 2D
A particle of mass is projected from the ground with an initial speed at an angle with the horizontal. At the highest point of its trajectory, it makes a completely inelastic collision with another identical particle, which was thrown vertically upward from the ground with the same initial speed . The angle that the composite system makes with the horizontal immediately after the collision is
- A
- B
- C
- D
Answer & Solution
Correct Answer
(A)
Step-by-step Solution
Detailed explanation
At the highest point

\(v_{1}=\frac{u_{0} \cos \alpha}{2} \quad\) (by applying momentum conservation in horizontal direction)
\(\mathrm{v}_{2}=\frac{\mathrm{u}_{0} \cos \alpha}{2} \quad\) (by applying momentum conservation in vertical direction)
\(\theta=45^{\circ}\left(\mathrm{H}=\frac{\mathrm{u}_{0}^{2} \sin ^{2} \alpha}{2 \mathrm{~g}}\right)\)

\(v_{1}=\frac{u_{0} \cos \alpha}{2} \quad\) (by applying momentum conservation in horizontal direction)
\(\mathrm{v}_{2}=\frac{\mathrm{u}_{0} \cos \alpha}{2} \quad\) (by applying momentum conservation in vertical direction)
\(\theta=45^{\circ}\left(\mathrm{H}=\frac{\mathrm{u}_{0}^{2} \sin ^{2} \alpha}{2 \mathrm{~g}}\right)\)
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