Animated Solution for Physics - Laws of Motion: A solid horizontal surface is covered with a thin layer of oil. A rectangular block of mass $m=0.4\mathrm{kg}$ is at rest on this surface. An impulse of $1.0\mathrm{Ns}$ is applied to the block at time $t=0$, so that it starts moving along the $X$-axis with a velocity $v(t)=v_0{e}^{-\frac{t}{\tau }}$, where $v_0$ is a constant and $\tau =4\mathrm{s}$. The displacement of the block, in metres, at $t=\tau$ is .............. . (Take, $e^{-1}=0.37$).

Have you ever tried sliding a heavy box across a floor covered in spilled oil? It doesn't just stop abruptly like it would on dry concrete, nor does it glide forever like it would on perfect ice. Instead, it smoothly and continuously slows down, its speed fading away like a dying echo. This beautiful, continuous fading is the hallmark of viscous drag, and it is exactly what we are going to explore in this thrilling physics problem!

In this journey, we will uncover the mathematical elegance behind a block that is kicked into motion and then left to the mercy of an oily surface. We will bridge the gap between a sudden, violent impulse and the smooth, exponential decay of velocity that follows. Grab your mental toolkit, because we are about to dive deep into the mechanics of motion!

Awakening the Block

Our story begins with a rectangular block of mass $m=0.4\mathrm{kg}$ resting peacefully on a horizontal surface. But this peace is shattered at time $t=0$ when a sudden impulse $J=1.0\mathrm{Ns}$ is applied to it.

What exactly is an impulse? In physics, an impulse is a massive force applied over a tiny fraction of a second—think of a golf club striking a ball or a bat hitting a puck. Mathematically, the Impulse-Momentum Theorem tells us that an impulse delivers a direct, instantaneous change to an object's momentum:

Since our block was initially at rest, its initial momentum was zero. The impulse entirely transforms into the block's new starting momentum:

We can easily rearrange this to find the block's initial velocity, $v_0$, right after the kick:

Let's plug in the numbers provided in the problem:

Boom! The block is now sliding along the X-axis at $2.5m/s$. If the surface were frictionless, it would maintain this speed forever. But remember that thin layer of oil? It's about to make things very interesting.

An Exponential Fading

As the block slides, the oil exerts a viscous drag force on it. Unlike solid kinetic friction, which is usually constant, viscous drag is proportional to the object's velocity. The faster the block moves, the harder the oil pushes back. As the block slows down, the drag force weakens.

This feedback loop—where the rate of slowing down depends on the current speed—always leads to an exponential relationship in nature. The problem graciously provides us with the exact equation for this decaying velocity:

Here, $\tau$ (tau) is a constant called the "time constant," given as $4\mathrm{s}$. It dictates how quickly the velocity fades. A smaller $\tau$ would mean the oil is very thick and stops the block quickly, while a larger $\tau$ would mean the oil is very thin.

From Velocity to Displacement

Our ultimate goal is to find the total displacement of the block at a specific time, $t=\tau$. We have the velocity as a function of time, but how do we get to displacement?

We must call upon the fundamental definition of velocity in kinematics. Velocity is the rate of change of position with respect to time:

By equating our two expressions for velocity, we set the stage for our mathematical heavy lifting:

To solve this differential equation, we use a technique called "separation of variables." We want all the $x$ terms on one side and all the $t$ terms on the other. Multiplying both sides by $dt$ gives us:

Now, we are ready to integrate! We want to find the displacement $x$ from the starting moment $t=0$ to the target moment $t=\tau$. We set up our definite integrals accordingly:

Crunching the Calculus

The left side of the equation is trivial; the integral of $dx$ is simply $x$. The right side requires a bit more care. We need to integrate an exponential function.

Remember the standard calculus rule: $\int e^{kt}dt=\frac{e^{kt}}{k}$. In our equation, the constant $k$ in the exponent is $-\frac{1}{\tau }$. Therefore, when we integrate, we must divide by $-\frac{1}{\tau }$, which is mathematically identical to multiplying by $-\tau$.

Let's perform the integration:

Pulling the constants out front makes it cleaner:

Now, we apply the Fundamental Theorem of Calculus by substituting our upper limit ($\tau$) and subtracting the expression with our lower limit ($0$):

Simplify the exponents:

Since any non-zero number to the power of zero is $1$, we have:

To get rid of that pesky negative sign in the front, we can distribute it through the parentheses, flipping the terms inside:

Plugging in the Numbers

We have successfully derived the master equation for the displacement! All that remains is to substitute the numerical values we gathered along the way.

We know: - $v_0=2.5m/s$ - $\tau =4\mathrm{s}$ - $e^{-1}=0.37$ (as generously provided by the problem statement)

Let's plug them in:

First, multiply the constants outside the parentheses:

Next, perform the subtraction inside the parentheses:

And finally, multiply by $10$:

Take a moment to appreciate what we just did. We took a physical scenario involving a sudden impact and a complex, continuously changing drag force, translated it into a differential equation, solved it using calculus, and arrived at a clean, precise numerical answer. This is the true power and beauty of physics!