Engineering Mechanics Miscellaneous

Engineering Mechanics Miscellaneous

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Engineering Mechanics

Engineering Mechanics Miscellaneous
  1. Match 4 correct pairs between List-I and List-ll for questions.
    List-I
    A. Collision of particles
    B. Stability
    C. Satellite motion
    D. Spinning top
    List-II
    1. Euler's equation of motion
    2. Minimum kinetic energy
    3. Minimum potential energy
    4. Impulse momentum principle
    5. Conservation of moment of momentum
    Codes:








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    (a) -4, (b) –3, (c) –2, (d) –1

    Correct Option: A

    (a) -4, (b) –3, (c) –2, (d) –1

  1. A ball of mass 3 kg moving with a velocity of 4 m/s undergoes a perfectly-elastic directcentral impact with a stationary ball of mass m. After the impact is over, the kinetic energy of the 3 kg ball is 6 J. The possible value(s) of m is/are








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    Conservation of linear momentum,
    m1u1 + m2u2 = m1v1 + m2v2
    ⇒ u2 = 0
    3 × 4 = 3 × v1 + mv2 ...(i)
    ∵ Coefficient of restitution, e = 1 for perfectly elastic collision,
    v2 – v1 = (u1 – u2)
    ⇒ v2 – v1 = u1 = 4
    ⇒ v2 – v1 = 4 ...(ii)
    Conservation of energy,

    1
    		m1u²1 +
    1
    		m2u²2 =
    1
    		m1v²1 +
    1
    		m2v²2
    2222

    ⇒
    1
    		 × 3 × (4)² + 0 = 6 +
    1
    		 m2v²2
    22

    ⇒ (24 - 6)2 = m2v²2
    ⇒ mv²2 = 36 ...(iii)
    ∵ v1 = v2 – 4
    Putting in equation (i)
    ⇒ 12 = 3v2 – 12 + mv2
    ⇒ v2 (3 + m) = 24 ...(iv)
    Putting in equation (iii),
    &RArr;
    24
    		² = 36
    3 + m

    ⇒ (3 + m)² = 16m
    ⇒ 9 + m² + 6m – 16m = 0
    ⇒ m² – 10m + 9 = 0
    ⇒ m² – 9m – m + 9 = 0
    (m – 9)(m – 1) = 0
    m = 1,9 kg

    Correct Option: B


    Conservation of linear momentum,
    m1u1 + m2u2 = m1v1 + m2v2
    ⇒ u2 = 0
    3 × 4 = 3 × v1 + mv2 ...(i)
    ∵ Coefficient of restitution, e = 1 for perfectly elastic collision,
    v2 – v1 = (u1 – u2)
    ⇒ v2 – v1 = u1 = 4
    ⇒ v2 – v1 = 4 ...(ii)
    Conservation of energy,

    1
    		m1u²1 +
    1
    		m2u²2 =
    1
    		m1v²1 +
    1
    		m2v²2
    2222

    ⇒
    1
    		 × 3 × (4)² + 0 = 6 +
    1
    		 m2v²2
    22

    ⇒ (24 - 6)2 = m2v²2
    ⇒ mv²2 = 36 ...(iii)
    ∵ v1 = v2 – 4
    Putting in equation (i)
    ⇒ 12 = 3v2 – 12 + mv2
    ⇒ v2 (3 + m) = 24 ...(iv)
    Putting in equation (iii),
    &RArr;
    24
    		² = 36
    3 + m

    ⇒ (3 + m)² = 16m
    ⇒ 9 + m² + 6m – 16m = 0
    ⇒ m² – 10m + 9 = 0
    ⇒ m² – 9m – m + 9 = 0
    (m – 9)(m – 1) = 0
    m = 1,9 kg

  1. A small ball of mass 1 kg moving with a velocity of 12 m/s undergoes a direct central impact with a stationary ball of mass 2 kg. The impact is perfectly elastic. The speed (in m/s) of 2 kg mass ball after the impact will be ______.








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    For elastic collision m1u1 + m1u2 = m1v1 + m2v2 ...(i)
    By Moment conservation principle,
    m1 = 1 kg u1 = 12 m/s
    m2 = 2 kg u2 = 0 m/s

    1
    		m1u²1 +
    1
    		m2u²2 =
    1
    		m1v²1 +
    1
    		m2v²2.........(II)
    2222

    By Energy conservation principle,
    From equation (i)
    12 = v1 + 2v2 ...(iii)
    From equation (ii)
    1
    		 × 1 × 144 +
    1
    		 × 2 × 0 =
    1
    		 × 1 × v²1 +
    1
    		 × 2 × v²2
    2222

    ⇒ 144 = v²1 + 2v²2...(iv)
    From equations (iii) and (iv)
    144 = 144 + 4v²2 - 48v2 + 2²2
    ⇒ 6v²2 - 48v2 = 0
    ⇒ 6v2 (v2 – 8) = 0
    ⇒ v2 = 8 m/s

    Correct Option: C

    For elastic collision m1u1 + m1u2 = m1v1 + m2v2 ...(i)
    By Moment conservation principle,
    m1 = 1 kg u1 = 12 m/s
    m2 = 2 kg u2 = 0 m/s

    1
    		m1u²1 +
    1
    		m2u²2 =
    1
    		m1v²1 +
    1
    		m2v²2.........(II)
    2222

    By Energy conservation principle,
    From equation (i)
    12 = v1 + 2v2 ...(iii)
    From equation (ii)
    1
    		 × 1 × 144 +
    1
    		 × 2 × 0 =
    1
    		 × 1 × v²1 +
    1
    		 × 2 × v²2
    2222

    ⇒ 144 = v²1 + 2v²2...(iv)
    From equations (iii) and (iv)
    144 = 144 + 4v²2 - 48v2 + 2²2
    ⇒ 6v²2 - 48v2 = 0
    ⇒ 6v2 (v2 – 8) = 0
    ⇒ v2 = 8 m/s

  1. A ball of mass 0.1 kg, initially at rest, dropped from height of 1 m. Ball hits the ground and bounces off the ground. Upon impact with the ground, the velocity reduces by 20%. The height (in m) to which the ball will rise is _______.








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    v = √2gh = √2 × 9.01 × 1 ⇒ 4.4294 m/s
    v' = 0 × v = 3.5432 m/s

    h =
    v²
    		 = 0.64 m
    2g

    Correct Option: C

    v = √2gh = √2 × 9.01 × 1 ⇒ 4.4294 m/s
    v' = 0 × v = 3.5432 m/s

    h =
    v²
    		 = 0.64 m
    2g

  1. A 1 kg mass of clay, moving with a velocity of 10 m/s, strikes a stationary wheel and sticks to it. The solid wheel has a mass of 20 kg and a radius of 1 m. Assuming that the wheel is set into pure rolling motion, the angular velocity of the wheel immediately after the impact is approximately








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    Taking both clay & wheel as system
    Li = Lf (about A)

    Li = mvr = 1 × 10 × 1 = 10
    Lf = mvcm r + Iω

    = 20 × rω × r +
    mr²ω
    		 = 20ω + 10ω
    2

    Lf = 30ω
    Li = Lf
    10 = 30ω
    ω = 1/3 rad/sec

    Correct Option: B

    Taking both clay & wheel as system
    Li = Lf (about A)

    Li = mvr = 1 × 10 × 1 = 10
    Lf = mvcm r + Iω

    = 20 × rω × r +
    mr²ω
    		 = 20ω + 10ω
    2

    Lf = 30ω
    Li = Lf
    10 = 30ω
    ω = 1/3 rad/sec