# Question-1

**What do you mean by the term â€˜bankingâ€™? If**

**Î¸**

**is the banking angle, write a formula for the angle of banking in terms of optimum speed (v), radius of the curve (R) and acceleration due to gravity (g).**

**Solution:**

The term 'banking' implies that the outer part of the road is raised a little above the inner side so that the road is sloping towards the centre of the curve.

Î¸ = tan

^{-1}or tan Î¸ = .

# Question-2

**Two bodies of masses 1 kg and 1.5 kg respectively are tied to the ends of a light string which passes over a light frictionless pulley of diameter 10 cm. The masses are initially at the same horizontal level and then released. Find the direction of motion of the centre of mass and its acceleration.**

**Solution:**

Let the initial position of masses m

_{1}= 1 kg and m

_{2}= 1.5 kg be A and B respectively. Let C be the centre of mass of the system in the horizontal line AB. Taking moments about C, we have

m

_{1}AC = m

_{2}BC or

or or

or

âˆ´ BC = 4 cm

The acceleration of the system (a) is given by

a = g = g =

Also, a

_{c}=

= or a

_{C}=

Thus, the acceleration of the centre of mass is given by a

_{C}= .

# Question-3

**A car moves at a speed of 36 km/h on a level road. The coefficient of friction between the tyres and the road is 0.8. The car negotiates a curve of radius 10 m at this speed. Will the car skid while negotiating the curve? (g = 10 m/s**

^{2}).**Solution:**

The maximum centripetal force that the friction can provide is given by,

f

_{r}= Î¼ mg = or r

_{min}= = = 12.5 m

It is the minimum radius, the curve must have for the car to negotiate it at 10 m/s. Since the radius of the curve is only 10 m, therefore, the car will skid while negotiating the curve.

# Question-4

**Calculate the angle, which the bicycle and the rider must make with the vertical while going round a curve of 10 m radius with 18 km/hr. What is the best value of the coefficient of friction between the tyres and the road, which would prevent slipping? If the bicycle and the rider have a mass of 100 kg, what frictional force must the ground exert on the wheel?**

**Solution:**

We know that tan Î¸ = = 5 m/s

tan Î¸ = = 0.255

Î¸ = 14Â° 19â€²

The force of friction supplies the necessary centripetal force, i.e.

f

_{r}= Î¼ R = Î¼ mg ( R = mg)

âˆ´ Î¼ mg =

or Î¼ =

Frictional force (f_{r}) is given by

f_{r} = Î¼ R = Î¼ mg = 0.255 Ã— 100 Ã— 9.8 = 250 N.

# Question-5

**Why does a cyclist lean to one side while going along a curve? Also deduce the conditions under which skidding will occur.**

**Solution:**

Consider a cyclist negotiating a curve of radius 'r' to the left at a speed v. Suppose the cyclist leans inwards at an angle Î¸ . The various forces acting on the system are

(i) Weight (mg) acting vertically downwards through the centre of gravity G.

(ii) normal reaction (R) of the ground passing through G and inclined at an angle Î¸ to the vertical.

(iii) force of friction (f) between the tyre and the ground at O. The horizontal component (R sin Î¸ ) provides the necessary centripetal force , where m is mass of the system (cyclist + cycle), v is the speed and r is radius of the curved path. The vertical component (R cos Î¸ ) supports the weight (mg). The friction between the tyres and the ground soon makes the tyres wear out.

âˆ´ R sin Î¸ =

R cos Î¸ = mg or tan Î¸ = or Î¸ = tan^{-1}

If the cyclist were to remain vertical, mg would act through O, and the cyclist will tapple under the action of the forces f and .

But f = Î¼ R = Î¼ mg

âˆ´ The cyclist will skid when > Î¼ mg or v^{2} > Î¼ rg

In other words, skidding will occur if (i) v is large (ii) Î¼ is small (iii) r is small (i.e. sharp curve).

# Question-6

**The radius of curvature of a railway line at a place when the train is moving with a speed of 36 km/h is 1000 m, the distance between the two rails being 1.5 m. Calculate the elevation of the outer rail above the inner rails so that there may be no side pressure on the rails.**

**Solution:**

Velocity v = 36 km/h = m/s = 10 m/s

Radius of curvature r = 1000 m

tan Î¸ =

Let h be the height through which outer rail is raised. Let l be the distance between the two rails. Distance between the two rails l = 1.5 m.

