Engineering Mechanics Miscellaneous

Engineering Mechanics Miscellaneous

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Engineering Mechanics

Engineering Mechanics Miscellaneous
  1. A bullet of mass m travels at a very high velocity V (as shown in the figure) and gets embedded inside the block of mass M initially at rest on a rough horizontal floor. The block with the bullet is seen to move a distance 's' along the floor. Assuming μ to be the coefficient of kinetic friction between the block and the floor and g the acceleration due to gravity. What is the velocity V of the bullet?








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    Let, m = mass of the bullet
    v = velocity of bullet
    M = mass of the block
    u = final velocity of system (block + bullet)

    From conservation of linear momentum
    mv = (M + m)u

    ⇒ u =
    mv
    m + M

    For horizontal equilibrium of body (M + m)g = N
    Fr = Frictional force = μN = u(M + m)g
    Frictional retardation =
    μ(M + m)g
    		 = -μg
    (M + m)


    (–ve sign shows that acceleration is opposite to motion)
    Let vf = final velocity of system = 0
    v²f = u² + 2as
    u² + 2as = 0
    u² – 2μgs = 0
    From equation (i)
    mv
    		² = 2μgs
    M + m

    mv
    		 = √2μgs
    M + m

    ⇒ v =
    M + m
    		√2μgs
    m

    Correct Option: A

    Let, m = mass of the bullet
    v = velocity of bullet
    M = mass of the block
    u = final velocity of system (block + bullet)

    From conservation of linear momentum
    mv = (M + m)u

    ⇒ u =
    mv
    m + M

    For horizontal equilibrium of body (M + m)g = N
    Fr = Frictional force = μN = u(M + m)g
    Frictional retardation =
    μ(M + m)g
    		 = -μg
    (M + m)


    (–ve sign shows that acceleration is opposite to motion)
    Let vf = final velocity of system = 0
    v²f = u² + 2as
    u² + 2as = 0
    u² – 2μgs = 0
    From equation (i)
    mv
    		² = 2μgs
    M + m

    mv
    		 = √2μgs
    M + m

    ⇒ v =
    M + m
    		√2μgs
    m

  1. A ball A of mass m falls under gravity from a height h and strikes another ball B of mass m which is supported at rest on a spring of stiffness k. Assume perfect elastic impact. Immediately after the impact








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    In a perfectly elastic collision between equal masses of two bodies, velocities exchange on impact.
    Velocity just before impect = Velocity immediately after impact.

    Correct Option: B

    In a perfectly elastic collision between equal masses of two bodies, velocities exchange on impact.
    Velocity just before impect = Velocity immediately after impact.

  1. A force of 100 N is applied to the centre of a circular disc, of mass 10 kg and radius 1 m, resting on a floor as shown in the figure. If the disc rolls without slipping on the floor, the linear acceleration (in m/s²) of the centre of the disc is _________ (correct to two decimal places).








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    m = 10 kg, r = 1 m
    100 – f = maCM = mr∝ ...(A) '
    [Tarque]CM = ICM’∝

    f × r =
    mr²
    		 × ∝
    2

    2f = mr∝
    2f = macm
    f =
    m
    		 × acm ...........(ii)
    2

    By (A) & (B) we get
    100 -
    macm
    		 = macm
    2

    we get on solving
    aCM = 6.67
    m
    s²

    Correct Option: B


    m = 10 kg, r = 1 m
    100 – f = maCM = mr∝ ...(A) '
    [Tarque]CM = ICM’∝

    f × r =
    mr²
    		 × ∝
    2

    2f = mr∝
    2f = macm
    f =
    m
    		 × acm ...........(ii)
    2

    By (A) & (B) we get
    100 -
    macm
    		 = macm
    2

    we get on solving
    aCM = 6.67
    m
    s²