Maharashtra Board Class 11 Physics Chapter 3 Motion in a Plane Solutions

1. Choose The Correct Option.

Question 1. An object thrown from a moving bus is on example of _____
(A) Uniform circular motion
(B) Rectilinear motion
(C) Projectile motion
(D) Motion in one dimension
Answer: (C) Projectile motion
In simple words: When an object is thrown from a moving vehicle, its motion is influenced by both its initial velocity from the vehicle and the force of gravity, resulting in a curved path characteristic of projectile motion.

🎯 Exam Tip: Understanding the fundamental characteristics of different types of motion (rectilinear, circular, projectile) is crucial for selecting the correct option in MCQs. Focus on real-world examples to solidify concepts.

Question 2. For a particle having a uniform circular motion, which of the following is constant
(A) Speed
(B) Acceleration
(C) Velocity
(D) Displacement
Answer: (A) Speed
In simple words: In uniform circular motion, a particle moves along a circular path at a constant rate, meaning its speed remains the same, but its velocity continuously changes direction.

🎯 Exam Tip: Differentiate clearly between speed (scalar) and velocity (vector) in uniform circular motion. While speed is constant, the continuous change in direction means velocity and acceleration are not constant.

Question 3. The bob of a conical pendulum undergoes
(A) Rectilinear motion in horizontal plane
(B) Uniform motion in a horizontal circle
(C) Uniform motion in a vertical circle
(D) Rectilinear motion in vertical circle
Answer: (B) Uniform motion in a horizontal circle
In simple words: A conical pendulum's bob moves in a horizontal circle while the string traces out a cone, indicating uniform circular motion in a horizontal plane.

🎯 Exam Tip: Visualize the motion of a conical pendulum to understand that the bob maintains a constant height and moves in a perfect horizontal circle, distinguishing it from simple pendulums.

Question 4. For uniform acceleration in rectilinear motion which of the following is not correct?
(A) Velocity-time graph is linear
(B) Acceleration is the slope of velocity time graph
(C) The area under the velocity-time graph equals displacement
(D) Velocity-time graph is nonlinear
Answer: (D) Velocity-time graph is nonlinear
In simple words: For uniform acceleration, velocity changes linearly with time, so its velocity-time graph is a straight line, not a curve.

🎯 Exam Tip: Remember that uniform acceleration implies a constant rate of change of velocity, which translates to a straight line (linear) on a velocity-time graph. Non-linear graphs indicate non-uniform acceleration.

Question 5. If three particles A, B and C are having velocities \(\vec{V}_A\), \(\vec{V}_B\) and \(\vec{V}_C\) which of the following formula gives the relative velocity of A with respect to B
(A) \(\vec{V}_A + \vec{V}_B\)
(B) \(\vec{V}_A - \vec{V}_C + \vec{V}_B\)
(C) \(\vec{V}_A - \vec{V}_B\)
(D) \(\vec{V}_C - \vec{V}_A\)
Answer: (C) \(\vec{V}_A - \vec{V}_B\)
In simple words: To find the relative velocity of object A with respect to object B, you subtract the velocity of B from the velocity of A.

🎯 Exam Tip: The concept of relative velocity is crucial. Always subtract the velocity of the reference frame (the "with respect to" object) from the velocity of the object being observed.

2. Answer The Following Questions.

Question 1. Separate the following in groups of scalar and vectors: velocity, speed, displacement, work done, force, power, energy, acceleration, electric charge, angular velocity.
Answer:
Scalars
Speed, work done, power, energy, electric charge.
Vectors
Velocity, displacement, force, acceleration, angular velocity (pseudo vector).
In simple words: Scalars have only magnitude (like speed), while vectors have both magnitude and direction (like velocity and force).

🎯 Exam Tip: Clearly distinguishing between scalar and vector quantities is fundamental in physics. Practice identifying common physical quantities as either scalar or vector based on whether direction is essential for their complete description.

Question 2. Define average velocity and instantaneous velocity. When are they same?
Answer:
Average velocity:
1. Average velocity \((\vec{v}_{av})\) of an object is the displacement \((\Delta x)\) of the object during the time interval \((\Delta t)\) over which average velocity is being calculated, divided by that time interval.
2. Average velocity = \((\frac{\text{Displacement}}{\text{Time interval}})\)
\(\vec{V}_{av} = \frac{x_2-x_1}{t_2-t_1} = \frac{\Delta x}{\Delta t}\)
3. Average velocity is a vector quantity.
4. Its SI unit is m/s and dimensions are \([M^0L^1T^{-1}]\)
5. For example, if the positions of an object are x +4 m and x = +6 m at times t = 0 and t = 1 minute respectively, the magnitude of its average velocity during that time is \(V_{av} = (6 – 4)/(1 – 0) = 2\) m per minute and its direction will be along the positive X-axis.
\(\therefore \vec{V}_{av} = 2 \hat{i} \text{ m/min}\)
Where, \(\hat{i}\) = unit vector along X-axis.
Instantaneous velocity:
1. The instantaneous velocity \((\vec{V})\) is the limiting value of the average velocity of the object over a small time interval \((\Delta t)\) around t when the value of time interval goes to zero.
2. It is the velocity of an object at a given instant of time.
3. \(\vec{v} = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = \frac{d\vec{x}}{dt}\)
where \(\frac{d\vec{x}}{dt}\) derivative of \(\vec{x}\) with respect to t.
In case of uniform rectilinear motion, i.e., when an object is moving with constant velocity along a straight line, the average and instantaneous velocity remain same.
In simple words: Average velocity is total displacement divided by total time, while instantaneous velocity is the velocity at a specific moment. They are the same when an object moves with constant velocity along a straight line.

🎯 Exam Tip: When asked to define, provide both conceptual and mathematical definitions. Highlight the condition for equality (uniform rectilinear motion) as it demonstrates a deeper understanding.

