Concept Card: Projectile Motion on an Inclined Plane & Uniform Circular Motion
Part A: Projectile Motion on an Inclined Plane
1. Coordinate Frame Transformation:
Orient the coordinate axes relative to the incline:
– $+x’$ axis: Directed up along the surface of the incline.
– $+y’$ axis: Directed perpendicular (normal) to the incline.
Let the inclination of the plane with the horizontal be $\beta$, and the angle of projection relative to the incline surface be $\theta$.
Resolving gravitational acceleration $\vec{g}$ along and perpendicular to the incline:
- Normal acceleration: $a_{y’} = -g\cos\beta$
- Parallel acceleration (upward launch): $a_{x’} = -g\sin\beta$
- Initial velocity components: $u_{x’} = u\cos\theta, \quad u_{y’} = u\sin\theta$
2. Time of Flight on the Incline ($T$):
Setting perpendicular displacement $y’ = 0$ upon landing on the incline:
$y’ = u_{y’} T – \frac{1}{2}(g\cos\beta) T^2 = 0 \implies T = \frac{2u_{y’}}{g\cos\beta} = \frac{2u\sin\theta}{g\cos\beta}$
Fundamental Insight: Flight duration depends solely on the component of velocity and acceleration perpendicular to the incline.
3. Maximum Height Above the Incline ($H_{\text{max}}$):
At maximum height above the surface, $v_{y’} = 0$:
$H_{\text{max}} = \frac{u_{y’}^2}{2g\cos\beta} = \frac{u^2\sin^2\theta}{2g\cos\beta}$
4. Range Along the Incline ($R$):
- Up the Incline:
$R_{\text{up}} = u_{x’} T – \frac{1}{2}(g\sin\beta) T^2 = \frac{u^2}{g\cos^2\beta} [\sin(2\theta + \beta) – \sin\beta]$
Maximum Range Condition: Occurs when $\sin(2\theta + \beta) = 1 \implies 2\theta + \beta = 90^\circ \implies \theta = 45^\circ – \frac{\beta}{2}$.
$R_{\text{up, max}} = \frac{u^2}{g(1 + \sin\beta)}$ - Down the Incline:
$R_{\text{down}} = \frac{u^2}{g\cos^2\beta} [\sin(2\theta – \beta) + \sin\beta]$
Maximum Range Condition: Occurs at $\theta = 45^\circ + \frac{\beta}{2}$.
$R_{\text{down, max}} = \frac{u^2}{g(1 – \sin\beta)}$
5. Condition for Perpendicular Landing:
The projectile lands perpendicularly to the incline if its parallel velocity component vanishes at impact ($v_{x’} = 0$):
$v_{x’} = u\cos\theta – (g\sin\beta) T = 0 \implies \cot\theta = 2\tan\beta \iff \tan\theta = \frac{1}{2}\cot\beta$
Part B: Uniform Circular Motion (UCM) & Angular Variables
1. Angular Kinematic Variables:
- Angular displacement: $\Delta\theta = \frac{\Delta s}{r}\text{ (rad)}$
- Angular velocity: $\omega = \frac{d\theta}{dt} = \frac{v}{r}\text{ (rad/s)}$
- Time period and frequency: $T = \frac{2\pi}{\omega} = \frac{2\pi r}{v}, \quad f = \frac{1}{T} = \frac{\omega}{2\pi}$
2. Vector Kinematics & Centripetal Acceleration:
- Position vector: $\vec{r}(t) = r(\cos\omega t\hat{i} + \sin\omega t\hat{j})$
- Velocity vector: $\vec{v}(t) = \vec{\omega} \times \vec{r} = \omega r(-\sin\omega t\hat{i} + \cos\omega t\hat{j})$
Speed $v = \omega r$ is constant, but the direction of $\vec{v}$ rotates continuously. - Centripetal acceleration: $\vec{a}_c = \frac{d\vec{v}}{dt} = -\omega^2\vec{r}$
Magnitude: $a_c = \frac{v^2}{r} = \omega^2 r = v\omega$
Direction: Directed radially inward toward the center ($\vec{a}_c \perp \vec{v} \implies \vec{a}_c \cdot \vec{v} = 0$).
Because $\vec{a}_c \perp d\vec{r}$, the work done by centripetal force is identically zero ($W_c = 0$).
