Acceleration of the solid sphere while rolling down without slipping
\(a_{1}=\frac{g \sin \theta}{1+\frac{k^{2}}{r^{2}}}\)
Acceleration developed while slipping down \(a_{2}=g \sin \theta\)
\(\text { Required ratio } \frac{a_{1}}{a_{2}}=\frac{g \sin \theta}{1+\frac{k^{2}}{r^{2}}} / g \sin \theta\)
\(\frac{a_{1}}{a_{2}}=\frac{1}{1+\frac{k^{2}}{r^{2}}}\)
\(\text { For a solid sphere } \frac{k^{2}}{r^{2}}=\frac{2}{5}\)
\(\therefore \text { Ratio of accelerations } \frac{a_{1}}{a_{2}}=\frac{1}{1+\frac{2}{5}}\)
\(=\frac{1}{5+\frac{2}{5}}=\frac{1}{\frac{7}{5}}=\frac{5}{7}\)
\(\therefore a_{1}: a_{2}=5: 7 \)

\(\text { Kinetic energy } E_{k}=\frac{p^{2}}{2 m}\)
\(\frac{\Delta E_{k}}{E_{k}}=\frac{2 \Delta p}{p}\)
\(\text {Given that } \frac{\Delta p}{p}=0.1\)
∴ Increase in kinetic energy
\(\frac{\Delta E_{k}}{E_{k}}=2 \frac{\Delta p}{p} \)
\(\frac{\Delta E_{k}}{E_{k}}=2 \times 0.1=0.2 \% \)

Velocity and acceleration are in the same direction: So speed increases.

\(\text { Error in radius }=2 \%\)
\(\Delta r=\frac{2}{100}=0.02\)
\(\text {Volume of the sphere }=\frac{4}{3} \pi r^{3}\)
\(V =\frac{4}{3} \pi r^{3} \)
\(\frac{d V}{V} =\frac{4}{3} \pi \times 3 r^{2} d r \)
\(=3 d r=3(2 \%)=6 \%\)

Mass m = 10 kg
Force F = 16 N
time t = 10 s
\(a=F/m=\frac { 16N }{ 10 \ kg } =1.6 \ ms^{ -2 }\)
We know that, v = u + at = 0 + 1.6 \(\times\) 10 = 16 ms^-1
Kinetic energy K.E = \(\frac { 1 }{ 2 } { mv }^{ 2 }=\frac { 1 }{ 2 } \times 10\times 16\times 16=1280\ J\)

Let the mass of the first body be m which moves with an initial velocity, u₁ = 10 m s^-1.
Therefore, the mass of second body is 2m and its initial velocity is \({ u }_{ 2 }=\frac { 1 }{ 2 } { u }_{ 1 }=\frac { 1 }{ 2 } (10{ ms }^{ -1 })\)
\({ v }_{ 1 }=\left( \frac { { m }_{ 1 }-{ m }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } \right) { u }_{ 1 }+\left( \frac { 2m }{ { m }_{ 1 }+{ m }_{ 2 } } \right) { u }_{ 2 }\)
\({ v }_{ 1 }=\left( \frac { { m }_{ 1 }-2m }{ m+2m } \right) +10+\left( \frac { 2\times 2m }{ m+2m } \right) 5\)
\({ v }_{ 1 }=-\left( \frac { 1 }{ 3 } \right) 10+\left( \frac { 4 }{ 3 } \right) 5=\frac { -10+20 }{ 3 } =\frac { 10 }{ 3 } \)
v₁ = 3.33 ms^-1
\({ v }_{ 2 }=\left( \frac { 2{ m }_{ 1 } }{ { m }_{ 1 }+{ m }_{ 2 } } \right) { u }_{ 1 }+\left( \frac { { m }_{ 2 }-{ m }_{ 1 } }{ { m }_{ 1 }+{ m }_{ 2 } } \right) { u }_{ 2 }\)
\({ v }_{ 2 }=\left( \frac { 2m }{ m+2m } \right) 10+\left( \frac { 2m-m }{ m+2m } \right) 5\)
\({ v }_{ 2 }=\left( \frac { 2 }{ 3 } \right) 10+\left( \frac { 1 }{ 3 } \right) 5=\frac { 20+5 }{ 3 } =\frac { 25 }{ 3 } \)
v₂ = 8.33 ms^-1
As the two speeds v₁ and v₂ are positive, they move in the same direction with the velocities 3.33 ms^-1 and 8.33 ms^-1 respectively.

