Motion in a Plane


  1. A particle moves in a plane with constant acceleration in a direction different from the initial velocity. The path of the particle is​​









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    NA

    Correct Option: B

    NA


  1. A body of 3 kg moves in the XY plane under the action of a force given by 6t î + 4t ĵ . Assuming that the body is at rest at time t = 0, the velocity of the body at t = 3s is​​​









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    F = 6t î + 4t ĵ
    Fx = 6t , Fy = 4t

    ax =
    6t
    = 2t , ay =
    4t
    33

    vx = 0 + 2t . t = 18 , for t = 3s
    vy = 0 +
    4
    t.t = 12 for t = 3s
    3

    Velocity → 18î + 12ĵ

    Correct Option: C

    F = 6t î + 4t ĵ
    Fx = 6t , Fy = 4t

    ax =
    6t
    = 2t , ay =
    4t
    33

    vx = 0 + 2t . t = 18 , for t = 3s
    vy = 0 +
    4
    t.t = 12 for t = 3s
    3

    Velocity → 18î + 12ĵ



  1. Two particles A and B are connected by a rigid rod AB. The rod slides along perpendicular rails as shown here. The velocity of A to the left is 10 m/s. What is the velocity of B when angle α = 60° ?​​​










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    Let after 1 sec angle become 60°. When the end A moves by 10 m left, the end B moves upward by BB′ = 10 × √3 = 10 × 1.73 = 17.3 m / s

    Correct Option: D

    Let after 1 sec angle become 60°. When the end A moves by 10 m left, the end B moves upward by BB′ = 10 × √3 = 10 × 1.73 = 17.3 m / s


  1. The position vector of a particle is r = (a cos ωt)î + (a sin ωt)ĵ , The velocity of the particle is









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    r = (a cos ωt)î + (a sin ωt)ĵ

    v =
    dr
    =
    d
    { (a cos ωt)î + (a sin ωt)ĵ }
    dtdt

    = (-aω sin ωt)î + (aω cos ωt)ĵ
    = ω [ (-a sin ωt)î + (a cos ωt)ĵ ]
    Slope of position vector =
    a sin ωt
    = tan ωt
    a cos ωt

    & slope of velocity vector =
    -a cos ωt
    =
    -1
    a sin ωttan ωt

    ∴ velocity is perpendicular to the displacement.

    Correct Option: D

    r = (a cos ωt)î + (a sin ωt)ĵ

    v =
    dr
    =
    d
    { (a cos ωt)î + (a sin ωt)ĵ }
    dtdt

    = (-aω sin ωt)î + (aω cos ωt)ĵ
    = ω [ (-a sin ωt)î + (a cos ωt)ĵ ]
    Slope of position vector =
    a sin ωt
    = tan ωt
    a cos ωt

    & slope of velocity vector =
    -a cos ωt
    =
    -1
    a sin ωttan ωt

    ∴ velocity is perpendicular to the displacement.



  1. A bullet is fired from a gun with a speed of 1000 m/s in order to hit a target 100 m away. At what height above the target should the gun be aimed? (The resistance of air is negligible and g = 10 m/s2)









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    Speed of the bullet (v) = 1000 m/s and horizontal distance of the target (s) = 100 m. ​

    Time taken to cover the horizontal distance (t) =
    100
    = 0.1 sec
    1000

    During this time, the bullet will fall down vertically due to gravitational acceleration. ​
    ∴ height = ut +
    1
    gt2
    2

    = (0 × 0.1) +
    1
    10(0.1)2 = 0.05 m = 5 cm
    2

    Correct Option: A

    Speed of the bullet (v) = 1000 m/s and horizontal distance of the target (s) = 100 m. ​

    Time taken to cover the horizontal distance (t) =
    100
    = 0.1 sec
    1000

    During this time, the bullet will fall down vertically due to gravitational acceleration. ​
    ∴ height = ut +
    1
    gt2
    2

    = (0 × 0.1) +
    1
    10(0.1)2 = 0.05 m = 5 cm
    2