âˆ´ tan Î¸ =

h = l tan Î¸

h = 1.5 Ã— = 0.153 m.

# Question-7

**A string breaks under a load of 50 kg. A mass of 1 kg is attached to one end of the string 10 m long and is rotated in a horizontal circle. Calculate the greatest number of revolutions that the mass can make without breaking the string.**

**Solution:**

Length, r = 10 m

Mass, m = 1 kg

Centripetal force, F = 50 kg wt = 50 Ã— 9.8 N

Let n be the maximum number of revolutions made per second without breaking the string.

Then, F = mrÏ‰

^{2}= mr(2Ï€ n)

^{2}

âˆ´ 50 Ã— 9.8 = 1 Ã— 10

490 = 395.10n

^{2}

n

^{2}= 1.24

âˆ´ n = 1.11.

# Question-8

**A circular race track of radius 300 m is banked at an angle of 15**

**Â°**

**. If the coefficient of friction between the wheels of a race and the road is 0.2, what is the (a) optimum speed of the race car to avoid wear and tear on its tyres and the (b) maximum permissible speed to avoid slipping?**

**Solution:**

On a banked road, the horizontal component of the normal reaction and the frictional force contribute to provide centripetal force to keep the car moving on a circular turn without slipping. At the optimum speed, the component of the normal reaction is enough to provide the required centripetal force. In this case, the frictional force is not required. The optimum speed is given by

v

_{0}= (rg tan Î¸ )

^{1/2}

âˆ´ v

_{0}= (300 Ã— 9.8 tan 15Â° )

^{1/2}m/s = 28.1 m/s

The maximum permissible speed is given by

v

_{max}=

Substituting values and simplifying, we get v

_{max}= 38.1 m/s.

# Question-9

**Derive the maximum speed of car that travels on a circular level road.**

**Solution:**

Let us consider a car of weight mg moving on a circular level road of radius r with constant velocity v. While taking the round, the tyres of the car tend to leave the road and move away from the centre of curve. So the forces of friction f_{1} and f_{2} act inward to the two tyres. If R_{1} and R_{2} are the normal reactions of ground on the tyres, then

f_{1} = Î¼ _{s} R_{1} and f_{2} = Î¼ _{s} R_{2}

where Î¼ _{s} is the coefficient of static friction.

Total frictional force,

f = Î¼ _{s} R_{1} + Î¼ _{s} R_{2}

f = Î¼ _{s} (R_{1} + R_{2})

f = Î¼ _{s} R

where R is the reaction of the ground on the car.

Since the total frictional force f cannot exceed Î¼ _{s} R, therefore, we may say that

f â‰¤ Î¼ _{s} R

This frictional force will provide the necessary centripetal force f.

âˆ´ f =

Again, R = mg (weight of car)

Now, â‰¤ Î¼ _{s} R

â‰¤ Î¼ _{s} mg

v^{2} â‰¤ Î¼ _{s} rg

v â‰¤

âˆ´ Maximum speed, v_{max} =

If the car is driven at a speed greater than v_{max}, then the car will skid and go off the road in a circle of radius greater than r. This is because even the maximum available friction will be inadequate to provide the necessary centripetal force.

# Question-10

**The driver of a three-wheeler moving with a speed of 10 m/s sees a child standing in the middle of the road and brings his vehicle to rest in 4 s just in time to save the child. What is the average retarding force on the vehicle? The mass of the three-wheeler is 400kg and the mass of the driver is 65 kg.**

**Solution:**

Initial velocity u = 10 m/s

Final velocity v = 0

Time taken to travel t = 4s

Acceleration due to gravity a =?

Using, v = u + at

0 = 10 + a Ã— 4

or a =

Mass of the driver and three wheeler = 465 kg

âˆ´Magnitude of retarding force on the vehicle = 465 Ã— retardation = 465 Ã— = 1162.5 N.

# Question-11

**A bob of mass 0.1 kg hung from the ceiling of a room by a string 2m long is set into oscillation. The speeds of the bob at its mean position are 1 ms**

^{-1}. What is the trajectory of the bob if the string is cut when the bob is (a) at one of its extreme positions, (b) at its mean position.**Solution:**

(a) The velocity of the bob at its extreme position is zero. However, when the string is cut, it will fall down vertically.

(b) When the bob is in the mean position, its horizontal component of velocity = 1m/s. Therefore, the bob will trace a parabolic path as a projectile.

# Question-12

**A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei the products must be emitted in opposite directions.**

**Solution:**

If M is the mass of the nucleus at rest. Initial momentum of the nucleus = 0. Let and be the velocities of two disintegrated parts (masses M

_{1}and M

_{2}respectively) therefore 0 = M

_{1}+ M

_{2}. Since and have opposite signs therefore, M

_{1}and M

_{2}move in opposite directions.

# Question-13

**A force of 36 dyne is inclined to the horizontal at an angle of 60**

^{0}. Find the acceleration in a mass of 18g, which moves in a horizontal direction.**Solution:**

Force F = 36 dyne

Angle inclined to the horizontal Î¸ = 60

^{0}

âˆ´ Component of force along x-direction is given by

F_{x} = F cos 60^{0} = 36 Ã— dyne

But F_{x} = Ma_{x }

âˆ´ a_{x} = cm s^{-2}.

# Question-14

**Fuel is consumed at the rate of 100kg s**

^{-1}in a rocket. The exhaust gases are ejected at a speed of 4.5 Ã— 10^{4}ms^{-1}. What is the thrust experienced by the rocket?**Solution:**

Thrust = mass of fuel consumed per second Ã— speed of ejected gases

= (100 Ã— 4.5 Ã— 10^{4}) kg s^{-1} Ã— ms^{-1 }

= 4.5 Ã— 10^{6} kg ms^{-2} = 4.5 Ã— 10^{6} N.

# Question-15

**Two masses m**

_{1}and m_{2}are connected at the ends of a light inextensible string that passes over frictionless pulley. Find the acceleration, tension in the string and thrust on the pulley when the masses are released.**Solution:**

For mass m_{1}: T â€“ m_{1}g = m_{1}a -------(i)

For mass m_{2}: m_{2}g â€“ T = m_{2}a -------(ii)

Adding (i) and (ii), we have

(m_{2 }- m_{1})g = (m_{2} + m_{1}) a

or a = -------(iii)

Putting this value in (i) we get,

âˆ´T = m_{1}g + m_{1}

=

T =

Thrust on the pulley is given by

2T =

# Question-16

**A mass of 10kg is suspended from a string, the other end of which is held in hand. Find the tension in string when the hand is moved up with a uniform acceleration of 2 ms**

^{-2}. Take g = 10^{-2}.**Solution:**

Tension, T = m(g + a) = 10(10+2) = 120 N.

# Question-17

**A balloon of mass m is rising up with acceleration â€˜aâ€™ show that the fraction of weight of the balloon that must be detached in order to double its acceleration is assuming the upthrust of air to remain the same.**

**Solution:**

Suppose the upthrust of the air is R. The balloon rises with an acceleration 'a' such that

R = mg + ma = m(g + a)

(i) In the 2^{nd} case, let mâ€² be the mass detached from the balloon, such that acceleration becomes 2a

âˆ´ R = (m - mâ€² )g + (m - mâ€² ) 2a ........................... (ii)

From (i) and (ii), we get

(m - mâ€² )(g + 2a) = m(g + a)

or (m - mâ€² ) =

or mâ€² = m

=

Thus, the mass to be detached from the balloon is .

# Question-18

**A body placed on a rough inclined plane just begins to slide, when the slope of the plane is 1 in 4. Calculate the coefficient of friction.**

**Solution:**

From the Fig

sin Î¸ = âˆ´ tan Î¸ = =

âˆ´ = = 0.26.

# Question-19

**A pendulum is hanging from the ceiling of a train as an accelerometer. Derive the general expression relating the horizontal acceleration â€˜aâ€™ of the train to the angle**

**Î¸**

**made by the bob with the vertical.**

**Solution:**

The various forces acting on the bob are shown in the Fig. Î¸ is the angle which the pendulum makes with the vertical.

Vertical forces are balanced, therefore,

T cos Î¸ = mg -------(i)

T sin Î¸ = ma -------(ii)

From (i) and (ii) we have, tan Î¸ = or a = g tan Î¸ which is the required expression.

# Question-20

**Solution:**

Let a

_{1}be the acceleration of the 6kg mass relative to the table, and a

_{2}be the acceleration of the masses 4kg and 2kg with respect to the pulley, as shown in Fig.

The acceleration of 4 kg relative to the table = (a_{1 }+ a_{2})

The acceleration of 2 kg relative to the table = (a_{2 }- a_{1})

T_{1} = 2T_{2} ----(i)

T_{1} = 6a_{1} ----(ii)

Force equation for 4 kg and 2kg are given by

4g â€“ T_{2} = 4(a_{1 }- a_{2}) ----(iii)

and T_{2} â€“ 2g = 2(a_{2 }- a_{1}) ----(iv)

From (i) and (ii) we have,

6a_{1} = 2T_{2} or T_{2} = 3a_{1}

Putting T_{2} = 3 a_{1} in (ii) and (iv) we can write,

4g = 7a_{1 }+ 4a_{2} ----(v)

and 2g = 5a_{1 }+ 2 a_{2} ----(vi)

Multiply (vi) by 2 and adding (v) and (vi) we have,

8g = 17a_{1} or a_{1} =

Thus, the acceleration of 6 kg mass relative to the table is .

# Question-21

**Consider two bodies of mass m**

_{1}and m_{2}in contact placed on a frictionless table as shown in Fig. When force F is applied on mass m_{1}, calculate the acceleration produced, and the force of contact between the bodies. What will be the force of contact when the force F is applied on mass m_{2}?**Solution:**

Case 1: Let F be applied on m

_{1}and f is the force of contact between the two bodies.

âˆ´F -f = m_{1}a ----------(i)

Also, f = m_{2}a ----------(ii)

Adding (i) and (ii), we have

F = (m_{1 }+ m_{2})a or a =

Again, f = F - m_{1}a = F -

Case 2: Similarly, in the 2^{nd} case

F - f_{1} = m_{2}a and f_{1} = m_{1}a

âˆ´ F = (m_{1}+m_{2})a or a =

âˆ´ f_{1} =

# Question-22

**Solution:**

Let M be the mass of the body, 2l is the length of the inclined plane, and F is the force of friction as shown in the figure above.

Potential energy of the body at the top of the plane = Mgh

But h = 2l sin 30^{0} = 2l Ã—

âˆ´Potential energy of the body at the top (point C) = mgl

Work done against friction along SA = Fl

It is because the upper half of the plane is perfectly smooth (i.e. no friction).

As the body is brought to rest just when it reaches the bottom,

Potential energy at the top = work done against friction

Or mgl = Fl or

Thus, the required ratio is unity.

# Question-23

**Prove that the impulse received during an impact is equal to the total change in momentum produced during the impact.**

**Solution:**

We know, or

If the impact lasts for a small time t and the momentum of the body changes from to , then

or

is a measure of the impulse of the force.

Let be the constant force during the impact, then

âˆ´

Thus, the impulse received during an impact is equal to the total change in momentum produced during the impact.

# Question-24

**A ball of mass 0.1 kg moving with a velocity of 18 m/s hits a wall and rebounds along the same line with a velocity of 10 m/s. What is the impulse given to the ball? If the collision lasts for , what is the force exerted by the wall on the ball? What is the force exerted by the ball on the wall?**

**Solution:**

Mass of the ball m = 0.1kg,

Initial velocity u = +18 m/s,

Final velocity v = -10 m/s,

Time taken to travel t =

Impulse given to the ball = change in the momentum of the ball

= mv - mu

= m(v-u)

= 0.1 Ã— [(-10) - (18)]

= 0.1 Ã— (-28) = -2.8 Newton

Impulse = F Ã— t

âˆ´ F = =

âˆ´ Force exerted by the wall on the ball = -280 N

By Newton's third law of motion,

Force exerted by the ball on the wall = -F = -(-280N) = 280 N.

# Question-25

**What is the cause of friction?**

**Solution:**

Friction is a surface phenomenon. However, the surfaces are never perfectly smooth. Every surfaces reveals dents and irregularities. When two such bodies with irregular surfaces are placed in contact, there is some degree of interlocking of these irregularities. When some force is applied on one body to make it slide over the surface of another, the relative motion is resisted by an opposing force. This is known as the force of friction.