\(\vec v=\frac{\bar {dr}}{dt}=6t\hat i+5\hat j\)
The velocity at any time 't' is given by \(\vec v=6t\hat i+5\hat j\) . The magnitude of velocity is speed.
The speed at any time 't' is then given by
\(speed=\sqrt {(6t)^2+5^2}=\sqrt {6t^2+25}\)
Now the velocity at t = 3 sec is given by
\(\vec v=6(3)\hat i+5\hat j=18\hat i+5\hat j\)
and speed at t = 3 sec, is given by
\(speed=\sqrt {349}\ m/s\)

We know that speed is the magnitude of the velocity vector. Hence,
Speed of A = \(|\vec{V_A}|=\sqrt{(3)^2+(-5)^2+(2)^2}=\sqrt{9+25+4}=\sqrt{48}ms^{-1}\)
Speed of B = \(\vec{V_B}=\sqrt{(1)^2+(2)^2+(3)^2}=\sqrt{1+4+9}=\sqrt{14}ms^{-1}\)
Spedd of C = \(\vec{V_C}=\sqrt{(5)^2+(3)^2+(4)^2}=\sqrt{25+9+16}=\sqrt{50}ms^{-1}\)
The particle C has the greatest speed.
\(\sqrt{50}>\sqrt{38}>\sqrt{14}\)

Sum of the clockwise moments is equal to sum of the anticlockwise moments when a body is in rotational equilibrium or algebraic sum of moments at any point is zero.

S.No.	Conservative forces	Non-Conservative forces
1	Work done is independent of the path	Work done depends upon the path
2	Work done in a round trip is zero	Work done in a round trip is not zero.
3	Total energy remains constant	Energy is dissipated as heat energy
4	Work done is completely recoverable	Work done is not completely recoverable
5	Force is the negative gradient of potential energy	No such relation exists
6	Examples: Elastic spring force, electrostatic force, magnetic force, gravitational force, etc.	Eg: Frictional forces, viscous force

\(\overrightarrow { r } =3\hat { i } +2\hat { j } \)
\(\text { Unit vector } =\frac{\vec{r}}{|\vec{r}|}  \)
\(|\vec{r}| =\sqrt{3^{2}+2^{2}}=\sqrt{13}  \)
\(\therefore \text { Unit vector } =\frac{3 \hat{i}+2 \hat{j}}{\sqrt{13}}\)

Limitations of Dimensional analysis: 
(i) This method gives no information about the dimensionless constants in the formula like 1, 2,................ \(\pi\), e, etc.
(ii) This method cannot decide whether the given quantity is a vector or a scalar.
(iii) This method is not suitable to derive relations involving trigonometric, exponential and logarithmic functions.
(iv) It cannot be applied to an equation involving more than three physical quantities.
(v) It can only check on whether a physical relation is dimensionally correct but not the correctness of the relation.
For example, using dimensional analysis, s = ut + 1/3 at² is dimensionally correct whereas the correct relation is s = ut+1/2 at².

Since the coefficient of friction is \(\frac { 1 }{ \sqrt { 3 } } \)
tanθ = \(\frac { 1 }{ \sqrt { 3 } } \) ⇒ θ=30⁰.

We can resolve all the vectors in x, y and z components with respect to Cartesian coordinate system.
Once we resolve the components we can separately equate the x components on both sides, y components on both sides, and z components on both the sides of the equation, we then get
\(F_{1x}+F_{2x}+F_{3x}=F_{4x}\)
\(F_{1y}+F_{2y}+F_{3y}=F_{4y}\)
\(F_{1z}+F_{2z}+F_{3z}=F_{4z}\)

four

S.No	Sliding	Slipping
1.	In sliding the transitional motion is more than rotational motion.	In slipping the rotation is more than translational motion.
2.	It happens when sudden brake is applied in a moving vehicles or when the vehicle enters into a slippery road.	It happens when we suddenly start the vehicle from rest or the vehicle is stuck in mud.

Range of time scales: astronomical scales to microscopic scales, 10¹⁸ S to 10^-22s.

(a) A frame of reference which is at rest or which is moving with a uniform velocity along a straight line is called an inertial frame of reference.
(b) In the inertial frame of reference Newton's laws of motion holds good.
Example: The lift at rest, lift moving (up or down) with constant velocity, car moving with constant velocity on a straight road.