3. Velocity and Acceleration Changes Across Angular Interval $\Delta\theta$:
- Magnitude of change in velocity: $|\Delta\vec{v}| = 2v\sin\left(\frac{\Delta\theta}{2}\right)$
- Average acceleration over interval $\Delta t = \frac{r\Delta\theta}{v}$:
$|\vec{a}_{\text{avg}}| = \frac{|\Delta\vec{v}|}{\Delta t} = \frac{v^2}{r} \left[\frac{\sin(\Delta\theta / 2)}{\Delta\theta / 2}\right]$
Solved Examples
Example 1 (Incline Projectile: Flight Time, Range & Height):
A plane is inclined at $\beta = 30^\circ$ to the horizontal. A particle is projected from the base of the incline with speed $u = 20\text{ m/s}$ at an angle $\theta = 30^\circ$ relative to the inclined surface. Taking $g = 10\text{ m/s}^2$:
(a) Find the time of flight $T$ on the incline.
(b) Find the range $R_{\text{up}}$ along the incline.
(c) Find the maximum height $H_{\text{max}}$ reached above the inclined surface.
Solution:
(a) $T = \frac{2u\sin\theta}{g\cos\beta} = \frac{2(20)\sin(30^\circ)}{10\cos(30^\circ)} = \frac{40(0.5)}{10(\sqrt{3}/2)} = \frac{20}{5\sqrt{3}} = \frac{4}{\sqrt{3}}\text{ s} \approx 2.31\text{ s}$.
(b) $u_{x’} = 20\cos(30^\circ) = 10\sqrt{3}\text{ m/s}, \quad a_{x’} = -g\sin(30^\circ) = -5\text{ m/s}^2$.
$R_{\text{up}} = u_{x’} T + \frac{1}{2}a_{x’} T^2 = 10\sqrt{3}\left(\frac{4}{\sqrt{3}}\right) – \frac{1}{2}(5)\left(\frac{16}{3}\right) = 40 – \frac{40}{3} = \frac{80}{3}\text{ m} \approx 26.67\text{ m}$.
(c) $H_{\text{max}} = \frac{(u\sin\theta)^2}{2g\cos\beta} = \frac{(20 \times 0.5)^2}{2(10)(\sqrt{3}/2)} = \frac{100}{10\sqrt{3}} = \frac{10}{\sqrt{3}}\text{ m} \approx 5.77\text{ m}$.
Example 2 (Perpendicular Impact on an Inclined Plane):
A projectile is fired from the base of an incline of angle $\beta = 30^\circ$. Find the launch angle $\theta$ relative to the incline such that it strikes the incline perpendicularly. Also find the impact speed if $u = 10\text{ m/s}$.
Solution:
(a) For perpendicular impact: $\cot\theta = 2\tan\beta \implies \cot\theta = 2\tan(30^\circ) = \frac{2}{\sqrt{3}} \implies \tan\theta = \frac{\sqrt{3}}{2} \implies \theta = \arctan\left(\frac{\sqrt{3}}{2}\right) \approx 40.89^\circ$.
(b) At perpendicular impact, $v_{x’} = 0$, so the speed is purely normal: $|v_{y’}| = u\sin\theta$.
With $\tan\theta = \frac{\sqrt{3}}{2} \implies \sin\theta = \frac{\sqrt{3}}{\sqrt{(\sqrt{3})^2 + 2^2}} = \sqrt{\frac{3}{7}}$.
Impact speed $v = 10\sqrt{\frac{3}{7}}\text{ m/s} \approx 6.55\text{ m/s}$.
Example 3 (Uniform Circular Motion: Velocity Changes & Average Acceleration):
A particle moves around a circular path of radius $r = 50\text{ m}$ at a constant speed $v = 20\text{ m/s}$.
(a) Find the angular speed $\omega$ and centripetal acceleration $a_c$.
(b) Find the magnitude of change in velocity $|\Delta\vec{v}|$ across a quarter turn ($\Delta\theta = 90^\circ$) and half turn ($\Delta\theta = 180^\circ$).
(c) Find the magnitude of average acceleration during the quarter turn.
Solution:
(a) $\omega = \frac{v}{r} = \frac{20}{50} = 0.4\text{ rad/s}$.
$a_c = \frac{v^2}{r} = \frac{20^2}{50} = \frac{400}{50} = 8.0\text{ m/s}^2$.
(b) For $\Delta\theta = 90^\circ$: $|\Delta\vec{v}| = 2v\sin(45^\circ) = 2(20)\left(\frac{1}{\sqrt{2}}\right) = 20\sqrt{2}\text{ m/s} \approx 28.28\text{ m/s}$.
For $\Delta\theta = 180^\circ$: $|\Delta\vec{v}| = 2v\sin(90^\circ) = 2(20) = 40\text{ m/s}$.
(c) Time for quarter turn: $\Delta t = \frac{\pi r / 2}{v} = \frac{25\pi}{20} = \frac{5\pi}{4}\text{ s}$.
