Questions and Answers (Subjective)

Q1. State and prove the triangle law of vector addition. Find the magnitude and direction of the resultant of two vectors $\vec{A}$ and $\vec{B}$ in terms of their magnitudes and the angle $\theta$ between them.

Answer: Triangle Law of Vector Addition states that if two vectors can be represented both in magnitude and direction by the two sides of a triangle taken in the same order, then their resultant is represented completely, both in magnitude and direction, by the third side of the triangle taken in the opposite order.

Proof (Analytical Method): Consider two vectors $\vec{A}$ and $\vec{B}$ inclined at an angle $\theta$. Let them be represented by sides OP and PQ of a triangle OPQ. The resultant is $\vec{R} = \vec{A} + \vec{B}$, represented by OQ. Draw a perpendicular from Q to the extended line OP, meeting at N. In right-angled triangle QNP: $PN = B \cos\theta$ and $QN = B \sin\theta$.

In the right-angled triangle ONQ: $OQ^2 = ON^2 + QN^2$ $R^2 = (OP + PN)^2 + QN^2$ $R^2 = (A + B\cos\theta)^2 + (B\sin\theta)^2$ $R^2 = A^2 + B^2\cos^2\theta + 2AB\cos\theta + B^2\sin^2\theta$ $R^2 = A^2 + B^2(\cos^2\theta + \sin^2\theta) + 2AB\cos\theta$ $$R = \sqrt{A^2 + B^2 + 2AB\cos\theta}$$ This is the magnitude of the resultant.

For the direction, let $\alpha$ be the angle that $\vec{R}$ makes with $\vec{A}$. $$\tan\alpha = \frac{QN}{ON} = \frac{B\sin\theta}{A + B\cos\theta}$$

Q2. Derive the general equation of a projectile’s trajectory and prove that it is a parabola.

Answer: Let a particle be projected with an initial velocity $v_0$ at an angle $\theta_0$ with the horizontal x-axis. The components of the initial velocity are $v_{0x} = v_0 \cos\theta_0$ and $v_{0y} = v_0 \sin\theta_0$.

Horizontal Motion (no acceleration): The position at any time t is given by: $x = v_{0x} \cdot t = (v_0 \cos\theta_0)t$ From this, we can express time as: $$t = \frac{x}{v_0\cos\theta_0}$$ (Equation 1)

Vertical Motion (acceleration a_y = -g): The position at any time t is given by: $$y = v_{0y}t - \frac{1}{2}gt^2 = (v_0 \sin\theta_0)t - \frac{1}{2}gt^2$$ (Equation 2)

Deriving the Trajectory: To find the equation of the path, we eliminate the time variable 't' by substituting Equation 1 into Equation 2: $$y = (v_0 \sin\theta_0)\left(\frac{x}{v_0\cos\theta_0}\right) - \frac{1}{2}g\left(\frac{x}{v_0\cos\theta_0}\right)^2$$ Simplifying this expression: $$y = (\tan\theta_0)x - \left(\frac{g}{2v_0^2\cos^2\theta_0}\right)x^2$$ This equation is of the form $y = ax - bx^2$, where $a = \tan\theta_0$ and $b = \frac{g}{2v_0^2\cos^2\theta_0}$ are constants for a given projection. Since y is a quadratic function of x, this equation represents a parabola.

Q3. What is uniform circular motion? Derive the expression for centripetal acceleration.

Answer: Uniform circular motion describes the motion of an object traveling at a constant (uniform) speed on a circular path. While the speed is constant, the velocity is continuously changing because its direction is always changing (it's always tangent to the circle).

Derivation of Centripetal Acceleration: Consider a particle at two points, P and P', on a circle of radius R, with position vectors $\vec{r}$ and $\vec{r'}$, and velocities $\vec{v}$ and $\vec{v'}$. The angle between the position vectors is $\Delta\theta$. Since velocity is always perpendicular to the position vector, the angle between the velocity vectors is also $\Delta\theta$.

The change in velocity is $\Delta\vec{v} = \vec{v'} - \vec{v}$. For a small angle, the magnitude of this change can be approximated by the arc length on the velocity vector diagram: $|\Delta\vec{v}| \approx v \Delta\theta$. The time taken to move from P to P' is $\Delta t = \frac{|\Delta\vec{r}|}{v} = \frac{R\Delta\theta}{v}$.

The magnitude of the acceleration is: $$a = \lim_{\Delta t \to 0} \frac{|\Delta\vec{v}|}{\Delta t} = \lim_{\Delta t \to 0} \frac{v \Delta\theta}{R\Delta\theta/v} = \frac{v^2}{R}$$ The direction of this acceleration is always towards the center of the circle, hence it is called centripetal acceleration ($a_c$).