Question 3. Define free fall.
Answer:
The motion of any object under the influence of gravity alone is called as free fall.
In simple words: Free fall is when an object moves only under the force of gravity, with air resistance being negligible or absent.

🎯 Exam Tip: A concise and accurate definition is key. Emphasize "gravity alone" and imply negligible air resistance for a complete answer.

Question 4. If the motion of an object is described by x = f(t) write formulae for instantaneous velocity and acceleration.
Answer:
1. Instantaneous velocity of an object is given as, \(\vec{V} = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = \frac{dx}{dt}\)
2. Motion of the object is given as, x = f(t)
3. The derivative f '(t) represents the rate of change of the position f (t) at time t, which is the instantaneous velocity of the object. \(\therefore \vec{v} = \frac{dx}{dt} = f'(t)\)
4. Acceleration is defined as the rate of change of velocity with respect to time.
5. The second derivative of the position function f “(t) represents the rate of change of velocity i.e., acceleration. \(\therefore \vec{a} = \frac{\Delta v}{\Delta t} = \frac{d^2x}{dt^2} = f''(t)\)
In simple words: Instantaneous velocity is the first derivative of position with respect to time, and instantaneous acceleration is the second derivative of position with respect to time, or the first derivative of velocity.

🎯 Exam Tip: This question tests your understanding of calculus in kinematics. Clearly state the derivative relationships for position, velocity, and acceleration. Using vector notation where appropriate enhances clarity.

Question 5. Derive equations of motion for a particle moving in a plane and show that the motion can be resolved in two independent motions in mutually perpendicular directions.
Answer:
1. Consider an object moving in an x-y plane. Let the initial velocity of the object be \(\vec{u}\) at t = 0 and its velocity at time t be \(\vec{v}\).
2. As the acceleration is constant, the average acceleration and the instantaneous acceleration will be equal. \(\vec{a}_{av} = \frac{\vec{V}_2-\vec{V}_1}{t_2-t_1} = \frac{V_{2x}-V_{1x}}{t_2-t_1} \hat{i} + \frac{V_{2y}-V_{1y}}{t_2-t_1} \hat{j}\)
\(\therefore \vec{a} = \frac{\vec{v}-\vec{u}}{(t-0)}\)

\(\implies \vec{v} = \vec{u} + \vec{a}t\) ....(1) This is the first equation of motion in vector form.
3. Let the displacement of the object from time t = 0 to t be \(\vec{s}\)
For constant acceleration, \(\vec{V}_{av} = \frac{\vec{v} + \vec{u}}{2}\)
\(\therefore \vec{s} = \vec{V}_{av} t = \left(\frac{\vec{v}+\vec{u}}{2}\right) t = \left(\frac{\vec{u}+\vec{u}+\vec{a}t}{2}\right) t\)

\(\implies \vec{s} = \vec{u}t + \frac{1}{2}\vec{a}t^2\) ....(2) This is the second equation of motion in vector form.
4. Equations (1) and (2) can be resolved into their x and y components so as to get corresponding scalar equations as follows.
\(V_x = u_x + a_x t\) .......(3)
\(V_y = u_y + a_y t\) .......(4)
\(S_x = u_x t + \frac{1}{2} a_x t^2\) .......(5)
\(S_y = u_y t + \frac{1}{2} a_y t^2\) .......(6)
5. It can be seen that equations (3) and (5) involve only the x components of displacement, velocity and acceleration while equations (4) and (6) involve only the y components of these quantities.
6. Thus, the motion along the x direction of the object is completely controlled by the x components of velocity and acceleration while that along the y direction is completely controlled by the y components of these quantities.
7. This shows that the two sets of equations are independent of each other and can be solved independently.
In simple words: The vector equations of motion in a plane can be broken down into separate scalar equations for the x and y components. Since these component equations only involve variables from their respective directions, the horizontal and vertical motions are independent of each other.

🎯 Exam Tip: For derivations, clearly state assumptions (e.g., constant acceleration). Show vector forms first, then resolve them into scalar components along perpendicular axes. Emphasize the independence of these component motions as a key conclusion.

Question 6. Derive equations of motion graphically for a particle having uniform acceleration, moving along a straight line.
Answer:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक वेग-समय ग्राफ है जो एक कण की गति को दर्शाता है। क्षैतिज अक्ष (समय) पर 't' और ऊर्ध्वाधर अक्ष (वेग) पर 'v' दिखाया गया है। ग्राफ में एक चतुर्भुज OPRS और एक त्रिभुज PQR शामिल है, जिसमें प्रारंभिक वेग 'u' और अंतिम वेग 'v' है, तथा 't' समय के बाद कण की स्थिति 'Q' पर है।"
1. Consider an object starting from position x = 0 at time t = 0. Let the velocity at time (t = 0) and t be u and v respectively.
2. The slope of line PQ gives the acceleration. Thus
\(\therefore a = \frac{v-u}{t-0} = \frac{v-u}{t}\)

\(\implies v = u + at\) ...........(1) This is the first equation of motion.
3. The area under the curve in velocity-time graph gives the displacement of the object.
\(\therefore s =\) area of the quadrilateral OPQS = area of rectangle OPRS + area of triangle PQR.
\( = ut + \frac{1}{2} (v - u) t\)
But, from equation (1) at = v - u
\(\therefore s = ut + \frac{1}{2} at^2\)
This is the second equation of motion,
4. The velocity is increasing linearly with time as acceleration is constant. The displacement is given as,
\(s = V_{av}t = \left(\frac{v+u}{2}\right) t\)
\( = \frac{(v+u)(v-u)}{2a}\)
\(\therefore s = (v^2 – u^2) / (2a)\)
\(\therefore v^2 – u^2 = 2as\)
This is the third equation of motion.
In simple words: The three equations of motion can be derived from a velocity-time graph by analyzing its slope (acceleration) and the area under it (displacement). The first equation comes from the slope, the second from the area of a rectangle plus a triangle, and the third from substituting the first equation into the average velocity formula.