If a quantity depends on more than three factors having dimensional formula cannot be derived. Because on equating the powers of M, L & T on either side of the dimensional equation, three equations can be obtained, from which only three unknown dimensions can be calculated.

One radian is the angle subtended at the center of a circle by an arc that is equal in length to the radius of the circle.
1 rad = \(\frac{180}{\pi}\) degree = 57.295^o

A body is said to be in equilibrium if both the linear momentum and angular momentum of the rigid body remain constant with time. Hence for a body in equilibrium, the linear acceleration of its centre of mass would be zero and also the angular acceleration of the rigid body about any axis would be zero.
The different types of equilibrium of a body are
1. Stable equilibrium
2. Unstable equilibrium
3. Neutral equilibrium
Equilibrium is thus stable, unstable neutral according, to whether potential energy is minimum, maximum or instant.

Let us consider the motion of a marble along a curved surface of a bowls.
If a marble M is placed on a curved surface of a bowl S it rolls down and settles in equilibrium at the lowest point A as shown in figure (a).
If the marble is disturbed and displaced at B, its energy increases. When it is released, the marble rolls back at A. Thus the marble at the portion A is said to be in stable equilibrium. In this case, the body possess minimum potential energy.
Suppose, now that bowl S is inverted and the marble is placed at its top point at A as shown in Fig (b).
If the marble is displaced slightly to the point C, its potential energy is lowered and tends to move further away from the equilibrium position to one of lowest energy. Thus the marble is said to be in unstable equilibrium.
Consider that the marble M is placed on the plane surface as shown in figure (c). If it is displaced slightly, its potential energy does not change. In this case, the marble is said to be in neutral equilibrium.
Translational equilibrium:
The resultant of all the external forces acting on the body must be zero.
\(\sum \vec{F}_{ext} =0 \ or \sum F_x=0\sum F_y=0\sum F_2=0\)
\(\sum \vec{F}_{ext} =M\vec{\alpha}_{CM}=M \frac{d\vec{v}_{CM}}{dt}=0\)
or \(\frac{d\vec{v}_{CM}}{dt}\)=0 or \(\vec{v}_{cm}\) = constant
This implies that a body in translational equilibrium, will be either at rest (v = 0) or in uniform motion. If the body is in uniform motion along a straight path, it is in dynamic equilibrium
Rotational equilibrium:
For rotational equilibrium \(\sum \vec{\tau}_{ext}=\sum \vec{r}.x\vec{F}_{ext}=0\)
If the total torque is zero about any point, then it will be zero about any other point when the body is in equilibrium

i. Every object continues to be in the state of rest or of uniform motion unless there is external force acting on it.
ii. The force acting on an object is equal to the rate of change of its momentum.
iii. For every action there is an equal and opposite reaction.
Discussion:
i) Newton's laws are vector laws. The equation \(\vec{F}=\mathrm{m} \vec{a}\) can be written in cartesian coordinates as
\(\mathrm{F}_{x} \hat{\mathrm{i}}+\mathrm{f}_{\mathrm{y}} \hat{\mathrm{j}}+\mathrm{F}_{z} \hat{\mathrm{k}}=\operatorname{ma}_{x} \hat{\mathrm{i}}+\operatorname{ma}_{\mathrm{y}} \hat{\mathrm{j}}+\operatorname{ma}_{\mathrm{z}} \hat{\mathrm{k}}\)
On comparing both sides,
\(\mathrm{F}_{x}=m \mathrm{~m}_{x}  \)
\(\mathrm{F}_{\mathrm{y}}=m \mathrm{ma}_{\mathrm{y}}  \)
\(\mathrm{F}_{\mathrm{z}}=m \mathrm{ma}_{\mathrm{z}}\)
From the above equations, we can infer that the force acting along y direction cannot alter the acceleration along n direction. In the same way, F₂ can not affect a_y and a_n
ii) The acceleration experienced by the body at a time t depends on the force which acts on the body at that instant of time. Thus, \(\vec{F}(\mathrm{t})=\mathrm{m} \vec{a}(\mathrm{t})\)
when a bowler throws the ball to a batsman the acceleration of the ball is determined by the gravitational and air frictional forces and not by the speed which it is thrown.
iii) The direction of force may be different from the direction of force. The following motions are possible.
a) Force and motion in the same direction.
Example: Falling of an apple from the tree.
b) Force and motion are not in the same direction.
Example: The Moon experiences a force towards the Earth but Moon moves in the elliptical path.
c) Force and motion are in the opposite direction.
Example: The motion of a body thrown vertically upward.
d) Zero net force, but there is motion.
Example: The falling of rain drops.
4) If multiple forces \(\vec{F}_{1}, \vec{F}_{2}, \vec{F}_{3} \ldots\) act on the same body then total force is equal to the vectorial sum of the individual forces \(\vec{F}_{\text {net }}=\vec{F}_{1}+\vec{F}_{2}+\vec{F}_{3}+\ldots\)
5) Newton's second law can be written in the second derivative of position vector as  \(\vec{F}=\mathrm{m} \frac{\mathrm{d}^{2}\overrightarrow{\mathrm{r}}}{\mathrm{dt}^{2}}\). It means that whenever the second derivative of position vector is not zero, there must be a force acting on the body.
6) If no force acts on the body then \(m \frac{d \vec{v}}{\mathrm{dt}}=0\). It implies that \(\vec{v}\) is constant. Thus second law is consistent with the first law even though they are independent to each other.
7) Newton's second law is cause and effect relation force is the cause and acceleration is effect.

