Laws of Motion

  • Question 561
    CBSEENPH11020864

    A skier starts from rest at point A and slides down the hill without turning or breaking. The friction coefficient is μ. When he stops at point B, his horizontal displacement is S. what is the height difference between points A and B?

    (The velocity of the skier is small so that the additional pressure on the snow due to the curvature can be neglected. Neglect also the friction of air and the dependence of μ on the velocity of the skier.)

    • h = μS

    • h = μ/S

    • h = 2μS

    • h = μS2

    Solution

    A.

    h = μS

    According to the question, the condition is shown in the figure

    For a sufficiently safe-horizontal displacement △S can be considered straight. If the corresponding length of path element is △L, the friction force is given by μmg(△S/△L ).△L =  μmg△S

    Adding up, we find that along the whole path the total work done by the friction force is μmgs. By energy conservation, this must equal the decrease mgh in potential energy of skier.

    Hence, h = μS

     

    Question 562
    CBSEENPH11020865

    A bicycle wheel rolls without slipping on the horizontal floor. Which one of the following is true about the motion of points on the rim of the wheel, relative to the axis at the wheel's centre?

    • Points near the top move faster than points near the bottom

    • Points near the bottom move faster than points near the top

    • All points on the rim move with the same speed

    • All points have the velocity vectors that are pointing in the radial direction towards the centre of the wheel

    Solution

    A.

    Points near the top move faster than points near the bottom

    We have, VA = 2v sinθ/2

    Hence, the velocity of a point on rim increases with θ for 0° < θ < 180° and decreases with θ for 180° < θ < 30°.

    Question 563
    CBSEENPH11020869

    The upper half of an inclined plane of inclination θ is perfectly smooth while the lower half rough. A block starting from rest at the top of the plane will again come to  rest at the bottom if the coefficient of friction between the block and the lower half of the plane is given by

    • μ = 2 tanθ

    • μ = tanθ

    • μ = 2/(tan θ)

    • μ = 1/ tan θ

    Solution

    A.

    μ = 2 tanθ

    suppose the length of plane is L. When block descends the plane, rise in kinetic energy = fall in potential energy = mg L sinθ i

    Work done by friction = μ x (reaction) x distance 

    0 + μ (mg cos θ) x (L/2)

    =  μ (mg cos θ) x (L/2)

    Now, work done = change in KE

    mgL sin θ = μ (mg cos θ) x (L/2)

    ⇒ tan θ  = μ/2 

     μ = 2 tan θ

    Question 564
    CBSEENPH11020870

    Two masses 10 kg and 20 kg respectively are connected by a massless springs as shown in the figure. A force of 200 N acts on the 20 kg mass. At the instant shown is a figure the 10 kg mass has an acceleration of 12 m/s2. The value of the acceleration of 20 kg mass is

    • 4 m/s2

    • 10 m/s2

    • 20 m/s2

    • 30 m/s2

    Solution

    A.

    4 m/s2

    The equation of motion of m1 = 10 kg mass is

    F1 = m1a1 = 10 x 12  = 120 N

    Force on 10 kg mass is 120 N to the right. As action and reaction are equal and opposite, the reaction force F- on 20 kg mass F = 120 N to the left.

    therefore, equation of motion of mass m2 = 20 kg is

    200 - F = 20 a2

    200-120 = 20a2

    80 = 20a2

    a2 = 80 /20 = 4 m/s2

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