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A block of mass M is kept on a smooth ho...

A block of mass M is kept on a smooth horizontal surface. Another block of mass m, moving with velocity v, collides with it and coalesces. What is the resultant velocity of the system?

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Knowledge Check

  • Total KE of sphere of mass M rolling with velocity v is

    A
    `(7)/(10)Mv^(2)`
    B
    `(5)/(6)Mv^(2)`
    C
    `(7)/(5) Mv^(2)`
    D
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  • A uniform rod of length l and mass 2m rests on a smooth horizontal table. A point mass m moving horizontally at right angle to the rod with velocity v collides with one end of the rod and sticks to it, then

    A
    angular velocity of the system after collision is v/l
    B
    angular velocity of the system after collision is v/2l
    C
    the loss in kinetic energy of the system as a whole as a result the collision is `mv^2//6`
    D
    the loss in kinetic energy of the system as a whole as a result of the collision is `7mv^2//24`

  • A circular hoop of mass m and radius R rests flat on a horizontal smooth surface. A bullet of same mass m and moving with velocity v strikes the hoop tangentially and gets embedded in it. Neglect the thickness of hoop in comparison to radius of hoop. Answer the following question based on above passage: Mark the correct option

    A
    Energy of system (hoop + bullet) is conserved
    B
    Momentum of system is conserved
    C
    Angular momentum of system is conserved about any point in the plane of hoop
    D
    Angular momentum is not conserved due to torque of gravitational force

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