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JEE Main 201212 MayOnline
Exam
JEE Main
Year
2012
Date
12 May
Mode
Online
Subject
Physics
Chapter
Center of Mass, Momentum and Collision

JEE Main 2012 · 12 May · Online · Physics · Center of Mass, Momentum and Collision

Two bodies A and B of mass m and 2 m respectively are placed on a...

Two bodies AA and BB of mass mm and 2m2 m respectively are placed on a smooth floor. They are connected by a spring of negligible mass. AA third body CC of mass mm is placed on the floor. The body CC moves with a velocity v0v_0 along the line joining AA and BB and collides elastically with AA. At a certain time after the collision it is found that the instantaneous velocities of AA and BB are same and the compression of the spring is x0x_0. The spring constant kk will beTwo bodies A and B of mass m and 2 m respectively are placed on a smooth floor. They are connected by a spring ofTwo bodies A and B of mass m and 2 m respectively are placed on a smooth floor. They are connected by a spring of

Options

  1. Amv02x02m \frac{v_0^2}{x_0^2}
  2. Bmv02x0m \frac{v_0}{2 x_0}
  3. C2mv0x02 m \frac{v_0}{x_0}
  4. D23m(v0x0)2\frac{2}{3} m\left(\frac{v_0}{x_0}\right)^2

Correct answer

D. 23m(v0x0)2\frac{2}{3} m\left(\frac{v_0}{x_0}\right)^2

Step-by-step solution

Two bodies A and B of mass m and 2 m respectively are placed on a smooth floor. They are connected by a spring of
Initial momentum of the system block (C)=mv0(C)=m v_0. After striking with AA, the block CC comes to rest and now both block A and B moves with velocity v\mathrm{v} when compression in spring is x0.Bythelawofconservationofx_0. By the law of conservation oflinear momentum m v_0=(m+2 m) v \Rightarrow v=v03\frac{v_0}{3}By the law of conservation of energy K.E. of block C=C= K.E. of system + P.E. of system 12mv02=12(3m)(v03)2+12kx02⇒12mv02=16mv02+12kx02⇒12kx02=12mv02−16mv02=mv023\begin{aligned}& \frac{1}{2}m v_0^2=\frac{1}{2}(3 m)\left(\frac{v_0}{3}\right)^2+\frac{1}{2}k x_0^2 \\ & \Rightarrow \quad \frac{1}{2}m v_0^2=\frac{1}{6}m v_0^2+\frac{1}{2}k x_0^2 \\ & \Rightarrow \quad \frac{1}{2}k x_0^2=\frac{1}{2}m v_0^2-\frac{1}{6}m v_0^2=\frac{m v_0^2}{3}\\ & \end{aligned}⇒12kx02=12mv02−16mv02=mv023∴k=23m(v0x0)2\begin{aligned}& \Rightarrow \quad \frac{1}{2}k x_0^2=\frac{1}{2}m v_0^2-\frac{1}{6}m v_0^2=\frac{m v_0^2}{3}\\ & \therefore \quad k=\frac{2}{3}m\left(\frac{v_0}{x_0}\right)^2 \end{aligned}
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