A particle of mass M originally at rest is subjected to a force whose direction is constant but magn

Q: Why do we integrate from $0$ to $2T$?

Because the problem states that the force acts *only* for the time interval $2T$. Before $t=0$ and after $t=2T$, the force is zero, so the velocity doesn't change outside this window.

Q: Can we use the equations of motion like $v = u + at$ here?

No! Those equations are strictly for *constant* acceleration. Here, the force (and thus acceleration) depends on time $t$, so we must use calculus.

Q: How did we integrate $(t-T)^2$?

We used the power rule combined with the chain rule (or substitution). Since the derivative of $(t-T)$ with respect to $t$ is $1$, $\int (t-T)^2 dt = \frac{(t-T)^3}{3}$.

Q: What is the physical meaning of the area under the $F-t$ graph?

The area under a Force-Time graph represents Impulse, which is exactly equal to the change in momentum ($\Delta p$) of the object.

Q: Is there a way to avoid the messy integration?

Yes! The force equation represents a downward-opening parabola. The area of a parabolic segment is $\frac{2}{3} \times \text{base} \times \text{height}$. Using this geometric property gives the impulse directly!

Question

A particle of mass $M$ originally at rest is subjected to a force whose direction is constant but magnitude varies with time according to the relation
$F = F_0 \left[ 1 - \left( \frac{t - T}{T} \right)^2 \right]$
where, $F_0$ and $T$ are constants. The force acts only for the time interval $2T$. The velocity $v$ of the particle after time $2T$ is

Options:

(A) $\frac{2F_0 T}{M}$

(B) $\frac{F_0 T}{2M}$

(C) $\frac{4F_0 T}{3M}$

(D) $\frac{F_0 T}{3M}$

Sigma Percentile · Accepted Answer

Imagine a block of mass $M$ resting peacefully on a smooth, frictionless surface. Suddenly, a force begins to push it. But this isn't your everyday constant force. It's dynamic. It starts at zero, smoothly ramps up to a maximum value of $F_0$ at time $T$, and then gracefully drops back to zero at time $2T$.

Our mission? To find the final velocity of this block after this entire push is over. Let's dive into the physics and mathematics of this beautiful problem.

### Analyzing the Setup

We are given a time-varying force:
$$F(t) = F_0 \left[ 1 - \left( \frac{t - T}{T} ight)^2 ight]$$

How do we connect this force to the velocity of the block? Newton's Second Law is our bridge. We know that force equals mass times acceleration, and acceleration is simply the rate of change of velocity, $\frac{dv}{dt}$.

So, we can write:
$$F = M \frac{dv}{dt}$$

### The Master Equation

By substituting our given force expression into Newton's Second Law, we get a differential equation:
$$M \frac{dv}{dt} = F_0 \left[ 1 - \left( \frac{t - T}{T} ight)^2 ight]$$

To solve this, we need to separate the variables. We keep the velocity terms on the left and move all the time terms to the right. Multiplying both sides by $dt$ and dividing by $M$, we get:
$$dv = \frac{F_0}{M} \left[ 1 - \left( \frac{t - T}{T} ight)^2 ight] dt$$

Now, we are ready to integrate. The particle starts from rest, so at $t = 0$, the velocity $v = 0$. The force acts until time $2T$, and let's call the final velocity $v$. Setting up the definite integral:
$$\int_0^v dv = \int_0^{2T} \frac{F_0}{M} \left[ 1 - \left( \frac{t - T}{T} ight)^2 ight] dt$$

### Executing the Integration

Integrating the left side is straightforward; it simply gives us $v$. For the right side, we integrate term by term. The integral of $1$ is $t$. For the second term, we use the power rule combined with the chain rule. The integral of $x^2$ is $\frac{x^3}{3}$.

Applying this, we get:
$$v = \frac{F_0}{M} \left[ t - \frac{(t - T)^3}{3T^2} ight]_0^{2T}$$

Now comes the critical part: substituting the limits. We must be very careful with the negative signs!

First, substitute the upper limit $2T$:
$$	ext{Upper Limit} = 2T - \frac{(2T - T)^3}{3T^2} = 2T - \frac{T^3}{3T^2} = 2T - \frac{T}{3}$$

Next, substitute the lower limit $0$:
$$	ext{Lower Limit} = 0 - \frac{(0 - T)^3}{3T^2} = -\frac{-T^3}{3T^2} = \frac{T}{3}$$

Subtracting the lower limit from the upper limit:
$$v = \frac{F_0}{M} \left[ \left( 2T - \frac{T}{3} ight) - \left( \frac{T}{3} ight) ight]$$

### Final Calculation

Let's wrap up the algebra. 
$$v = \frac{F_0}{M} \left[ 2T - \frac{2T}{3} ight]$$

Taking the LCM, $6T - 2T$ gives us $4T$. 
$$v = \frac{F_0}{M} \left[ \frac{4T}{3} ight]$$

So, our final velocity is:
**$$v = \frac{4F_0 T}{3M}$$**

### The Pro-Tip: Graphical Method

Is there a way to solve this without the messy integration? Absolutely!

The integral of force over time is simply the area under the Force-Time graph, which represents the **Impulse ($J$)**. Our force equation represents a downward-opening parabola.

A beautiful geometric property states that the area of a parabolic segment is exactly $\frac{2}{3} 	imes 	ext{base} 	imes 	ext{height}$.

Looking at our graph, the base is $2T$ and the maximum height is $F_0$. 
$$J = 	ext{Area} = \frac{2}{3} 	imes (2T) 	imes F_0 = \frac{4}{3} F_0 T$$

Since Impulse equals the change in momentum ($J = Mv - 0$), we can directly write:
$$v = \frac{J}{M} = \frac{4F_0 T}{3M}$$

Boom! We arrive at the exact same answer in just one line. Mastering both the rigorous calculus approach and the elegant graphical shortcut is what makes a true physics champion!

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Visualized Solution

$The Physical Setup$

$Newton’s Second Law$

$Substituting the Force$

$Separating Variables$

$Setting up the Integral$

$Integration$

$Applying Limits$

$Final Velocity$

$Graphical Shortcut$

The Sigma Insight: Newton's Laws of Motion

Analyzing the Setup

The Master Equation

Executing the Integration

Final Calculation

The Pro-Tip

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Physical Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Newton’s Second Law

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting the Force

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Separating Variables

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Setting up the Integral

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Integration

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Applying Limits

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Velocity

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Graphical Shortcut