$|\vec{a}_{\text{avg}}| = \frac{|\Delta\vec{v}|}{\Delta t} = \frac{20\sqrt{2}}{5\pi / 4} = \frac{16\sqrt{2}}{\pi}\text{ m/s}^2 \approx 7.20\text{ m/s}^2$.
Example 4 (Ratio of Maximum Downhill to Uphill Range):
A projectile can be launched with speed $u$ at any angle on an incline of angle $\beta$.
(a) Prove that the ratio of maximum downhill range to maximum uphill range is $\frac{1 + \sin\beta}{1 – \sin\beta}$.
(b) If $\beta = 30^\circ$, evaluate this ratio.
Solution:
(a) $R_{\text{down, max}} = \frac{u^2}{g(1 – \sin\beta)}$ and $R_{\text{up, max}} = \frac{u^2}{g(1 + \sin\beta)}$.
$\text{Ratio} = \frac{R_{\text{down, max}}}{R_{\text{up, max}}} = \frac{u^2 / [g(1 – \sin\beta)]}{u^2 / [g(1 + \sin\beta)]} = \frac{1 + \sin\beta}{1 – \sin\beta}$.
(b) For $\beta = 30^\circ$: $\sin(30^\circ) = 0.5$.
$\text{Ratio} = \frac{1 + 0.5}{1 – 0.5} = \frac{1.5}{0.5} = 3$.
The maximum downhill range is three times the maximum uphill range.
Worksheet: 10 Practice Problems
Problem 1 (JEE Main – Single Correct):
A projectile is launched from the base of an incline of angle $\beta$ with speed $u$ at angle $\theta$ to the incline. The time of flight is:
(A) $\frac{2u\sin\theta}{g\cos\beta}$
(B) $\frac{2u\sin\theta}{g\sin\beta}$
(C) $\frac{2u\cos\theta}{g\cos\beta}$
(D) $\frac{2u\sin(\theta + \beta)}{g}$
Problem 2 (JEE Main – Single Correct):
For a given launch speed $u$ up an inclined plane of slope $\beta$, the maximum range is achieved when the launch angle $\theta$ relative to the incline surface is:
(A) $45^\circ$
(B) $45^\circ – \frac{\beta}{2}$
(C) $45^\circ + \frac{\beta}{2}$
(D) $90^\circ – \beta$
Problem 3 (JEE Main – Single Correct):
A body moves in a circular path of radius $r$ with uniform speed $v$. Its centripetal acceleration is:
(A) $\frac{v}{r}$
(B) $\frac{v^2}{r}$
(C) $v \cdot r$
(D) Zero
Problem 4 (JEE Main – Single Correct):
A particle moves in a circle of radius $r$ with constant speed $v$. What is the magnitude of the change in velocity when the particle turns through an angle of $120^\circ$?
(A) $v$
(B) $\sqrt{2}v$
(C) $\sqrt{3}v$
(D) $2v$
Problem 5 (JEE Main – Single Correct):
A particle executes uniform circular motion of radius $r$ with constant speed $v$. The ratio of the magnitude of average acceleration during a quarter revolution to the instantaneous centripetal acceleration is:
(A) $\frac{2\sqrt{2}}{\pi}$
(B) $\frac{\sqrt{2}}{\pi}$
(C) $\frac{2}{\pi}$
(D) $1$
Problem 6 (JEE Advanced – One or More Correct):
A projectile is fired up an inclined plane of slope $\beta$ at angle $\theta$ to the incline. It strikes the inclined plane perpendicularly. Which of the following statements is/are correct?
(A) The condition for perpendicular impact is $\cot\theta = 2\tan\beta$.
(B) The time of flight is $\frac{2u\sin\theta}{g\cos\beta}$.
(C) The speed of impact is $u\sin\theta$.
(D) The horizontal component of velocity in the ground frame is zero at impact.
Problem 7 (JEE Advanced – One or More Correct):
For a particle executing uniform circular motion with radius $r$ and speed $v$:
(A) The acceleration vector is perpendicular to the velocity vector at every instant.
(B) The net work done by the resultant force on the particle in any time interval is zero.
(C) The linear momentum of the particle is constant.
(D) The angular momentum of the particle about the center of the circle is conserved.
Problem 8 (JEE Advanced – One or More Correct):
For projectile motion down an inclined plane of inclination $\beta$ with launch angle $\theta$ relative to the incline:
(A) The effective acceleration parallel to the incline is $g\sin\beta$ downhill.
(B) The maximum range is obtained when $\theta = 45^\circ + \frac{\beta}{2}$.