🎯 Exam Tip: When deriving graphically, clearly label your velocity-time graph. Explain how slope relates to acceleration and area relates to displacement for each equation. Use standard notations consistently.

Question 7. Derive the formula for the range and maximum height achieved by a projectile thrown from the origin with initial velocity \(\vec{u}\) at an angle \(\theta\) to the horizontal.
Answer:
Expression for range:
1. Consider a body projected with velocity \(\vec{u}\), at an angle \(\theta\) of projection from point O in the co-ordinate system of the XY- plane, as shown in figure.
2. The initial velocity \(\vec{u}\) can be resolved into two rectangular components:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक प्रक्षेप्य की गति को दर्शाता है, जिसे क्षैतिज अक्ष पर 'X' और ऊर्ध्वाधर अक्ष पर 'Y' के साथ XY-समतल में दर्शाया गया है। प्रारंभिक वेग 'u' को इसके क्षैतिज (\(u_x\)) और ऊर्ध्वाधर (\(u_y\)) घटकों में विभाजित किया गया है, और प्रक्षेपवक्र के विभिन्न बिंदुओं पर वेग घटकों (\(V_x\), \(V_y\)) को भी दर्शाया गया है।"
\(u_x = u \cos \theta\) (Horizontal component)
\(u_y = u \sin \theta\) (Vertical component)
3. The horizontal component remains constant throughout the motion due to the absence of any force acting in that direction, while the vertical component changes according to
\(V_y = u_y + a_y t\)
with \(a_y = -g\) and \(u_y = u \sin \theta\)
4. Thus, the components of velocity of the projectile at time t are given by,
\(V_x = u_x = u \cos \theta\)
\(V_y = u_y – gt = u \sin \theta – gt\)
5. Similarly, the components of displacements of the projectile in the horizontal and vertical directions at time t are given by,
\(s_x = (u \cos \theta) t\)
\(S_y = (u \sin \theta)t – \frac{1}{2} gt^2\)
6. At the highest point, the time of ascent of the projectile is given as,
\(t_A = \frac{u \sin \theta}{g}\) .....(2)
7. The total time in air i.e., time of flight is given as, \(T = 2t_A = \frac{2u \sin \theta}{g}\) .....(3)
8. The total horizontal distance travelled by the particle in this time T is given as,
\(R = u_x \cdot T\)
\(R = u \cos \theta \cdot (2t_A)\)
\(R = u \cos \theta \cdot \frac{2u \sin \theta}{g}\) .............[From (3)]
\(R = \frac{u^2 (2 \sin \theta \cos \theta)}{g}\)
\(R = \frac{u^2 \sin 2\theta}{g}\) .....[\(\because \sin 2\theta = 2\sin \theta \cos \theta\)]
This is required expression for horizontal range of the projectile.
Expression for maximum height of a projectile:
The maximum height H reached by the projectile is the distance travelled along
the vertical (y) direction in time \(t_A\).
Substituting \(s_y = H\) and \(t = t_A\) in equation (1), we have,
\(H = (u \sin \theta)t_A – \frac{1}{2} gt_A^2\)
\(\therefore H = u \sin \theta \left(\frac{u \sin \theta}{g}\right) - \frac{1}{2} g \left(\frac{u \sin \theta}{g}\right)^2\)
....[From (2)]
\(H = \frac{u^2 \sin^2 \theta}{2g} = \frac{u_y^2}{2g}\)
This equation represents maximum height of projectile.
In simple words: The range of a projectile is the total horizontal distance it covers, derived from its horizontal velocity component and total flight time. Maximum height is the highest vertical point reached, calculated from its initial vertical velocity and the time to reach that peak. Both formulas depend on initial velocity, projection angle, and gravity.

🎯 Exam Tip: Break down projectile motion into independent horizontal and vertical components. Clearly define variables, use appropriate kinematic equations for each component, and show step-by-step derivation for both range and maximum height.

Question 8. Show that the path of a projectile is a parabola.
Answer:
1. Consider a body projected with velocity initial velocity \(\vec{u}\), at an angle \(\theta\) of projection from point O in the co-ordinate system of the XY-plane. as shown in figure.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक प्रक्षेप्य की गति को दर्शाता है, जिसमें प्रारंभिक वेग 'u' और प्रक्षेप्य कोण '\(\theta\)' है। XY-समतल में प्रक्षेप्य का पथ और उसके वेग के घटकों (\(u_x\), \(u_y\), \(V_x\), \(V_y\)) को दर्शाया गया है, जिससे प्रक्षेप्य का परवलयिक मार्ग स्पष्ट होता है।"
2. The initial velocity \(\vec{u}\) can be resolved into two rectangular components:
\(u_x = u \cos \theta\) (Horizontal component)
\(u_y = u \sin \theta\) (Vertical component)
3. The horizontal component remains constant throughout the motion due to the absence of any force acting in that direction, while the vertical component changes according to,
\(V_y = U_y + a_y t\)
with \(a_y = -g\) and \(u_y = u \sin \theta\)
4. Thus, the components of velocity of the projectile at time t are given by,
\(V_x = u_x = u \cos \theta\)
\(V_y = u_y - gt = u \sin \theta – gt\)
5. Similarly, the components of displacements of the projectile in the horizontal and vertical directions at time t are given by,
\(s_x = (u \cos \theta)t\) .....(1)
\(S_y = (u \sin \theta)t – \frac{1}{2} gt^2\) .....(2)
6. As the projectile starts from x = O, we can use \(s_x = x\) and \(s_y = y\).
Substituting \(s_x = x\) in equation (1),
\(x = (u \cos \theta) t\)
\(\therefore t = \frac{x}{(u \cos \theta)}\) .....(3)
Substituting, \(s_y\) in equation (2),
\(y = (u \sin \theta)t – \frac{1}{2} gt^2\) .....(4)
Substituting equation (3) in equation (4), we have,
\(y = u \sin \theta \left(\frac{x}{u \cos \theta}\right) - \frac{1}{2} g \left(\frac{x}{u \cos \theta}\right)^2\)
\(\therefore y = x (\tan \theta) – \left(\frac{g}{2u^2 \cos^2 \theta}\right)x^2\) .....(5)
Equation (5) represents the path of the projectile.
7. If we put \(\tan \theta = A\) and \(\frac{g}{2u^2\cos^2\theta} = B\) then equation (5) can be written as \(y = Ax – Bx^2\) where A and B are constants. This is equation of parabola. Hence, path of projectile is a parabola.
In simple words: By expressing the horizontal and vertical displacements of a projectile as functions of time and then eliminating time, we arrive at an equation that matches the standard form of a parabola (\(y = Ax - Bx^2\)), thus proving its path is parabolic.

