Numerical/Problem-solving

1. TITLE

How to solve Area under the F-t Graph numericals in NEET — Step-by-step

2. PROBLEM TYPE INTRO

This numerical falls under the "Impulse and Momentum" category of Dynamics. You will need the Impulse-Momentum Theorem, which states that the area under a Force-time (F-t) graph equals the change in momentum (Impulse = $\Delta p$). A typical mistake students make is forgetting to convert time units (like milliseconds to seconds) or equating the area directly to final momentum instead of the change in momentum.

3. MCQ

Question:
A particle of mass $1\text{ kg}$ is initially moving with a velocity of $2\text{ m/s}$ along the positive x-axis. A variable force acts on the particle along the direction of motion. The Force-time (F-t) graph of this force forms a triangle starting from $t = 0$ to $t = 200\text{ ms}$, with a peak force of $50\text{ N}$ at $t = 100\text{ ms}$. What is the final velocity of the particle at $t = 200\text{ ms}$?

(A) $7\text{ m/s}$
(B) $12\text{ m/s}$
(C) $-3\text{ m/s}$
(D) $5002\text{ m/s}$

4. STEP-BY-STEP SOLUTION

SOLUTION
Given:
* Mass ($m$) = $1\text{ kg}$
* Initial velocity ($u$) = $2\text{ m/s}$
* Peak Force ($F_{max}$ or height) = $50\text{ N}$
* Time duration ($\Delta t$ or base) = $200\text{ ms} = 0.2\text{ s}$

To Find:
* Final velocity ($v$) of the particle at $t = 200\text{ ms}$.

Formula:
* $\text{Impulse} = \text{Area under F-t graph} = \Delta p$
Use when a variable force acts over a time interval and you need to find the change in momentum.
* $\Delta p = m(v - u)$
Use when linking the change in momentum to the initial and final velocities.

Step 1: Calculate the area under the F-t graph to find Impulse.
Since the graph is a triangle, $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$
$\text{Impulse} = \frac{1}{2} \times 0.2\text{ s} \times 50\text{ N}$
$\text{Impulse} = 5\text{ N s}$ (or $5\text{ kg m/s}$)

Step 2: Relate the Impulse to the change in momentum.
$\text{Impulse} = p_{final} - p_{initial}$
$5 = m \cdot v - m \cdot u$

Step 3: Substitute the known values and solve for final velocity.
$5 = (1\text{ kg})(v) - (1\text{ kg})(2\text{ m/s})$
$5 = v - 2$
$v = 5 + 2 = 7\text{ m/s}$

Answer: $7\text{ m/s}$ (Option A)

5. WHERE STUDENTS GO WRONG

  • ❌ Mistake 1: Forgetting to convert ms to s. If you leave the base as 200, the area becomes $5000\text{ N s}$, leading to a massive and incorrect final velocity of $5002\text{ m/s}$ (Option D).
  • ❌ Mistake 2: Missing the $\frac{1}{2}$ in the triangle area formula. Calculating area as base × height gives $10\text{ N s}$, resulting in an incorrect final velocity of $12\text{ m/s}$ (Option B).
  • ❌ Mistake 3: Subtracting instead of adding. Treating the impulse as a retarding force by mistake leads to $v = 2 - 5 = -3\text{ m/s}$ (Option C).

6. QUICK FORMULA REFERENCE

Formulas Used:
* Formula 1 — $\text{Impulse} = \text{Area under F-t curve}$ — Use when determining the total effect of a variable force plotted against time.
* Formula 2 — $\text{Impulse} = m(v - u)$ — Use when you need to calculate the final or initial velocity of an object after an impulse is applied.

7. SIMILAR PROBLEM HINT

If NEET changes the values, the method stays the same. Practice with: A rectangular F-t graph where the force drops below the x-axis, representing a retarding force (negative area) that reduces the initial momentum.

8. NCERT REFERENCE

Based on: NCERT Class 11, Chapter 4, Section 4.5 (Newton's Second Law of Motion & Impulse)

9. CTA

More solved numericals → Laws of Motion Complete Guide
Practice problems → Mock exams