(C) The maximum range down the incline is $\frac{u^2}{g(1 – \sin\beta)}$.
(D) The time of flight is $\frac{2u\sin\theta}{g\cos\beta}$.
Problem 9 (JEE Main / Advanced – Numerical Value Type):
A particle revolves in a horizontal circle of radius $r = 4.0\text{ m}$ with a constant speed $v = 8.0\text{ m/s}$. The magnitude of its centripetal acceleration in $\text{m/s}^2$ is $a_c$. Find $a_c$.
Problem 10 (JEE Main / Advanced – Numerical Value Type):
A projectile is launched from the base of an incline of slope $\beta = 30^\circ$ with an initial speed $u = 20\text{ m/s}$. Taking $g = 10\text{ m/s}^2$, the maximum range up the incline in meters is $R_{\text{max}}$. Find the value of $R_{\text{max}}$ to the nearest integer.
Solutions & Explanations
Answer Key Summary:
1. (A) | 2. (B) | 3. (B) | 4. (C) | 5. (A) | 6. (A, B, C) | 7. (A, B, D) | 8. (A, B, C, D) | 9. 16 | 10. 27
Solution 1:
Perpendicular displacement upon landing: $y’ = u_{y’} T – \frac{1}{2}a_{y’} T^2 = 0 \implies u\sin\theta \cdot T – \frac{1}{2}(g\cos\beta) T^2 = 0 \implies T = \frac{2u\sin\theta}{g\cos\beta}$.
Correct Answer: (A)
Solution 2:
$R_{\text{up}} = \frac{u^2}{g\cos^2\beta} [\sin(2\theta + \beta) – \sin\beta]$. Maximize by setting $\sin(2\theta + \beta) = 1 \implies 2\theta + \beta = 90^\circ \implies \theta = 45^\circ – \frac{\beta}{2}$.
Correct Answer: (B)
Solution 3:
Centripetal acceleration magnitude: $a_c = \frac{v^2}{r} = \omega^2 r$.
Correct Answer: (B)
Solution 4:
$|\Delta\vec{v}| = 2v\sin\left(\frac{\Delta\theta}{2}\right) = 2v\sin(60^\circ) = 2v\left(\frac{\sqrt{3}}{2}\right) = \sqrt{3}v$.
Correct Answer: (C)
Solution 5:
For a quarter circle ($\Delta\theta = \pi/2$): $|\Delta\vec{v}| = \sqrt{2}v$.
$\Delta t = \frac{\pi r / 2}{v} = \frac{\pi r}{2v}$.
$|\vec{a}_{\text{avg}}| = \frac{\sqrt{2}v}{\pi r / (2v)} = \frac{2\sqrt{2}}{\pi} \frac{v^2}{r} = \frac{2\sqrt{2}}{\pi} a_c$. The ratio is $\frac{2\sqrt{2}}{\pi}$.
Correct Answer: (A)
Solution 6:
– (A) Condition $v_{x’} = 0 \implies u\cos\theta – g\sin\beta T = 0 \implies \cot\theta = 2\tan\beta$ (True).
– (B) $T = \frac{2u\sin\theta}{g\cos\beta}$ (True).
– (C) Impact speed is purely normal: $|v_{y’}| = u\sin\theta$ (True).
– (D) Parallel component to the incline is zero, not the horizontal component in ground frame (False).
Correct Answer: (A, B, C)
Solution 7:
– (A) $\vec{a}_c$ is radial and $\vec{v}$ is tangential, so $\vec{a}_c \perp \vec{v}$ (True).
– (B) Since $\vec{F}_c \perp d\vec{r}$, $W = 0$ (True).
– (C) Velocity vector rotates, so linear momentum $\vec{p} = m\vec{v}$ varies (False).
– (D) $\vec{\tau}_{\text{center}} = \vec{r} \times \vec{F}_c = \vec{0}$, so angular momentum about center is conserved (True).
Correct Answer: (A, B, D)
Solution 8:
All statements (A, B, C, D) are established results for projectile motion down an inclined plane.
Correct Answer: (A, B, C, D)
Solution 9:
$a_c = \frac{v^2}{r} = \frac{8.0^2}{4.0} = \frac{64}{4} = 16\text{ m/s}^2$.
Correct Answer: 16
Solution 10:>
Correct Answer: 16
Solution 10:
$R_{\text{up, max}} = \frac{u^2}{g(1 + \sin\beta)} = \frac{20^2}{10(1 + \sin 30^\circ)} = \frac{400}{10(1.5)} = \frac{400}{15} = \frac{80}{3} \approx 26.67\text{ m}$. To the nearest integer: $27$.
Correct Answer: 27