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 sharp impulse so that it aquires a velocity v 0 towards right-A- Velocity of centre of mass will be m 2 V 0- m 1- m 0B- Velocity of centre of mass will be m1-m2 v0-m1-m2C- Maximum elongation that the spring will suffer x max- m 2- k m 1- m 2- v 0D- Maximum elongation that the spring will suffer x max- m 1 m 2- k m 1- m 2- v 0

Name: JEE - 11 - Physics - WAVES - Session 05 - W11
Uploaded: 2022-07-04T10:36:42+05:30
Description: Given, nbsp;Initial velocity of m_2 = v_0 Initial velocity of m_1 = 0 As we know, V_com = m_1 v_1 + m_2 v_2m_1 + m_2 V_com = m_1 (0) + m_2 (v_0)m_1 + m_2 = m_ ...

Given, nbsp;Initial velocity of m_2 = v_0 Initial velocity of m_1 = 0 As we know, V_com = m_1 v_1 + m_2 v_2m_1 + m_2 V_com = m_1 (0) + m_2 (v_0)m_1 + m_2 = m_ ...

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Byju's Answer

Standard XII

Physics

Work Energy Theorem Application

Two blocks of...

Question

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 sharp impulse so that it aquires a velocity $v_{o}$ towards right.

Velocity of centre of mass will be

\frac{m_{2} v_{0}}{m_{1} + m_{2}}

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Velocity of centre of mass will be

\frac{m_{1} v_{0}}{m_{1} + m_{2}}

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Maximum elongation that the spring will suffer

x_{m a x} = \sqrt{\frac{m_{2}}{k (m_{1} + m_{2})}} . v_{0}

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Maximum elongation that the spring will suffer

x_{m a x} = \sqrt{\frac{m_{1} m_{2}}{k (m_{1} + m_{2})}} . v_{0}

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Solution

The correct options are
A Velocity of centre of mass will be $\frac{m_{2} v_{0}}{m_{1} + m_{2}}$
D Maximum elongation that the spring will suffer $x_{m a x} = \sqrt{\frac{m_{1} m_{2}}{k (m_{1} + m_{2})}} . v_{0}$
Given, Initial velocity of $m_{2} = v_{0}$
Initial velocity of $m_{1} = 0$
As we know, $V_{c o m} = \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}}$
$V_{c o m} = \frac{m_{1} (0) + m_{2} (v_{0})}{m_{1} + m_{2}} = \frac{m_{2} v_{0}}{m_{1} + m_{2}}$

The spring will have maximum elongation when velocities of both blocks will attain the velocity of COM of the two blocks.
As we know,
Change in KE = Potential energy stored in the spring.
Assume $x_{m a x} =$ maximum elongation in the spring
Hence, $K E_{i} - K E_{f}$ = Potential energy stored
$\frac{1}{2} m_{2} v_{0}^{2} - [\frac{1}{2} (m_{1} + m_{2}) v_{c o m}^{2}] = \frac{1}{2} k x_{m a x}^{2}$
$\Rightarrow \frac{1}{2} m_{2} v_{0}^{2} - [\frac{1}{2} (m_{1} + m_{2}) {(\frac{m_{2} v_{0}}{m_{1} + m_{2}})}^{2}] = \frac{1}{2} k x_{m a x}^{2}$
$\Rightarrow m_{2} v_{0}^{2} [1 - \frac{m_{2}}{m_{1} + m_{2}}] = k x_{m a x}^{2}$
$\Rightarrow x_{m a x} = \sqrt{\frac{m_{1} m_{2}}{k (m_{1} + m_{2})}} v_{0}$

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Two blocks of masses $m_{1}$ and $m_{2}$ are connected by a spring of constant k (given figure).The block of mass $m_{2}$ is given a sharp impulse so that it acquires a velocity $v_{0}$ towards right.Find (a) the velocity of the centre of mass,(b) the maximum elongation that the spring will suffer.