Name: Variable Mass System
Uploaded: 2019-07-19
Duration: 12 min 25 s
Description: In this video, we will look at how the laws of motion and their mathematical formulation apply to a variable mass system like a rocket. Don't miss it.

Question

Three particles $A, B, C$ of mass $m$ each are joined to each other by massless rigid rods to form an equilateral triangle of side $a$ . Another particle of mass $m$ hits $B$ with a velocity $v_{0}$ directed along $B C$ as shown. The colliding particle stops immediately after impact.

(a) Calculate the time required by the triangle $A B C$ to complete half revolution in its subsequent motion.

(b) What is the net displacement of point $B$ during this time interval?

Accepted Answer

(a) From conservation of angular momentum about cm (external torque is zero) of the system. $L_{i} = L_{f} m v_{0} \times \frac{a}{2 \sqrt{3}} = I ω I_{cm} = 3 m {(\frac{a}{\sqrt{3}})}^{2}, A D = \frac{\sqrt{3} a}{2} \Rightarrow m v_{0} \times \frac{a}{2 \sqrt{3}} = m a^{2} ω \Rightarrow ω = \frac{v_{0}}{2 \sqrt{3} a}$

Therefore, time required to complete half circle,

$t = \frac{π}{ω} \Rightarrow t = \frac{2 π \sqrt{3} a}{v_{0}}$

(b) From conservation of linear momentum.

$p_{i} = p_{f} m v_{0} = 3 m v_{cm} \Rightarrow v_{cm} = \frac{v_{0}}{3}$

To get net displacement of $B$ with respect to ground, we have to get displacement of $cm$ with respect to ground and displacement of $B$ with respect to $cm$ .

$x = x_{B / cm} + x_{cm / grd} = \frac{a}{2} + \frac{v_{0}}{3} \times \frac{2 π \sqrt{3} a}{v_{0}} \Rightarrow x = \frac{a (\sqrt{3} + 2 π)}{2 \sqrt{3}} y = y_{B / cm} + y_{cm / grd} y_{cm / grd} = 0, there is no force in y direction . \Rightarrow y = \frac{a}{\sqrt{3}} + 0 \Rightarrow y = \frac{a}{\sqrt{3}}$

Therefore, net displacement of $B$ ,

$s_{B} = \sqrt{x^{2} + y^{2}} \Rightarrow s_{B} = \sqrt{{|\frac{a (\sqrt{3} + 2 π)}{2 \sqrt{3}}|}^{2} + {[\frac{a}{\sqrt{3}}]}^{2}} \Rightarrow s_{B} = \frac{a}{\sqrt{3}} \sqrt{\frac{{(\sqrt{3} + 2 π)}^{2}}{4}} + 1$

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