🎯 Exam Tip: The key to this derivation is to establish equations for x and y coordinates as functions of time, then eliminate time (t) to get an equation relating y and x. Recognizing the final equation as a parabolic form is essential for the conclusion.

Question 9. What is a conical pendulum? Show that its time period is given by \(2\pi \sqrt{\frac{l \cos \theta}{g}}\), where \(l\) is the length of the string, \(\theta\) is the angle that the string makes with the vertical and g is the acceleration due to gravity.
Answer:
A simple pendulum, which is given such a motion that the bob describes a horizontal circle and the string making a constant angle with the vertical describes a cone, is
called a conical pendulum.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक शंक्वाकार पेंडुलम को दर्शाता है, जिसमें एक द्रव्यमान 'm' की गेंद एक धागे 'l' से बंधी हुई है। धागा ऊर्ध्वाधर के साथ '\(\theta\)' का कोण बनाता है, और गेंद 'r' त्रिज्या के एक क्षैतिज वृत्त में घूमती है। तनाव 'T' को इसके घटकों (\(T \cos \theta\), \(T \sin \theta\)) में और गुरुत्वाकर्षण बल 'mg' को नीचे की ओर दर्शाया गया है।"
O : rigid support
T : tension in the string
l : length of string
h : height of support from bob
v : velocity of bob
r : radius of horizontal circle
\(\theta\) : semi vertical angle
mg : weight of bob
1. Consider a bob of mass m tied to one end of a string of length 'l' and other end is fixed to rigid support.
2. Let the bob be displaced from its mean position and whirled around a horizontal circle of radius 'r' with constant angular velocity \(\omega\), then the bob performs U.C.M.
3. During the motion, string is inclined to the vertical at an angle \(\theta\) as shown in the figure above.
4. In the displaced position, there are two forces acting on the bob.
\(\bullet\) The weight mg acting vertically downwards.
\(\bullet\) The tension T acting upward along the string.
5. The tension (T) acting in the string can be resolved into two components:
\(\bullet\) T \(\cos \theta\) acting vertically upwards.
\(\bullet\) T \(\sin \theta\) acting horizontally towards centre of the circle.
6. Since, there is no net force, vertical component T \(\cos \theta\) balances the weight and horizontal component T \(\sin \theta\) provides the necessary centripetal force.
\(\therefore T \cos \theta = mg\) ...........(1)
\(T \sin \theta = \frac{mv^2}{r} = mr\omega^2\) ...........(2)
7. Dividing equation (2) by (1),
\(\tan \theta = \frac{v^2}{rg}\) ...........(3)
Therefore, the angle made by the string with the vertical is \(\theta = \tan^{-1} \left(\frac{v^2}{rg}\right)\)
8. Since we know \(v = \frac{2\pi r}{T}\)

\(\implies \tan \theta = \frac{4\pi^2 r^2/T^2}{rg} = \frac{4\pi^2 r}{T^2 g}\) ....[From (3)]

\(\implies T^2 = \frac{4\pi^2 r}{g \tan \theta}\)

\(\implies T = 2\pi \sqrt{\frac{r}{g \tan \theta}}\)
We also know that \(r = l \sin \theta\)

\(\implies T = 2\pi \sqrt{\frac{l \sin \theta}{g \tan \theta}}\) .....(\(\because r = l \sin \theta\))

\(\implies T = 2\pi \sqrt{\frac{l \sin \theta}{g \frac{\sin \theta}{\cos \theta}}}\)

\(\implies T = 2\pi \sqrt{\frac{l \cos \theta}{g}}\)
We also know that \(h = l \cos \theta\)

\(\implies T = 2\pi \sqrt{\frac{h}{g}}\) .....(\(\because h = l \cos \theta\))
where \(l\) is length of the pendulum and h is the vertical distance of the horizontal circle from the fixed point O.
In simple words: A conical pendulum's bob moves in a horizontal circle, creating a cone with the string. Its time period is derived by resolving the string tension into vertical (balancing gravity) and horizontal (providing centripetal force) components, and then relating these forces to the angular velocity and geometric parameters.

🎯 Exam Tip: Begin with a clear definition and a well-labelled diagram showing forces and components. Apply Newton's second law in both vertical (equilibrium) and horizontal (centripetal force) directions. Systematically derive the time period by eliminating intermediate variables like tension and velocity.

