Laws of Motion

  • Question 573
    CBSEENPH11026254

    Three concurrent co-planar forces 1 N, 2 N and 3 N acting along different directions on a body

    • can keep the body in equilibrium if 2 N and 3 N act at right angle

    • can keep the body in equilibrium if 1 N and 2 N act at right angle

    • cannot keep the body in equilibrium

    • can keep the body in equilibrium in 1 N and 3 N act at an acute angle

    Solution

    C.

    cannot keep the body in equilibrium

    If we keep 1 N and 2 N forces act in the same direction then these are balanced by 3 N force, but this is against statement of question.

    Hence, options (c) is correct.

    Question 574
    CBSEENPH11026268

    Two rectangular blocks A and B of masses 2 kg and 3 kg respectively are connected by a spring of constant 10.8 Nm-1 and are placed on a frictionless horizontal surface. The block A was given an initial velocity of 0.15 ms-1 in the direction shown in the figure. The maximum compression of the spring during the motion is

            

    • 0.01 m

    • 0.02 m

    • 0.05 m

    • 0.03 m

    Solution

    C.

    0.05 m

    As the block A moves with velocity 0.15 ms-1  it compresses the spring which pushes B towards right. A goes on compressing the spring till the velocity acquired by B becomes equal to the velocity of A, i.e., 0.15 ms-1.Let this velocity be v.

    Now, spring is in a state of maximum compression. Let x be the maximum compression at this stage.

               

    According to the law of conservation of linear momentum, we get

                 mA u = mA + mB v

    ⇒               v = mA umA + mB        

    ⇒                   = 2 × 0.152 + 3

                      v = 0.06 ms-1

    According to the law of conservation of energy

             12 m A u2 = 12 mA +  mB v2 + 12 k x2

              12 mA u2 - 12 mA  +  mB v2 = 12 k x2   

       12× 2 × 0.152 - 12 2 + 3 0.062  = 12  k x2

                             0.0225 - 0.009 = 12 k x2

                              0.0135 = 12 k x2

    ⇒                              x = 0.027k

    ⇒                                 = 0.02710.8

                               x = 0.05 m

    Question 575
    CBSEENPH11026274

    A boy on a cycle pedals around a circle of 20 metres radius at a speed of 20 m/s. The combined mass of the body and the cycle makes with the vertical so that it may not fall is (g = 9.8 m/s2 )

    • 60.25o

    • 63.90o

    • 26.12o

    • 30.00o

    Solution

    B.

    63.90o

    A body that travels an equal distance in equal amounts of time along a circular path has a constant speed but not constant velocity. This is because velocity is a vector and thus it has magnitude as well as direction.

                   v = ( Rg tanθ )1/2

                   tan θ = v2r g

                   = 40020 × 900

             tan θ = 63.70o 

             tanθ ≈ 63.90o

    Question 576
    CBSEENPH11026281

    A car of mass 1000 kg moves on a circular track of radius 40 m. If the coefficient of friction is 1.28. The maximum velocity with which the car can be moved, is

    • 22.4 m/s

    • 112 m/s

    • 0.64 × 401000 × 100m/s

    • 1000 m/s

    Solution

    A.

    22.4 m/s

    A car moves in a circular track so it perform circular motion. 

    According to second law the force providing this acceleration is

               fcmv2R

    But according to static friction

             fs ≤ μs N

         f = m v2R  μs N

         vmax = μ R g

    Given:-  μ = 1.28 

    where μ is  the coefficient of friction 

      R = 40 m 

    The maximum velocity

              vmaxμ R g 

                       = 1.28 × 40 × 9.8

                 vmax = 22.4 m/s 

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