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Two blocks of mass m and M are lying on ...

Two blocks of mass m and M are lying on a smooth table. A spring is attached with the block of mass M (see fig). Block of mass m is given a velocity towards the other block. Find the value of `(M)/(m)` for which the kinetic energy of the system will never fall below one third of the initial kinetic energy imparted to the block of mass m.

Text Solution

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The correct Answer is:
`(M)/(m) le 2`
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Knowledge Check

  • Two blocks of masses m_1 and m_2 are connected as shown in the figure. The acceleration of the block m_2 is :

    A
    `(m^2g)/(m_1+m_2)`
    B
    `(m_1g)/(m_1+m_2)`
    C
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    D
    `(m_2g)/(m_1+4m_2)`

  • Three blocks of masses m , m and M are kept on a frictionless floor as shown in the figure .The leftmost block is given velocity v towards the right . All the collision between the blocks are perfectly inelastic . The loss in kinetic energy after all the collision is 5//6th of initial kinetic energy. The ratio of M//m will be

    A
    `1/4`
    B
    `1/8`
    C
    2
    D
    4

  • A block of mass m_(1) rests on a horizontal table. A string tied to the block is passed on a frictionless pulley fixed at the end of the table and to the other end of string is hung another block of mass m_(2) . The acceleration of the system is

    A
    `(m_(2)g)/(m_(1)+m_(2))`
    B
    `(m_(1)g)/(m_(1)+m_(2))`
    C
    g
    D
    `(m_(2)g)/(m_(1))`

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