Verified Solution • 2025–26 Session moderate

QJEEM20Ph07d0: JEE Main 2020 Physics - Class 11 JEEM20Ph07d0 Solution: A block of mass m = 1 kg slides with velocity v = 6 m/s on a…

A block of mass m = 1 kg slides with velocity v = 6 m/s on a frictionless horizontal surface and collides with a uniform vertical rod and sticks to it as shown. The rod is pivoted about O and swings as a result of the collision making angle θ\theta before momentarily coming to rest. If the rod has mass M = 2 kg, and length ll = 1 m, the value of θ\theta is approximately :
(take g = 10 m/s2) JEE Main 2020 (Online) 3rd September Morning Slot Physics - Center of Mass and Collision Question 70 English

Multiple Choice Options

  • A 63o
  • B 69o
  • C 55o
  • D 49

Step-by-Step Verified Solution

JEE Main 2020 (Online) 3rd September Morning Slot Physics - Center of Mass and Collision Question 70 English Explanation

Applying Angular momentum conservation

mv ll = (Ml23+ml2)\left( {{{M{l^2}} \over 3} + m{l^2}} \right) ω\omega

⇒\Rightarrow ω=1×6×123+1\omega = {{1 \times 6 \times 1} \over {{2 \over 3} + 1}} = 185{{18} \over 5}

Now, using energy conservation

12Iω2=(m+M)gh{1 \over 2}I{\omega ^2} = \left( {m + M} \right)gh

⇒\Rightarrow 12(Ml23+ml2)ω2=(m+M)gxCOM(1−cos⁡θ){1 \over 2}\left( {{{M{l^2}} \over 3} + m{l^2}} \right){\omega ^2} = \left( {m + M} \right)g{x_{COM}}\left( {1 - \cos \theta } \right)

⇒\Rightarrow 12(23+1)(185)2=(m+M)g×M×l2+ml(M+m)(1−cos⁡θ){1 \over 2}\left( {{2 \over 3} + 1} \right){\left( {{{18} \over 5}} \right)^2} = \left( {m + M} \right)g \times {{M \times {l \over 2} + ml} \over {\left( {M + m} \right)}}\left( {1 - \cos \theta } \right)

⇒\Rightarrow 12×53×18×1825=10×(2×12+1)(1−cos⁡θ){1 \over 2} \times {5 \over 3} \times {{18 \times 18} \over {25}} = 10 \times \left( {2 \times {1 \over 2} + 1} \right)\left( {1 - \cos \theta } \right)

⇒\Rightarrow 56×18×1825=20(1−cos⁡θ){5 \over 6} \times {{18 \times 18} \over {25}} = 20\left( {1 - \cos \theta } \right)

⇒\Rightarrow (1−cos⁡θ)=185×320\left( {1 - \cos \theta } \right) = {{18} \over 5} \times {3 \over {20}}

⇒\Rightarrow cos⁡θ=1−2750{\cos \theta = 1 - {{27} \over {50}}}

⇒\Rightarrow cos⁡θ=2350{\cos \theta = {{23} \over {50}}}

⇒\Rightarrow θ\theta ≃\simeq 63o

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring