Solution:

(i) Friction - It is an opposing force acting tangentially on a body.
(ii) Plotting a graph between applied force and frictional force (see figure), we have

`O L =` static friction `= F_1`
`OP =` limiting friction `= F_2`
`MN =` kinetic friction `= F_3`

`F_1` goes on increasing with applied force at L the static friction is maximum beyond L the frictional force decreases slightly. As the portion MN of the curve is parallel to OX therefore, kinetic friction does not change with applied force. Also kinetic friction is always slightly less than limiting friction as once motion actually starts irregularities of one surface are not able to get locked into irregularities of another surface.
(iii) Magnitude and direction of the static friction force adjust themselves according to applied force. When we change the applied force, force of static friction changes accordingly in the direction opposite to the applied force.

(iv) Laws of limiting friction:
(a) Magnitude of force of limiting friction (F) between any two bodies in contact is directly proportional to normal reaction (R), i.e. `F \ \ alpha \ \ R`.
(b) Direction of force of limiting friction is always opposite to the one in which the body starts moving over another.
(c) As long as normal reaction between two bodies in contact remains same the force of limiting friction is independent of area of contact.As long as normal reaction between two bodies in contact remains same the force of limiting friction is independent of area of contact.
(d) Force of limiting friction only depends upon the material and nature of the surfaces in contact.

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Solution:

The change in momentum associated with a body can be produced by a large force for small time or a small force for large time.

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Solution:

Consider of a rocket with initial mass `M_i` burning `d M` amount of fuel in a time `d t`, to be released with a constant velocity `u` in the form of fumes. If the mass at any instant of the rocket is `M` then applying the conservation of momentum, we have,
(change in momentum of burnt fuel)
`= -` (Change in momentum of rocket and its contents.)
`:. u d M = - M d v`, where `dv` refers to change in velocity of rocket.

(i) ` u d M= - M d u \ \ :. \ \ dv = - u (dM)/M`.

Integrating,

` int_o^v dv = - u int_(M_i)^M (dM)/M`

`v = - u | log_e M |_(M_i)^M`

` = - u (log_e M - log_e M_i)`

`:. v = u [ log_e M_i/M ]`

(ii) ` u d M= - M d v ,`

`:. (dv)/(dt) = - (u dM//dt)/M`

` => a = - (u dM//dt)/M`

Therefore, the force or thrust experienced

` = - u (dM)/(dt)` .

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Solution:

When no external force acts on the body, the momentum remains conserved.
According to third law, for every action there is an equal and opposite reaction. So if `dp_1` and `dp_ 2` are change in momentum of two masses `m_1` and `m_2` then,

` (dp_1)/(dt) = - (dp_2)/(dt) `, since `F_1 = - F_2`

`:. - d/(dt) (p_1 + p_2) = 0`

i.e., `p_1 + p_2 =` constant.

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Solution:

When no external force acts on a body, there is no change in momentum or momentum will remain conserved.

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Solution:

(i) A body at rest or uniform motion will continue to maintain the status, till an unbalanced force acts on it.
(ii) The rate of change of momentum is a measure of the force acting on the body.
or
The total unbalanced external force acting on a body is the product of its mass and acceleration.
(iii) For every action there exists an equal and opposite reaction.

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Solution:

When no external force acts on a system the momentum will remain conserved. Consider a system of `n` bodies of masses `m_1 , m_2, m_3 ..... m_n` . If `p_1 , p_2 , p_3 ..... p_n` are the momentum associated then the rate of change of momentum with the system,

` (dp)/(dt) = (dp_1)/(dt) + (dp_2)/(dt) + (dp_3)/(dt) + .......... + (dp_n)/(dt) `

` = d/(dt) (p_1 + p_2 + p_3 + ··········· + p_n)`

If no external force acts,

` (dp)/(dt) = 0 \ \ :. p = ` constant

i.e., `p_1 + p_2 + ..... + p_n =` constant.

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Solution:

The force parallel to the surface of contact which opposes relative motion between the two surfaces. By using ball bearings, the wheel rolls on balls instead of sliding on axle. So the sliding friction is converted into rolling friction which is much less.

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Solution:

From the forces acting on the vehicle in a banked curve `(theta)`.

`N cos theta - F_f sin theta = mg`

`N sin theta + F_f cos theta = mv^2//r . F_f = mu N`.

Dividing the equations, we have,

` v^2/(rg) = ( N sin theta + mu N cos theta)/( N cos theta - mu N sin theta)`

[dividing each term of right side by `N cos theta`]

`v^2 = r g ( (tan theta + mu)/(1 - mu tan theta) ) ; v = sqrt ( rg ( (mu + tan theta)/(1 - mu tan theta)) )`

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Solution:

(i) Ball bearings are used to reduce friction under rolling as it is easier to roll a body than to slide as `mu _r < mu _k`·

(ii) Horse presses the ground in slanting direction. The reaction `vec R_h` of the ground on the horse acts in opposite direction. `vec R_h` can be resolved into two rectangular component.

(a) `vec V =` vertical component.

(b) `vec H =` horizontal component. The horse moves forward if `H > T`, in that case
net force acting on horse `= H - T`
If the acceleration of horse is 'a' and 'm' its mass
`H - T = ma` ............(i)

The cart moves forward if `T > f`, in that case
net force acting on cart `= T - f`
`T - f = Ma` ... (ii)
Adding (i) and (ii)
`a = (H - f)/(M + m)` ... (iii)

'a' is positive if `(H - f)` is positive if `H - f > 0` or if `H > f` thus the system moves if `H > f`
So the horse has to apply more force to start.
(iii) Banking is necessary because :
(a) Force of friction providing centripetal force is not reliable one.
(b) It prevents skidding.
(c) The speed can be more without skidding.
(d) Wear and tear can be reduced.

(iv) Advantages of friction :
♦ Had there been no friction it would have been impossible for us to walk.
♦ It would have been impossible to transmit power with help of belts.
Disadvantages of friction :
♦ It causes wear and tear of machinery.
♦ Reduces speed of vehicles.

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Solution:

The angle of repose is defined as the maximum inclination of the plane for which the mass kept over it can stay at rest. Let `phi` be the angle of repose. Then,

`N = m g \ \ cos phi , F_f = mg \ \ sin phi`

Also, `F_f = mu N`

`:. mu \ \ m g \ \ cos \ \ phi = mg sin phi,`

i.e., ` mu = tan phi`
Since the body is at rest, the friction is static friction and the coefficient is for static case.

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