Question 10. Define angular velocity. Show that the centripetal force on a particle undergoing uniform circular motion is \(-m\omega^2\vec{r}\).
Answer:
Angular velocity of a particle is the rate of change of angular displacement.
Expression for centripetal force on a particle undergoing uniform circular motion:
i) Suppose a particle is performing U.C.M in anticlockwise direction.
The co-ordinate axes are chosen as shown in the figure.
Let,
A = initial position of the particle which lies on positive X-axis
P = instantaneous position after time t
\(\theta\) = angle made by radius vector
\(\omega\) = constant angular speed
\(\vec{r}\) = instantaneous position vector at time t
ii) From the figure,
\(\vec{r} = \hat{i}x + \hat{j}y\)
where, \(\hat{i}\) and \(\hat{j}\) are unit vectors along X-axis and Y-axis respectively.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक कण को वृत्ताकार पथ में वामावर्त दिशा में गति करते हुए दिखाता है। मूल बिंदु O पर केंद्रित वृत्त, जिसमें एक कण P(x,y) पर स्थित है। तात्कालिक स्थिति सदिश '\(\vec{r}\)' और वेग सदिश '\(\vec{v}\)' दर्शाए गए हैं, साथ ही कोण '\(\theta\)' और कोणीय वेग '\(\omega\)' भी हैं।"
iii. Also, \(x = r \cos \theta\) and \(y = r \sin \theta\)
\(\therefore \vec{r} = [\hat{i}r \cos \theta + \hat{j}r \sin \theta]\)
But \(\theta = \omega t\)
\(\therefore \vec{r} = [\hat{i}r \cos \omega t + \hat{j}r \sin \omega t]\) ....(1)
iv. Velocity of the particle is given as rate of change of position vector.
\(\therefore \vec{v} = \frac{d\vec{r}}{dt} = \frac{d}{dt} [\hat{i}r \cos \omega t + \hat{j}r \sin \omega t]\)
\( = r \left[ \frac{d}{dt} (\cos \omega t) \hat{i} + \frac{d}{dt} (\sin \omega t) \hat{j} \right]\)

\(\implies \vec{v} = -r\omega \hat{i} \sin \omega t + r\omega \hat{j} \cos \omega t\)

\(\implies \vec{v} = r\omega (-\hat{i} \sin \omega t + \hat{j} \cos \omega t)\)
v. Further, instantaneous linear acceleration of the particle at instant t is given by,
\(\vec{a} = \frac{d\vec{v}}{dt} = \frac{d}{dt} [r\omega (-\hat{i} \sin \omega t + \hat{j} \cos \omega t)]\)
\( = r\omega \frac{d}{dt} (-\hat{i} \sin \omega t + \hat{j} \cos \omega t)\)
\( = r\omega \left[ \frac{d}{dt} (-\sin \omega t) \hat{i} + \frac{d}{dt} (\cos \omega t) \hat{j} \right]\)
\( = r\omega (-\omega \hat{i} \cos \omega t – \omega \hat{j} \sin \omega t)\)
\( = -r\omega^2 (\hat{i} \cos \omega t + \hat{j} \sin \omega t)\)

\(\implies \vec{a} = -\omega^2 (\hat{i}r \cos \omega t + \hat{j}r \sin \omega t)\) ....(2)
vi. From equation (1) and (2),
\(\vec{a} = -\omega^2 \vec{r}\) ....(3)
Negative sign shows that direction of acceleration is opposite to the direction of position vector. Equation (3) is the centripetal acceleration.
vii) Magnitude of centripetal acceleration is given by \(a = \omega^2r\)
viii) The force providing this acceleration should also be along the same direction, hence centripetal.
\(\therefore \vec{F} = m\vec{a} = -m\omega^2\vec{r}\)
This is the expression for the centripetal force on a particle undergoing uniform circular motion.
ix) Magnitude of \(F = m\omega^2r = \frac{mv^2}{r} = m\omega v\)
[Note: The definition of angular velocity is not mentioned in this chapter but is in Ch.2 Mathematical Methods.]
In simple words: Angular velocity describes how fast an object rotates or revolves. Centripetal force, responsible for uniform circular motion, is derived by taking successive derivatives of the position vector with respect to time to find acceleration, then applying Newton's second law (\(\vec{F} = m\vec{a}\)). The negative sign indicates the force is directed opposite to the position vector, i.e., towards the center.

🎯 Exam Tip: For deriving centripetal force, start by defining the position vector in polar coordinates or components, then differentiate it twice with respect to time to find velocity and acceleration. Clearly show the vector operations and explain the significance of the negative sign in the final expression.

3. Solve The Following Problems.

Question 1. An aeroplane has a run of 500 m to take off from the runway. It starts from rest and moves with constant acceleration to cover the runway in 30 sec. What is the velocity of the aeroplane at the take off ?
Answer:
Given: Length of runway (s) = 500 m, t = 30 s
To find: Velocity (v)
Formulae. i) \(s = ut + \frac{1}{2} at^2\)
ii) \(v = u + at\)
Calculation: As the plane was initially at rest, \(u = 0\)
From formula (i),
\(500 = 0 + \frac{1}{2} \times a \times (30)^2\)
\(\therefore 500 = 450 a\)
\(\therefore a = \frac{10}{9} \text{ m/s}^2\)
From formula (ii),
\(v = 0 + \frac{10}{9} \times 30\)
\(\therefore V = \frac{100}{3} \text{ m/s} = \left(\frac{100}{3} \times \frac{18}{5}\right) \text{ km/hr}\)
\(\therefore v = 120 \text{ km/hr}\)
The velocity of the aeroplane at the take off is 120 km/hr.
In simple words: To find the final velocity, first calculate the acceleration using the given distance, initial velocity (zero), and time. Then, use this acceleration with the initial velocity and time to find the final velocity.

🎯 Exam Tip: Identify the knowns and unknowns correctly. Choose the appropriate kinematic equations. Ensure unit consistency throughout the calculation and for the final answer. Show all steps clearly.

