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QJEEM25Ph8a19: JEE Main 2025 Physics - Class 11 JEEM25Ph8a19 Solution: Three equal masses are kept at vertices of an equilateral…

JEE Main 2025 (Online) 29th January Evening Shift Physics - Center of Mass and Collision Question 5 English

Three equal masses mm are kept at vertices (A,B,C)(A, B, C) of an equilateral triangle of side a in free space. At t=0t=0 , they are given an initial velocity VA→=V0AC→,VB→=V0BA→\overrightarrow{V_A}=V_0 \overrightarrow{A C}, \overrightarrow{V_B}=V_0 \overrightarrow{B A} and VC→=V0CB→\overrightarrow{V_C}=V_0 \overrightarrow{C B} . Here, AC→,CB→\overrightarrow{A C}, \overrightarrow{C B} and BA→\overrightarrow{B A} are unit vectors along the edges of the triangle. If the three masses interact gravitationally, then the magnitude of the net angular momentum of the system at the point of collision is :

Multiple Choice Options

  • A 12\frac{1}{2} a mV0\mathrm{mV}_0
  • B 3 a mV0\mathrm{mV}_0
  • C 32\frac{3}{2} a mV0\mathrm{mV}_0
  • D 32\frac{\sqrt{3}}{2} a m0\mathrm{m}_0

Step-by-Step Verified Solution

JEE Main 2025 (Online) 29th January Evening Shift Physics - Center of Mass and Collision Question 5 English Explanation

Here, r=a2−a24r = \sqrt {{a^2} - {{{a^2}} \over 4}}

⇒r=3a2\Rightarrow r = {{\sqrt 3 a} \over 2}

Here, as τext=0{\tau_{ext}} = 0

we know, τ=dLdt\tau = {{dL} \over {dt}}

⇒dLdt=0\Rightarrow {{dL} \over {dt}} = 0 ⇒\Rightarrow L = Constant

So, we can apply angular momentum conservation. i.e. angular momentum of the system about any point (Let's take point C) is conserved.

⇒Lf=Li\Rightarrow {L_f} = {L_i}

⇒Lf→=o×PA→+r→×mV0B^A+o×Pc→\Rightarrow \overrightarrow {{L_f}} = o \times \overrightarrow {{P_A}} + \overrightarrow r \times m{V_0}\widehat BA + o \times \overrightarrow {{P_c}} (as PB=mv0BA^{P_B} = m{v_0}\widehat {BA} and L→=r→×P→\overrightarrow L = \overrightarrow r \times \overrightarrow P )

⇒Lf→=(3a2)mv0sin⁡90∘\Rightarrow \overrightarrow {{L_f}} = \left( {{{\sqrt 3 a} \over 2}} \right)m{v_0}\sin 90^\circ (As ∣r→∣=3a2\left| {\overrightarrow r } \right| = {{\sqrt 3 a} \over 2} )

⇒Lf→=32amv0\Rightarrow \overrightarrow {{L_f}} = {{\sqrt 3 } \over 2}am{v_0} (outwards & ⊥\bot to plane)

⇒Lf=32amv0\Rightarrow {L_f} = {{\sqrt 3 } \over 2}am{v_0}

Hence, option D is correct.

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