Q4. Two tall buildings face each other and are at a distance of 180 m from each other. With what velocity must a ball be thrown horizontally from a window 55 m above the ground in one building, so that it enters a window 10.9 m above the ground in the second building? [JEE Main]

Answer:

1. Analyze Vertical Motion: The ball undergoes free fall in the vertical direction. The initial vertical velocity is $v_{0y} = 0$. The vertical displacement is $y = y_f - y_i = 10.9 - 55 = -44.1$ m. The acceleration is $a_y = -g = -9.8\\ m/s^2$. We can find the time of flight using the kinematic equation: $y = v_{0y}t + \frac{1}{2}a_y t^2$. $$-44.1 = 0 \cdot t + \frac{1}{2}(-9.8)t^2$$ $$-44.1 = -4.9t^2$$ $$t^2 = \frac{44.1}{4.9} = 9 \implies t = 3\\ s$$

2. Analyze Horizontal Motion: The ball travels at a constant horizontal velocity, $v_x$, for the duration of its flight. The horizontal distance to cover is $x = 180$ m. The time taken is $t = 3$ s (calculated from the vertical motion).

3. Calculate Horizontal Velocity: $$v_x = \frac{\text{distance}}{\text{time}} = \frac{180\\ m}{3\\ s} = 60\\ m/s$$

   The ball must be thrown horizontally with a velocity of 60 m/s.

Q5. A boat is crossing a river flowing at 4 m/s. The boat’s speed in still water is 5 m/s. Find the shortest time to cross a 100 m wide river. Also find the drift. [JEE Main 2019]

Answer: Shortest time: when boat moves perpendicular to river: $$t = \frac{100}{5} = 20\ s$$

Drift (distance carried by river): $$d = 4 \cdot 20 = 80\ m$$

Q6. A ball is projected horizontally from a height of 45 m with a speed of 10 m/s. How far from the base of the building will the ball strike the ground? [NEET 2020]

Answer: Time to fall: $h = \frac{1}{2}gt^2 \Rightarrow t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \cdot 45}{9.8}} \approx 3.03$ s

Horizontal distance = $u_x \cdot t = 10 \cdot 3.03 = 30.3$ m

Q7. Show that in projectile motion, the path is symmetrical about the peak.

Answer: Time to reach max height = $\frac{u\sin\theta}{g}$; total time of flight = $\frac{2u\sin\theta}{g}$.

The time taken to ascend and descend are equal. Horizontal velocity is constant. Hence, range is equally divided before and after peak ⇒ symmetric path.

Q8. Find the angle of projection at which the horizontal range is maximum for a given initial velocity.

Answer: Range is given by: $$R = \frac{u^2 \sin 2\theta}{g}$$ It is maximum when $\sin 2\theta = 1 \Rightarrow 2\theta = 90^\circ \Rightarrow \theta = 45^\circ$

Q9. Two projectiles are thrown with the same initial speed but different angles: 30° and 60°. Show that they have the same range. [JEE Main 2016]

Answer: Range $R = \frac{u^2 \sin 2\theta}{g}$

For $\theta = 30^\circ$: $\sin 2\theta = \sin 60^\circ$ For $\theta = 60^\circ$: $\sin 2\theta = \sin 120^\circ = \sin 60^\circ$ ⇒ Same range.

Q10. What is relative velocity? Derive expression for relative velocity of two bodies moving at angle to each other.

Answer: Let velocities be $\vec{v}_1$ and $\vec{v}_2$. Then: $$\vec{v}_{rel} = \vec{v}_1 - \vec{v}_2$$

If $\theta$ is the angle between $\vec{v}_1$ and $\vec{v}_2$: $$|\vec{v}_{rel}| = \sqrt{v_1^2 + v_2^2 - 2v_1v_2\cos\theta}$$

Q11. A cyclist moves in a circular track of radius 50 m with constant speed of 10 m/s. Find his acceleration and time to complete one round.

Answer: Centripetal acceleration: $$a = \frac{v^2}{r} = \frac{100}{50} = 2\ m/s^2$$

Time for one round: $$T = \frac{2\pi r}{v} = \frac{2 \cdot 3.14 \cdot 50}{10} = 31.4\ s$$

Q12. Explain vector resolution. Resolve a vector of magnitude 10 N at 60° into its horizontal and vertical components.

Answer: Horizontal component: $$F_x = 10 \cos 60^\circ = 10 \cdot 0.5 = 5\ N$$

Vertical component: $$F_y = 10 \sin 60^\circ = 10 \cdot 0.866 = 8.66\ N$$

Q13. Show that time of flight of a projectile is proportional to initial velocity and sine of angle of projection.

Answer: Time of flight: $$T = \frac{2u \sin\theta}{g}$$

Hence, $T \propto u \sin\theta$

Q14. A plane is flying horizontally with speed 200 m/s at a height of 490 m. It drops a bomb. How far from the release point will the bomb hit the ground? [JEE Advanced 2018]

Answer: Time to fall: $$t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \cdot 490}{9.8}} = 10\ s$$

Horizontal distance: $$x = 200 \cdot 10 = 2000\ m$$

Q15. What is angular displacement and angular velocity? Derive relation between linear and angular velocity.

Answer: Angular displacement $\theta$ is angle swept in radians. Angular velocity $\omega = \frac{d\theta}{dt}$.

Relation: $$v = r \omega$$