Question 2. A car moving along a straight road with a speed of 120 km/hr, is brought to rest

Question 3. A car travels at a speed of 50 km/hr for 30 minutes, at 30 km/hr for next 15 minutes and then 70 km/hr for next 45 minutes. What is the average speed of the car?
Answer:
Given: \(v_1\) = 50 km/hr, \(t_1\) = 30 minutes = 0.5 hr,
\(v_2\) = 30 km/hr, \(t_2\) = 15 minutes = 0.25 hr,
\(v_3\) = 70 km/hr, \(t_3\) = 45 minutes 0.75 hr
To find: Average speed of car (\(v_{av}\))
Formula \(v_{av} = \frac{\text{total path length}}{\text{total time interval}}\)
Calculation:
Path length,
\(x_1 = v_1 \times t_1\) = 50 \(\times\) 0.5 = 25km
\(x_2 = v_2 \times t_2\) = 30 \(\times\) 0.25 = 7.5 km
\(x_3 = v_3 \times t_3\) = 70 \(\times\) 0.75 = 52.5 km
From formula,
\(v_{av} = \frac{x_1+x_2+x_3}{t_1+t_2+t_3}\)
\(v_{av} = \frac{25+7.5+52.5}{0.5+0.25+0.75} = \frac{85}{1.5}\)
\(v_{av}\) = 56.66 km/hr
In simple words: To find the average speed, we calculate the total distance covered by summing up the distances for each segment, and then divide this total distance by the total time taken for the entire journey.

🎯 Exam Tip: Always convert all time units to a consistent format (e.g., hours) before performing calculations to avoid errors in average speed problems.

Question 4. A velocity-time graph is shown in the adjoining figure.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक वेग-समय ग्राफ है जो एक कार की गति को दर्शाता है। X-अक्ष पर समय (सेकंड में) और Y-अक्ष पर वेग (मीटर/सेकंड में) दिखाया गया है। ग्राफ में O से A तक बढ़ती हुई गति, A से B तक स्थिर गति और B से C तक घटती हुई गति (मंदता) प्रदर्शित होती है। OABC एक ट्रैपीज़ॉइडल आकृति बनाता है।
Determine:
1. initial speed of the car
2. maximum speed attained by the car
3. part of the graph showing zero acceleration
4. part of the graph showing constant retardation
5. distance travelled by the car in first 6 sec.
Answer:
1. Initial speed is at origin i.e. 0 m/s.
2. Maximum speed attained by car, \(v_{max}\) = speed from A to B = 20 m/s.
3. The part of the graph which shows zero acceleration is between t = 3 s and t = 6 s i.e., AB. This is because, during AB there is no change in velocity.
4. The graph shows constant retardation from t = 6 s to t = 8 s i.e., BC.
5. Distance travelled by car in first 6 s
= Area of OABDO
= A(\(\triangle\)OAE) + A(rect. ABDE)
= \(\frac{1}{2}\) \(\times\) 3 \(\times\) 20 + 3 \(\times\) 20
= 30 + 60
\( \implies \) Distance travelled by car in first 6 s = 90 m
In simple words: The velocity-time graph visually represents motion, where the y-intercept indicates initial speed, the peak indicates maximum speed, a horizontal line shows zero acceleration (constant velocity), a downward sloping line shows retardation, and the area under the graph gives the distance travelled.

🎯 Exam Tip: Remember that the slope of a velocity-time graph gives acceleration, and the area under it gives displacement. Identify these features clearly on the graph for accurate analysis.

Question 5. A man throws a ball to maximum horizontal distance of 80 meters. Calculate the maximum height reached.
Answer:
Given: R = 80m
To find: Maximum height reached (\(H_{max}\))
Formula: \(R_{max} = 4H_{max}\)
Calculation: From formula,
\(H_{max} = \frac{R_{max}}{4} = \frac{80}{4}\) = 20 m
The maximum height reached by the ball is 20m.
In simple words: For a projectile launched to achieve maximum horizontal range, the maximum height it reaches is one-fourth of that maximum range.

🎯 Exam Tip: This relationship (\(R_{max} = 4H_{max}\)) is specific to projection at 45 degrees, which gives the maximum horizontal range. Always double-check the projection angle if it's not explicitly stated as "maximum horizontal distance".

Question 6. A particle is projected with speed \(v_0\) at angle \(\theta\) to the horizontal on an inclined surface making an angle \(\Phi\) (\(\Phi\) < \(\theta\)) to the horizontal. Find the range of the projectile along the inclined surface.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक प्रक्षेप्य गति का आरेख है जहाँ एक कण को प्रारंभिक वेग \(v_0\) से क्षैतिज के साथ \(\theta\) कोण पर फेंका जाता है। एक झुकी हुई सतह को \(\Phi\) कोण पर दिखाया गया है, जिस पर प्रक्षेप्य गिरता है। R प्रक्षेप्य की झुकी हुई सतह पर परास को दर्शाता है।
i) The equation of trajectory of projectile is given by,
\(y(\tan \theta)x - (\frac{g}{2u^2 \cos^2 \theta})x^2\) ...............(1)
ii) In this case to find R substitute,
y = R sin\(\Phi\) ................. (2)
x = R cos\(\Phi\) ................. (3)
iii) From equations (1), (2) and (3),
we have,
R sin \(\Phi\) = tan \(\theta\) (R cos \(\Phi\)) - \(\frac{g}{2v_0^2 \cos^2 \theta}\) \(R^2 \cos^2 \Phi\)
\( \implies \) (\(\because\) u = \(v_0\))
iv. So, sin \(\Phi\) = tan \(\theta\) cos \(\Phi\) - \(\frac{gR \cos^2 \Phi}{2v_0^2 \cos^2 \theta}\)
\( \implies \) \(\frac{gR \cos^2 \Phi}{2v_0^2 \cos^2 \theta}\) = tan \(\theta\) cos \(\Phi\) - sin \(\Phi\)
v. Hence,
R = \(\frac{2v_0^2 \cos^2 \theta}{g \cos^2 \Phi}\) [tan\(\theta\) cos \(\Phi\) – sin \(\Phi\)]
= \(\frac{2v_0^2 \cos^2 \theta}{g \cos^2 \Phi}\) [\(\frac{\sin \theta}{\cos \theta}\) cos \(\Phi\) – sin \(\Phi\)]
= \(\frac{2v_0^2 \cos \theta}{g \cos^2 \Phi}\) [sin \(\theta\) cos \(\Phi\) – cos \(\theta\) sin \(\Phi\)]
vi. So, R = \(\frac{2v_0^2 \cos \theta}{g \cos^2 \Phi}\) sin (\(\theta\)-\(\Phi\))
\( \implies \) [(\(\because\) sin (\(\theta\)-\(\Phi\)) = sin \(\theta\) cos \(\Phi\) – cos \(\theta\) sin \(\Phi\))]
In simple words: To find the range of a projectile on an inclined plane, we use the equation of trajectory and substitute the coordinates of the landing point on the incline, then solve for the horizontal distance along the incline.