When the applied force is increased, the opposing force also increases, but at a certain stage the body begins to slide over the surface of another.

# Question-26

**What are concurrent forces? When are they said to be in equilibrium?**

**Solution:**

Forces acting at the same point on a body are called concurrent forces. Forces , , , , acting at a point O are called concurrent forces. The resultant of these forces is equal to their vector sum. If the magnitude of the resultant turns out to be zero, the concurrent forces are said to be in equilibrium, and no net force acts on the body.

# Question-27

**Why does a cricket player lower his hands while catching a ball?**

**Solution:**

By lowering his hands, the cricket player increases the interval in which the catch is taken. This increase in time interval results in the lesser rate of change of momentum. Therefore, in accordance with Newtonâ€™s 2

^{nd}law of motion, lesser force acts on his hands, and the player saves himself from being hurt.

# Question-28

**When a man jumps out of a boat, the boar is pushed away. Why?**

**Solution:**

This is due to Newtonâ€™s third law of motion. When man jumps out of a boat, he pushes the boat in the backward direction and in turn, the reaction of the boat on the man pushes him out of the boat.

# Question-29

**Proper inflation of tyres saves fuel. Explain.**

**Solution:**

Proper inflated tyres roll without sliding. Since roiling is less than riding friction, therefore, there is less dissipation of energy against the friction. Hence proper inflation of tyres saves fuel.

# Question-30

**The outer rail of a curved railway track is generally raised over the inner. Why?**

**Solution:**

When the outer rail of a curved railway back is raised, the weight of the train gives a component of its weight along the radius of the curved track. This component of the weight provides the train the necessary centripetal force so as to enable it to move along curved path.

# Question-31

**Why do we pull the rope downwards for clamping up?**

**Solution:**

When we pull the rope downwards, an upward reaction helps us to rise up.

# Question-32

**Why does a heavy gun not kick so strong as a light gun using the same cartridges?**

**Solution:**

The recoil speed of the gun is inversely proportional to its mass. Therefore the recoil speed of heavy gun is less than that of light gun.

# Question-33

**Automobile tyres are generally provided with irregular projections over their surface. Why?**

**Solution:**

Irregular projections are provided over the surface of automobile tyres to increase the friction between the tyres and the road.

# Question-34

**A train moves on an unbanked circular bane of rails. Which rail will wear out faster?**

**Solution:**

The inner rail will wear out faster. This is because of the inward pressure on the inner rail is more than that on the outer rail.

# Question-35

**Ball bearings are used in machinery. Why?**

**Solution:**

Since the rolling friction is much less than the sliding friction, we convert the sliding friction into rolling friction. This is done by using ball bearings arrangement. Ball bearings are placed in between axle and hub of the wheel. The ball bearings tend to roll round the axle as the wheel turns and as such the frictional force is diminished.

# Question-36

**Why are lubricants used in machines?**

**Solution:**

A lubricant spreads like a thin layer between the two surfaces. The motion is now between the surface and the lubricant layer. Now the friction between them is very less and hence the force of friction between the various parts of the machine is also reduced.

# Question-37

**Polishing beyond a certain limit may increase the friction between the surfaces. Explain why?**

**Solution:**

Polishing increases the area of contact of the surfaces. Polishing beyond a certain limit increases the area of contact of the surface to such an extent that the cohesive force between the surface becomes more than the force of friction.

# Question-38

**It is reasonable to expect the coefficient of friction to exceed unity?**

**Solution:**

In case of normal plane surfaces, the coefficient of friction is less than unity. But, when the surface is irregular with sharp minute projections and cavities existing on the surface, the coefficient of friction may exceed unity.

# Question-39

**Why does a cyclist lean to one side while going along a curve? In which direction does he lean?**

**Solution:**

The cyclist leans to one side so as to provide himself the necessary centripetal force. He leans towards the center of the circular curve.

# Question-40

**A stone tied at the end of a string is whirled in a horizontal circle. When the string breaks, the stone flies away tangentially. Explain why?**

**Solution:**

When the stone moves in a circular path, its velocity is always tangential to the point of the circle. When the string breaks, the force acts on the stone. Hence according to Newtonâ€™s first law of motion, the stone flies away in the direction of motion.