Question

A block of mass m = 1 kg slides with velocity v = 6 m/s on a friction less 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 before momentarily coming to rest. If the rod has mass M = 2 kg and length l = 1 m, the value of θ is approximately (take g = 10 m/s2)

• A. 30°
• B. 45°
• C. 60°
• D. 90°

Answer 1

Answer: (C)

60°

Answer 2

1. Conservation of Angular Momentum (Collision):

   • Initial angular momentum of the block about O: $L_{initial} = mv \times l$.
   • Moment of inertia of the rod about O: $I_{rod} = \frac{1}{3}Ml^2$.
   • Moment of inertia of the block about O: $I_{block} = ml^2$.
   • Total moment of inertia after collision: $I_{total} = I_{rod} + I_{block} = \frac{1}{3}Ml^2 + ml^2 = l^2(\frac{M}{3} + m)$.
   • Just after collision, angular momentum $L_{final} = I_{total}\omega$.
   • Equating $L_{initial} = L_{final}$: $mvl = l^2(\frac{M}{3} + m)\omega \implies \omega = \frac{mv}{l(\frac{M}{3} + m)}$.
   • Plugging in values: $m=1, v=6, M=2, l=1$. $\omega = \frac{1 \times 6}{1(\frac{2}{3} + 1)} = \frac{6}{5/3} = \frac{18}{5} = 3.6$ rad/s.

2. Conservation of Mechanical Energy (Swing):

   • Initial kinetic energy (just after collision): $KE_{initial} = \frac{1}{2}I_{total}\omega^2$. $I_{total} = 1^2(\frac{2}{3} + 1) = \frac{5}{3}$ kg m$^2$. $KE_{initial} = \frac{1}{2} \times \frac{5}{3} \times (3.6)^2 = \frac{5}{6} \times 12.96 = 10.8$ J.
   • Potential energy gained when rod swings by angle $\theta$: Center of mass of rod rises by $\frac{l}{2}(1-\cos\theta)$. Block rises by $l(1-\cos\theta)$. $PE_{final} = Mgh_{rod} + mgh_{block} = Mg\frac{l}{2}(1-\cos\theta) + mgl(1-\cos\theta) = (\frac{M}{2} + m)gl(1-\cos\theta)$.
   • Plugging in values: $M=2, m=1, g=10, l=1$. $PE_{final} = (\frac{2}{2} + 1) \times 10 \times 1 \times (1-\cos\theta) = 2 \times 10 \times (1-\cos\theta) = 20(1-\cos\theta)$.
   • Equating $KE_{initial} = PE_{final}$: $10.8 = 20(1-\cos\theta)$. $1-\cos\theta = \frac{10.8}{20} = 0.54$. $\cos\theta = 1 - 0.54 = 0.46$.
   • $\theta = \arccos(0.46) \approx 62.63^\circ$.
   • This value is closest to $60^\circ$.