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JIPMER 2015
Exam
JIPMER
Year
2015
Subject
Physics
Chapter
Work, Power and Energy

JIPMER 2015 · Physics · Work, Power and Energy

Two blocks of masses m_1 and m_2 are connected by a spring of...

Two blocks of masses m1m_1 and m2m_2 are connected by a spring of spring constant kk. The block of mass m2m_2 is given a sharp empulse so that it acquires a velocity v0v_0 towards right. Find the maximum elongation that the spring will suffer. Two blocks of masses m_1 and m_2 are connected by a spring of spring constant k . The block of mass m_2 is given a

Options

  1. A[m1m2m1+m2]12v0\left[\frac{m_1 m_2}{m_1+m_2}\right]^{\frac{1}{2}} v_0
  2. B(m1+m2m1−m2)v0\left(\frac{m_1+m_2}{m_1-m_2}\right) v_0
  3. C[m1+m2m1−m2]12v0\left[\frac{m_1+m_2}{m_1-m_2}\right]^{\frac{1}{2}} v_0
  4. D[2m1+m2m1m2]12v0\left[\frac{2 m_1+m_2}{m_1 m_2}\right]^{\frac{1}{2}} v_0

Correct answer

A. [m1m2m1+m2]12v0\left[\frac{m_1 m_2}{m_1+m_2}\right]^{\frac{1}{2}} v_0

Step-by-step solution

The velocity of centre of mass vcm=m1v1+m2v2m1+m2v_{\mathrm{cm}}=\frac{m_1 v_1+m_2 v_2}{m_1+m_2} When v1=0v_1=0 and v2=v0v_2=v_0, then vcm=m2v0m1+m2v_{c m}=\frac{m_2 v_0}{m_1+m_2} Now, let xx be the elongation in the spring. Change in potential energy = potential energy stored in spring ⇒12m2v02−12(m1+m2)(m2V0m1+m2)2=12kx2\begin{gathered} \Rightarrow \quad \frac{1}{2} m_2 v_0^2-\frac{1}{2}\left(m_1+m_2\right) \\ \left(\frac{m_2 V_0}{m_1+m_2}\right)^2=\frac{1}{2} k x^2 \end{gathered} This gives, x=(m1m2m1+m2)12v0x=\left(\frac{m_1 m_2}{m_1+m_2}\right)^{\frac{1}{2}} v_0
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