🎯 Exam Tip: When dealing with inclined planes, carefully resolve components of gravity and initial velocity parallel and perpendicular to the incline, or use the trajectory equation method as shown, being mindful of coordinate transformations.

Question 7. A metro train runs from station A to B to C. It takes 4 minutes in travelling from station A to station B. The train halts at station B for 20 s. Then it starts from station B and reaches station C in next 3 minutes. At the start, the train accelerates for 10 sec to reach the constant speed of 72 km/hr. The train moving at the constant speed is brought to rest in 10 sec. at next station.
(i) Plot the velocity- time graph for the train travelling from the station A to B to C.
(ii) Calculate the distance between the stations A, B and C.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक वेग-समय ग्राफ है जो एक मेट्रो ट्रेन की यात्रा को स्टेशन A से C तक दर्शाता है। इसमें त्वरण (बढ़ता वेग), स्थिर वेग, ठहराव (शून्य वेग), और मंदता (घटता वेग) के खंड शामिल हैं। X-अक्ष पर समय (सेकंड में) और Y-अक्ष पर वेग (मीटर/सेकंड में) है, जिससे ट्रेन की गति के विभिन्न चरणों को समझा जा सकता है।
The metro train travels from station A to station B in 4 minutes = 240 s.
The trains halts at station B for 20 s.
The train travels from station B' to station C in 3 minutes = 180 s.
\( \implies \) Total time taken by the metro train in travelling from station A to B to C
= 240 + 20 + 180 = 440 s.
At start, the train accelerates for 10 seconds to reach a constant speed of 72 km/hr = 20 m/s.
The train moving is brought to rest in 10 s at next station.
The velocity-time graph for the train travelling from station A to B to C is as follows:
Distance travelled by the train from station A to station B
= Area of PQRS
= A (\(\triangle\)PQQ') A (\(\square\)QRR') + A(SRR')
= (\(\frac{1}{2}\) \(\times\) 10 \(\times\) 20) + (220 \(\times\) 20) + (\(\frac{1}{2}\) \(\times\) 10 \(\times\) 20)
= 100 + 4400 + 100
= 4600m = 4.6km
Distance travelled by the train from station B' to station C
= Area of EFGD
= A(\(\triangle\)EFF') + A(\(\square\)F'FGG') + A(\(\triangle\)DGG')
= (\(\frac{1}{2}\) \(\times\) 10 \(\times\) 20) \(\times\) (160 \(\times\) 20) + (\(\frac{1}{2}\) \(\times\) 10 \(\times\) 20)
= 100 + 3200 + 100
= 3400m = 3.4km
In simple words: To calculate the distance covered by the train between stations, we divide the journey into segments of acceleration, constant velocity, and deceleration, and sum the areas under the velocity-time graph for each segment.

🎯 Exam Tip: When given different units for speed (km/hr) and time (seconds/minutes), convert all values to a consistent system (e.g., m/s for speed and seconds for time) before starting calculations to avoid errors.

Question 8. A train is moving eastward at 10 m/sec. A waiter is walking eastward at 1.2m/sec; and a fly is flying toward the north across the waiter's tray at 2 m/s. What is the velocity of the fly relative to Earth.
Answer:
Given
velocity of train w.r.t Earth, \(\vec{V}_{TE}\) = 10\(\hat{i}\)
velocity of waiter w.r.t train, \(\vec{V}_{WT}\) = 1.2\(\hat{i}\)
velocity of fly w.r.t waiter, \(\vec{V}_{FW}\) = 2\(\hat{j}\)
\( \implies \) Velocity of fly with respect to Earth
\(\vec{V}_{FE} = \vec{V}_{FT} - \vec{V}_{ET}\)
= (\(\vec{V}_{FW}\) - \(\vec{V}_{TW}\)) - \(\vec{V}_{ET}\)
= 2\(\hat{j}\)-(-1.2 \(\hat{i}\))-(-10\(\hat{i}\))
= 2\(\hat{j}\) + 11.2\(\hat{i}\)
....(considering north along +y axis)
Magnitude = \(\sqrt{11.2^2 + 2^2}\)
= 11.38 m/s \(\approx\) 11.4 m/s
Direction of velocity,
\(\theta\) = \(\tan^{-1}\) (\(\frac{2}{11.2}\)) \(\approx\) 10° towards north of east.
In simple words: To find the velocity of the fly relative to Earth, we sum the velocity vectors of the train, the waiter relative to the train, and the fly relative to the waiter, ensuring all directions are accounted for correctly.

🎯 Exam Tip: In relative velocity problems, correctly identify all reference frames and express velocities as vectors. Use vector addition to combine velocities and vector subtraction to find relative velocities between moving objects.

