Physics Solution: Projectile on Incline

Solution to Question 27

Kinematics of a Beetle and Food-Grain on an Incline

Step 1: Geometry and Coordinates

We define a coordinate system aligned with the slope to simplify the equations of motion:

X-axis: Parallel to the slope (positive upwards).
Y-axis: Perpendicular to the slope (positive outwards).

The platform is at a vertical height $h$ above the slope. Converting this to the perpendicular distance ($d$) from the slope:

$$ d = h \cos \theta $$

Step 2: Motion of the Food-Grain (Time $t_1$)

The food-grain is released from the platform. Relative to the slope frame:

Initial velocity: $u_y = 0$, $u_x = u$.
Acceleration: $a_y = -g \cos \theta$, $a_x = -g \sin \theta$.
Displacement: The perpendicular distance to cover is $y = -h \cos \theta$.

Using the equation of motion for the perpendicular direction:

-h \cos \theta = 0 \cdot t_1 – \frac{1}{2} (g \cos \theta) t_1^2 \implies t_1 = \sqrt{\frac{2h}{g}}

Calculating Separation ($S$):
At time $t_1$, the separation along the slope between the platform’s current position and the grain’s landing spot is determined by the relative acceleration (since initial relative velocity in x is 0):

S = x_{platform} – x_{grain} = \frac{1}{2} g \sin \theta t_1^2 $$ Substitute $t_1^2 = \frac{2h}{g}$: $$ S = \frac{1}{2} g \sin \theta \left( \frac{2h}{g} \right) = h \sin \theta

Step 3: Motion of the Beetle (Time $t_2$)

The beetle jumps at $t_1$ with a relative velocity of $3u$ horizontally backwards. We must resolve this vector into our slope coordinate system.

Vector Resolution:
The angle between “Horizontal (Left)” and the “Slope (Up-Right)” is $180^\circ – \theta$.

$v_{rel, x} = 3u \cos(180^\circ – \theta) = -3u \cos \theta$ (Down slope)
$v_{rel, y} = 3u \sin(180^\circ – \theta) = 3u \sin \theta$ (Perpendicular)

Net Initial Velocity of Beetle ($v_b$):
Adding the platform’s velocity ($u$ up the slope):

v_{bx} = u – 3u \cos \theta $$ $$ v_{by} = 3u \sin \theta

Step 4: Meeting Condition

The beetle must catch the grain after time $t_2$.

1. Along the Slope ($x’_g = x’_b$):
The grain slides down from rest (relative to landing spot). The beetle starts at distance $S$ up the slope.

-\frac{1}{2} g \sin \theta t_2^2 = S + (u – 3u \cos \theta) t_2 – \frac{1}{2} g \sin \theta t_2^2 $$ $$ S = (3u \cos \theta – u) t_2

Substituting $S = h \sin \theta$:

t_2 = \frac{h \sin \theta}{u(3 \cos \theta – 1)} \quad \dots(1)

2. Perpendicular to Slope ($y$):
The beetle lands on the slope, covering displacement $-h \cos \theta$.

-h \cos \theta = (3u \sin \theta) t_2 – \frac{1}{2} g \cos \theta t_2^2 $$ Rearranging signs: $$ h \cos \theta = \frac{1}{2} g \cos \theta t_2^2 – 3u \sin \theta t_2

Step 5: Solving for $h$ (Detailed)

We now solve the system of equations step-by-step.

A. Normalize the Equation Divide the vertical motion equation by $h$: $$ \cos \theta = \frac{1}{2} g \cos \theta \frac{t_2^2}{h} – 3u \sin \theta \frac{t_2}{h} $$

B. Substitute Equation (1) Substitute $\frac{t_2}{h} = \frac{\sin \theta}{u(3 \cos \theta – 1)}$: $$ \cos \theta = \frac{1}{2} g \cos \theta \cdot h \left( \frac{\sin \theta}{u(3 \cos \theta – 1)} \right)^2 – 3u \sin \theta \left( \frac{\sin \theta}{u(3 \cos \theta – 1)} \right) $$

C. Rearrange Terms Move the second term to the LHS: $$ \cos \theta + \frac{3 \sin^2 \theta}{3 \cos \theta – 1} = h \cdot \frac{g \cos \theta \sin^2 \theta}{2u^2 (3 \cos \theta – 1)^2} $$

D. Simplify LHS Take the common denominator for the Left Hand Side: $$ \text{LHS} = \frac{\cos \theta(3 \cos \theta – 1) + 3 \sin^2 \theta}{3 \cos \theta – 1} $$ $$ \text{LHS} = \frac{3 \cos^2 \theta – \cos \theta + 3(1 – \cos^2 \theta)}{3 \cos \theta – 1} $$ $$ \text{LHS} = \frac{3 – \cos \theta}{3 \cos \theta – 1} $$

E. Equate and Solve Equate the simplified LHS to the RHS: $$ \frac{3 – \cos \theta}{3 \cos \theta – 1} = h \frac{g \cos \theta \sin^2 \theta}{2u^2 (3 \cos \theta – 1)^2} $$ Cancel $(3 \cos \theta – 1)$ from the denominator on both sides: $$ 3 – \cos \theta = h \frac{g \cos \theta \sin^2 \theta}{2u^2 (3 \cos \theta – 1)} $$

$$ h = \frac{2u^2 (3 \cos \theta – 1)(3 – \cos \theta)}{g \sin^2 \theta \cos \theta} $$