1.What are the laws of limiting friction?

There are three laws of limiting friction:

1. The magnitude of limiting friction depends upon the nature of the surfaces in contact and on their roughness. It does not depend upon the areas of the surfaces in contact.
2. The force of friction is tangential or parallel to the surfaces in contact, and its direction is opposite to the direction of motion of the body.
3. The value of limiting friction is directly proportional to the normal reaction between them that is

μ is constant and is called coefficient of limiting friction.

2.A man whirls a stone round his head on the end of a string 4m long. Can the string be in a horizontal plane? If the stone has a mass of 0.4 kg and the string will break if the tension in it exceeds 8 N, what is the smallest angle the string can make with the horizontal? What is the speed of the stone? Take g = 10 ms^-2 .

T cos θ = mg

As T cos θ = mg,

In case the string becomes horizontal, θ = 90°.

Thus, for making the string horizontal, the tension must be infinite which is impossible. Therefore, the string cannot be in horizontal plane.

The value of the maximum angle θ is given by the breaking tension of the string, such that,

                     T cos θ = mg, T_(max) = 8 N

or          θ = 60°

Therefore  angle with the horizontal

               = 90° - 60° = 30°

3.A person holding a box of 30 kg on his head is standing on a smooth surface. Calculate the force with which the surface pushes his feet. Given mass of the person is 80kg. Take g = 10 ms^-2

According to Newton's third las of motion the force acting on the feet of the person.

          F = Total force acting downward on the floor

             = weight of the person + weight of the box

             = Mg + mg = (M + m)g

             = (80 + 30) × 10 = 110 × 10

             = 1100 N.

4.A block of mass 3 kg hanged with a rope is going upward with an acceleration of 2 ms^-2. Calculate the tension in the rope.

The equation of motion of the block is given by,

                F = T - mg

      or      ma = T - mg

That is        T = mg + ma = m(g + a)

                     = 3(10 + 2) = 3 × 12 = 36 N

5.A ball of mass 0.1 kg is thrown against a wall. It strikes the wall normally with a velocity of 30 ms^-1 and rebounds with a velocity of 20 ms^-1. calculate the impulse of the force exerted by the ball on the wall.

Initial momentum of the ball,

p_i = mu = 0.1 × 30 = 3 kg ms^-1

Final momentum of the ball,

p_f = mv = 0.1 × 20 = 2 kg ms^-1

The direction of p_i and p_f are opposite to each other

That is       = 3 kg ms^-1   

and           = -2 kg ms^-1

Also, Impulse = change in momentum

                     =

                     = 3 kg ms^-1 - (-2 kg ms^-1)

                     = 5 kg ms^-1

                     = 5 N s.

6.A suitcase is placed on the floor of the compartment of a train which starts from rest. If static coefficient of friction between the floor of the compartment and the suitcase is 0.5, find the maximum acceleration attained by the train so that the suitcase does not move at all. Take g = 9.8 ms^-2 .

Force of friction is given by,

                F = μR

Since        R = mg (weight of suitcase)

Therefore   F = μmg

Let a be acceleration of the train so that the suitcase is at rest with respect to the train.

we have,    F = ma

Therefore  ma = μmg

or               a = μg = 0.5 × 9.8

                     = 4.90 ms^-2

7.Is a large brake on a bicycle wheel more effective than a small one? Explain.

No. The force of friction is independent of the area of contact. Therefore, a large brake and a small brake will have the same effect. However, the small brake may go out of order earlier because of faster wear and tear.

8.A body is rolling on the ground with a velocity of 1ms^-1 . After travelling a distance of 5 m, it comes to rest. Find the coefficient of friction. Take g = 10 ms^-2.

Given : u = 1 ms^-1, v = 0, S = 5 m

Using v² - u² = 2a S, we get,

Therefore Force pf friction,

Normal reaction,

R = mg = m × 10 = 10 mN

Therefore Coefficient of friction,

9.Why a horse cannot pull a cart and run in empty space?

In order to pull the cart, house needs the reaction (force) from the surface of the ground. In an empty space, there is no reaction, hence horse cannot pull the cart.

10.Vehicles stop on applying brakes. Does this phenomenon violate the principle of conservation of momentum?

When brakes are applied, some retarding force is being applied due to which the vehicle comes to rest such that the total loss of momentum of the vehicle is equal to the impulse of the applied force. Thus, the law of conservation of momentum does not get violated.

11.A car is moving with 72 km hr^-1 on a flat circular road of radius 10 m. Can the car negotiate the curve with this speed? Take μ = 0.3.

The car can negotiate the curve if,

Since speed of car = 72 km hr^-1

                            = 72 × = 20 ms^-1

so it cannot negotiate the curve Therefore, it goes off the road.

12.A fly wheel is rotating about its axle with angular speed ω₀. What will be the ratio of the linear velocities of two points on the wheel at a distance and R respectively? R is the radius of the flywheel.

We know, v = ωR

Now, linear velocity of the point at a distance , v₁ = and linear velocity of the point at a distance R,

                v₂ = ωR

13.The kinetic energy of a particle moving in a horizontal circle may remain the same everywhere. Is it true for the motion in a vertical circle?

No. For the motion in vertical circle, the speed of the particle at the bottom is and the speed at the top is . Therefore, K.E. of the particle in a vertical circle is maximum at the bottom and minimum at the top.

14.Two spheres of same size, one of mass 2kg another of mass 4kg are dropped simultaneously from the top of Qutab Minar (height = 72m). When they are 1 m above the ground, will the two spheres have the same acceleration?

Since both the spheres have different masses and the velocity of a falling body goes on increasing, hence momentum. K.E. and P.E. of both the spheres cannot be the same. Both the spheres fall with same acceleration equal to acceleration due to gravity.

15.A 100 kg ball moving with a speed of 20 ms^-1 returns in the opposite direction with a speed of 30 ms^-1 after being struck with a bat. What is the impulse?

Impulse = Change in momentum of the ball

            = mv₁ - (-mv₂) = m(v₁ + v₂)

            = 0.1 × 50 = 5 Ns