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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