JEE PYQ: Motion in a Plane - Question ID 853d78a2e33e (JEE Main 2021)

This problem tests the fundamental concept of relative motion and reference frames. The key idea is that the trajectory of an object depends on the observer's frame of reference.

When the bomb is dropped from the fighter plane:

• The bomb has an initial horizontal velocity equal to the plane's velocity ($v_{0x} = v_{\text{plane}}$) relative to the ground (inertial frame).
• The bomb is subject to gravitational acceleration ($g$) downward, causing vertical motion.
• In the ground frame, the bomb follows a parabolic trajectory due to the combination of horizontal motion (constant velocity) and vertical motion (accelerated by gravity).
• However, in the plane's frame (non-inertial, moving with the plane), the bomb has no initial horizontal velocity because both the plane and bomb are moving at the same horizontal speed initially. Thus, the bomb appears to fall straight down under gravity.

The relevant kinematic equations in the plane's frame are:

This results in a straight-line vertical trajectory as observed from the plane.

Let’s analyze the motion in two reference frames:

1. Ground Frame (Inertial Frame):
   • At $t = 0$, the plane is at position $(0, h)$ and moving horizontally with velocity $v_0$.
   • The bomb is released with initial velocity $v_{0x} = v_0$ (same as plane) and $v_{0y} = 0$.
   • Horizontal motion: $x(t) = v_0 t$ (constant velocity, no acceleration).
   • Vertical motion: $y(t) = h - \frac{1}{2} g t^2$ (free-fall under gravity).
   • Eliminating $t$: $t = \frac{x}{v_0}$, so $y = h - \frac{g x^2}{2 v_0^2}$ This is a parabola opening downward.
2. Plane's Frame (Non-Inertial Frame):
   • The plane is moving at $v_0$, so the bomb's initial velocity relative to the plane is: $\vec{v}_{\text{bomb, plane}} = \vec{v}_{\text{bomb, ground}} - \vec{v}_{\text{plane, ground}} = (v_0 \hat{i} + 0 \hat{j}) - v_0 \hat{i} = 0$
   • Thus, in the plane's frame, the bomb starts from rest and accelerates downward due to gravity.
   • Horizontal position relative to plane: $x'(t) = x_{\text{bomb}} - x_{\text{plane}} = v_0 t - v_0 t = 0$.
   • Vertical position relative to plane: $y'(t) = y_{\text{bomb}} - y_{\text{plane}} = \left(h - \frac{1}{2} g t^2\right) - h = -\frac{1}{2} g t^2$.
   • The trajectory is given by $x' = 0$ and $y' = -\frac{1}{2} g t^2$, which is a straight line vertically downward.

Thus, to an observer in the plane, the bomb falls straight down without any horizontal displacement.

Students often confuse the reference frames and assume the bomb follows the same trajectory in both the ground and plane frames. Common mistakes include:

• Choosing option B or D: Students may think the bomb follows a parabola in the plane's frame because they forget that the plane and bomb share the same initial horizontal velocity. This leads them to incorrectly apply the ground frame's parabolic trajectory to the plane's frame.
• Ignoring relative motion: Some students overlook the fact that the plane is moving, so they fail to subtract the plane's velocity when analyzing the bomb's motion relative to the plane.
• Misapplying projectile motion: Students may assume that any object dropped from a height follows a parabola, without considering the observer's frame of reference.

Exam Tip: Always clarify the reference frame before analyzing motion. If the observer is moving with the object (or in this case, the plane), the initial relative velocity is zero, and the trajectory simplifies significantly.

The correct answer is C: straight line vertically down the plane.