JEE PYQ: Motion in a Plane - Question ID 95b3b5edff5a (JEE Main 2020)

In this problem, we analyze the motion of raindrops relative to a moving car. The key concept is relative velocity. When the car is stationary, the driver observes the raindrops falling vertically downward. However, when the car moves, the driver perceives the raindrops as approaching at an angle due to the combination of the car's horizontal velocity and the raindrops' vertical velocity.

Let:

• $v_r$ = velocity of raindrops relative to the ground (vertical, downward).
• $v$ = speed of the car (horizontal, forward).
• $\theta$ = angle at which the driver sees the raindrops relative to the horizontal.

The relative velocity of the raindrops with respect to the car is: $\vec{v}_{r/c} = \vec{v}_r - \vec{v}_c$ Since $\vec{v}_r$ is vertical and $\vec{v}_c$ is horizontal, the magnitude of the relative velocity is: $|\vec{v}_{r/c}| = \sqrt{v_r^2 + v^2}$ The angle $\theta$ that the relative velocity makes with the horizontal is given by: $\tan \theta = \frac{v_r}{v}$ This is because the vertical component of the relative velocity is $v_r$ (downward), and the horizontal component is $v$ (opposite to the car's motion).

Step 1: Analyze the first scenario (car moving at speed $v$)
When the car moves at speed $v$, the driver sees the raindrops at an angle $60^\circ$ from the horizontal. Using the relative velocity concept: $\tan 60^\circ = \frac{v_r}{v}$ We know $\tan 60^\circ = \sqrt{3}$, so: $\sqrt{3} = \frac{v_r}{v} \implies v_r = v \sqrt{3}$

Step 2: Analyze the second scenario (car moving at speed $(1 + \beta)v$)
When the car's speed increases to $(1 + \beta)v$, the angle changes to $45^\circ$. Again, using the relative velocity concept: $\tan 45^\circ = \frac{v_r}{(1 + \beta)v}$ We know $\tan 45^\circ = 1$, so: $1 = \frac{v_r}{(1 + \beta)v} \implies v_r = (1 + \beta)v$

Step 3: Equate the two expressions for $v_r$
From Step 1 and Step 2, we have: $v \sqrt{3} = (1 + \beta)v$ The $v$ terms cancel out: $\sqrt{3} = 1 + \beta$ Solving for $\beta$: $\beta = \sqrt{3} - 1$

Step 4: Calculate the numerical value of $\beta$
We know $\sqrt{3} \approx 1.732$, so: $\beta = 1.732 - 1 = 0.732$ This value is close to $0.73$, which corresponds to option B.

1. Misidentifying the angle: Students often confuse whether the angle is measured from the vertical or horizontal. Here, the angle is given from the horizontal, so $\tan \theta = \frac{v_r}{v}$ is correct. If misread as from the vertical, the formula would incorrectly become $\tan \theta = \frac{v}{v_r}$.
2. Incorrect relative velocity direction: The relative velocity of the raindrops with respect to the car is $\vec{v}_r - \vec{v}_c$, not $\vec{v}_c - \vec{v}_r$. Mixing up the order leads to the wrong angle calculation.
3. Algebraic errors: When solving for $\beta$, students might incorrectly cancel terms or misapply the tangent function. Always double-check the trigonometric identities and algebraic manipulations.
4. Approximation errors: While $\sqrt{3} \approx 1.732$ is standard, some students might use a rougher approximation (e.g., $1.7$), leading to an incorrect $\beta$ value. Use precise values for such calculations.

Exam Tip: Always draw a velocity vector diagram to visualize the relative motion. Label the directions clearly (e.g., car moving right, raindrops falling down). This helps avoid confusion about the angle and the components of velocity.