JEE PYQ: Motion in a Plane - Question ID 5c3f79a5914c (JEE Main 2022)

Core Formula & Concept:

In projectile motion, an object is launched with an initial velocity \(\vec{v}_0 = v_{0x} \hat{i} + v_{0y} \hat{j}\) and moves under the influence of gravity (acting downward). The key concepts and formulas used here are:

• Horizontal motion: Since there is no acceleration in the horizontal direction (ignoring air resistance), the horizontal velocity remains constant: \[ x(t) = v_{0x} t \]
• Vertical motion: The vertical position as a function of time is given by: \[ y(t) = v_{0y} t - \frac{1}{2} g t^2 \] where \(g = 10\ \mathrm{m/s^2}\) is the acceleration due to gravity.
• Trajectory equation: By eliminating time \(t\) from the horizontal and vertical position equations, we get the trajectory \(y\) as a function of \(x\): \[ y = x \tan \theta - \frac{g x^2}{2 v_{0x}^2} \] where \(\theta\) is the launch angle.

In this problem, the initial horizontal velocity is given as a unit vector \(\hat{i}\), so \(v_{0x} = 1\ \mathrm{m/s}\). The trajectory is given as: \[ y = 5x(1 - x) = 5x - 5x^2 \] We are to find the \(y\)-component of the initial velocity, \(v_{0y}\).

Step-by-Step Derivation:

Step 1: Express \(y\) in terms of \(t\)

From horizontal motion: \[ x = v_{0x} t = 1 \cdot t = t \] So, \(t = x\).

Step 2: Write the vertical position as a function of time

The vertical position is: \[ y(t) = v_{0y} t - \frac{1}{2} g t^2 = v_{0y} t - 5 t^2 \]

Step 3: Substitute \(t = x\) into \(y(t)\)

\[ y = v_{0y} x - 5 x^2 \]

Step 4: Compare with the given trajectory

The given trajectory is: \[ y = 5x(1 - x) = 5x - 5x^2 \] Equate the two expressions for \(y\): \[ v_{0y} x - 5 x^2 = 5x - 5x^2 \]

Step 5: Simplify and solve for \(v_{0y}\)

Cancel \(-5x^2\) from both sides: \[ v_{0y} x = 5x \] Divide both sides by \(x\) (assuming \(x \neq 0\)): \[ v_{0y} = 5 \]

Step 6: Conclusion

The \(y\)-component of the initial velocity is \(5\ \mathrm{m/s}\), so the vector is \(5 \hat{j}\).

Common Traps & Exam Tip:

Students often make the following mistakes:

• Misinterpreting the trajectory equation: Some students try to directly compare coefficients without eliminating time, leading to incorrect results. Always express \(y\) in terms of \(x\) via time elimination.
• Ignoring units and vector directions: The problem gives \(v_{0x} = 1\) (unit vector \(\hat{i}\)), but students sometimes assume arbitrary values or confuse the direction of gravity.
• Algebraic errors: When simplifying \(y = 5x(1 - x)\), students may expand it incorrectly as \(y = 5x - x\) instead of \(y = 5x - 5x^2\).

Exam Tip: Always verify the trajectory equation by substituting back the derived initial velocity to ensure consistency.