i) Systematic errors
Systematic errors are reproducible inaccuracies that are consistently in the same direction. These occur often due to a problem that persists throughout the experiment. Systematic errors can be classified as follows
1) Instrumental errors
When an instrument is not calibrated properly at the time of manufacture, instrumental errors may arise. If a measurement is made with a meter scale whose end is worn out, the result obtained will have errors. These errors can be corrected by choosing the instrument carefully.
2) Imperfections in experimental technique or procedure
These errors arise due to the limitations in the experimental arrangement. As an example, while performing experiments with a calorimeter, if there is no proper insulation, there will be radiation losses. This results in errors and to overcome these, necessary correction has to be applied.
3) Personal errors
These errors are due to individuals performing the experiment, may be due to incorrect initial setting up of the experiment or carelessness of the individual making the observation due to improper precautions.
4) Errors due to external causes
The change in the external conditions during an experiment can cause error in measurement. For example, changes in temperature, humidity or pressure during measurements may affect the result of the measurement.
5) Least count error
Least count is the smallest value that can be measured by the measuring instrument, and the error due to this measurement is least count error. The instrument's resolution hence is the cause of this error. Least count error can be reduced by using a high precision instrument for the measurement.
ii) Random errors
Random errors may arise due to random and unpredictable variations in experimental conditions like pressure, temperature, voltage supply etc. Errors may also be due to personal errors by the observer who performs the experiment. Random errors are sometimes called "chance error". When different landings are obtained by a person every time he occurs. For example, consider the case of the thickness of a wire measured using a screw gauge. The readings taken may be different for different trials. In this case, a large number of measurements are made and then the arithmetic mean is taken.
If n number of trial readings are taken in an experiment, and the readings are a₁, a₂, a₃,...............a_n. The arithmetic mean is
\( a_{m}=\frac{a_{1}+a_{2}+a_{3}+\ldots \ldots \ldots \ldots \ldots \ldots a_{n}}{n} \text { or } \)
\(a_{m}=\frac{1}{n} \sum_{i=1}^{t=n} a_{i}\)
Usually this arithmetic mean is taken as the best possible true value of the quantity.
Certain procedures to be followed to minimize experimental errors, along with examples are shown in Table.
Type of error Example How to minimize it Random error Suppose you measure the mass of a ring three times using the same balance and get slightly different values. 15.46 g, 15.42 g, 15.44 g Take more data. Random errors can be evaluated through statistical analysis and can be reduced by averaging over a large number of observations. Systematic error Suppose the cloth tape measure that you use to measure the length of an object has been stretched out from years of use. (As a result all of the length measurements are not correct). Systematic errors are difficult to detect and cannot be analysed statistically, because all of the data is in the same direction. (Either too high or too low) iii) Gross Error
The error caused due to the sheer carelessness of an observer is called gross error.
For example
(i) Reading an instrument without setting it properly.
(ii) Taking observations in a wrong manner without bothering about the sources of errors and the precautions.
(iii) Recording wrong observations.
(iv) Using wrong values of the observations in calculations.
These errors can be minimized only when an observer is careful and mentally alert.