Question 9. A car moves in a circle at the constant speed of 50 m/s and completes one revolution in 40 s. Determine the magnitude of acceleration of the car.
Answer:
Given: v = 50 m/s, t = 40 s, s = 2\(\pi\)r
To find: acceleration (a)
Formulae: i) v = \(\frac{s}{t}\)
ii) a = \(\frac{v^2}{r}\)
Calculation: From formula (i),
50 = \(\frac{2\pi r}{40}\)
\( \implies \) r = \(\frac{50 \times 40}{2\pi}\)
\( \implies \) r = \(\frac{1000}{\pi}\) cm
From formula (ii),
a = \(\frac{v^2}{r} = \frac{50^2}{1000/\pi}\)
\( \implies \) a = \(\frac{5\pi}{2}\) = 7.85 m/s\(^2\)
The magnitude of acceleration of the car is 7.85 m/s.
Alternate method:
Given: v = 50 m/s, t = 40 s,
To find: acceleration (a)
Formula: a = r\(\omega^2\) = v\(\omega\)
Calculation: From formula,
a = v\(\omega\)
= v(\(\frac{2\pi}{T}\))
= 50(\(\frac{2 \times 3.142}{40}\))
= \(\frac{5}{2}\) \(\times\) 3.142
\( \implies \) a = 7.85m/s\(^2\)
In simple words: For an object moving in a circle at a constant speed, the centripetal acceleration can be found by dividing the square of its speed by the radius of the circular path, or by multiplying its speed by its angular velocity.

🎯 Exam Tip: Remember that for uniform circular motion, even if the speed is constant, there is always a centripetal acceleration directed towards the center, as the direction of velocity continuously changes. Be prepared to use either \(a = v^2/r\) or \(a = r\omega^2\) based on the given information.

Question 10. A particle moves in a circle with constant speed of 15 m/s. The radius of the circle is 2 m. Determine the centripetal acceleration of the particle.
Answer:
Given: v = 15 m/s, r = 2m
To find: Centripetal acceleration (a)
Formula: a = \(\frac{v^2}{r}\)
Calculation: From formula,
a = \(\frac{(15)^2}{2} = \frac{225}{2}\)
\( \implies \) a = 112.5m/s\(^2\)
The centripetal acceleration of the particle is 112.5 m/s\(^2\).
In simple words: Centripetal acceleration measures how quickly an object's direction changes while moving in a circle, calculated by squaring its speed and dividing by the circle's radius.

🎯 Exam Tip: Centripetal acceleration is always perpendicular to the velocity and points towards the center of the circle. This distinction is crucial for understanding its vector nature, even if only the magnitude is asked.

Question 11. A projectile is thrown at an angle of 30° to the horizontal. What should be the range of initial velocity (u) so that its range will be between 40m and 50 m? Assume g = 10 m s-2.
Answer:
Given: 40 < R \(\leq\) 50, \(\theta\) = 30°, g = 10 m/s\(^2\)
To find: Range of initial velocity (u)
Formula: R = \(\frac{u^2 \sin(2\theta)}{g}\)
Calculation: From formula,
The range of initial velocity,
40 \(\leq\) \(\frac{u^2 \sin (2\theta)}{g}\) \(\leq\) 50
40 \(\leq\) \(\frac{u^2 \sin (2 \times 30)}{10}\) \(\leq\) 50
\( \implies \) 40g \(\leq\) \(u^2\) \(\sin(2\theta)\) \(\leq\) 50g
\(\frac{40g}{\sin(2\theta)}\) \(\leq\) \(u^2\) \(\leq\) \(\frac{50g}{\sin(2\theta)}\)
\( \implies \) \(\sqrt{\frac{40g}{\sin(2\theta)}}\) \(\leq\) u \(\leq\) \(\sqrt{\frac{50g}{\sin(2\theta)}}\)
\( \implies \) \(\sqrt{\frac{40 \times 10}{\sin(60)}}\) \(\leq\) u \(\leq\) \(\sqrt{\frac{50 \times 10}{\sin(60)}}\)
\( \implies \) 21.49m/s \(\leq\) u \(\leq\) 24.03m/s
The range of initial velocity should be between 21.49 m/s \(\leq\) u \(\leq\) 24.03 m/s.
In simple words: To find the range of initial velocity for a projectile to fall within a specific horizontal range, we rearrange the projectile range formula to solve for initial velocity, then substitute the given minimum and maximum range values.

🎯 Exam Tip: Pay close attention to unit consistency and trigonometric functions. Ensure your calculator is in degree mode when calculating \(\sin(2\theta)\). Also, remember to take the square root to find 'u' after solving for \(u^2\).

Can You Recall? (Textbook Page No. 30)

Question 1. What is meant by motion?
Answer: The change in the position of an object with respect to its surroundings is called motion.
In simple words: Motion is simply an object changing its location relative to its environment over time.

🎯 Exam Tip: When defining motion, always emphasize the "change in position" and "with respect to its surroundings" aspects for a complete and accurate answer.

Question 2. What Is rectilinear motion?
Answer: Motion in which an object travels along a straight line is called rectilinear motion.
In simple words: Rectilinear motion is when something moves perfectly straight, without any curves or turns.

🎯 Exam Tip: The key characteristic of rectilinear motion is its straight-line path. Be ready to give examples like a car driving on a straight road or a ball falling directly downwards.

Question 3. What is the difference between displacement and distance travelled?
Answer:
(i) Displacement is the shortest distance between the initial and final points of movement.
(ii) Distance is the actual path followed by a body between the points in which it moves.
In simple words: Distance is the total length of the path an object takes, while displacement is just the straight-line measure from where it started to where it ended, including direction.

🎯 Exam Tip: Understand that distance is a scalar quantity (magnitude only), while displacement is a vector quantity (magnitude and direction). This distinction is fundamental to kinematics problems.

Question 4. What is the difference between uniform and non-uniform motion?
Answer:
(i) A body is said to have uniform motion if it covers equal distances in equal intervals of time.
(ii) A body is said to have non-uniform motion if it covers unequal distances in equal intervals of time.
In simple words: Uniform motion means moving at a steady pace, covering the same distance in the same amount of time, whereas non-uniform motion involves changing speeds, covering different distances in the same time intervals.

🎯 Exam Tip: Uniform motion implies constant velocity (zero acceleration), while non-uniform motion means the velocity is changing (non-zero acceleration). This conceptual clarity is important for numerical problems.