A small block is connected to one end of a massless spring of unstretched length 4.9 m. The other en

Q: Why did we use a cosine function instead of a sine function for the block's SHM?

We use $x(t) = A \cos(\omega t)$ because the block is released from rest at the maximum displacement ($t = 0 \implies x = A$). A sine function would imply the block starts at the mean position with maximum velocity, which contradicts the problem statement.

Q: How did we know the pebble travels $5\text{ m}$ horizontally?

The total distance from the origin $O$ to the launch point $P$ is $10\text{ m}$. At $t = 1\text{ s}$, the block is at $5\text{ m}$ from $O$. Since the pebble hits the block at this instant, it must have covered the remaining distance of $10 - 5 = 5\text{ m}$ horizontally.

Q: What if the vertical displacement at $t = 1\text{ s}$ was not zero?

If the vertical displacement wasn't zero, the pebble would either hit the ground before reaching the block or fly over it. The problem is beautifully designed so that the time of flight of the projectile is exactly $1\text{ s}$, ensuring it lands on the surface precisely when it reaches the block.

Q: Does the mass of the block affect the launch velocity of the pebble?

No, the mass of the block is already accounted for in the given angular frequency $\omega = \frac{\pi}{3}\text{ rad/s}$. Since $\omega$ is directly provided, we do not need to calculate it using $k$ and $M$.

Q: Can we solve this using relative motion?

While relative motion is powerful, here the block is accelerating (SHM) while the pebble is in free fall (projectile motion). Solving them in the ground frame by finding their positions at the common time $t = 1\text{ s}$ is much simpler and less prone to errors.

Question

A small block is connected to one end of a massless spring of unstretched length $4.9	ext{ m}$. The other end of the spring (see the figure) is fixed. The system lies on a horizontal frictionless surface. The block is stretched by $0.2	ext{ m}$ and released from rest at $t = 0$. It then executes simple harmonic motion with angular frequency $\omega = \frac{\pi}{3}	ext{ rad/s}$. Simultaneously at $t = 0$, a small pebble is projected with speed $v$ from point $P$ at an angle of $45^\circ$ as shown in the figure. Point $P$ is at a horizontal distance of $10	ext{ m}$ from $O$. If the pebble hits the block at $t = 1	ext{ s}$, the value of $v$ is (Take, $g = 10	ext{ m/s}^2$)

Options:

(A) $\sqrt{50}	ext{ m/s}$

(B) $\sqrt{51}	ext{ m/s}$

(C) $\sqrt{52}	ext{ m/s}$

(D) $\sqrt{53}	ext{ m/s}$

Sigma Percentile · Accepted Answer

### Introduction: The Symphony of Two Motions

Imagine a world where two completely independent physical phenomena—simple harmonic motion and projectile motion—are perfectly synchronized to meet at a single, precise point in space and time.

This is not just a textbook problem; it is a beautiful cosmic dance of a block sliding on a frictionless surface and a pebble soaring through the air.

Let us dive deep into the physics of this system and understand how we can orchestrate this perfect collision.

### Deconstructing the Block's SHM

First, let us focus on the block of mass $M$.

It is attached to a spring of natural length $L_0 = 4.9	ext{ m}$ and is initially pulled by $x_0 = 0.2	ext{ m}$ to the right before being released from rest.

Since the block is released from rest at its maximum displacement, its motion is described by a cosine function:

$$x(t) = A \cos(\omega t)$$

Here, the amplitude $A$ is $0.2	ext{ m}$, and the angular frequency $\omega$ is given as $\frac{\pi}{3}	ext{ rad/s}$.

Now, we want to find where the block is at the exact moment of impact, which is $t = 1	ext{ s}$.

Substituting $t = 1	ext{ s}$ into our equation:

$$x(1) = 0.2 \cos\left(\frac{\pi}{3} 	imes 1ight) = 0.2 	imes 0.5 = 0.1	ext{ m}$$

This means that at $t = 1	ext{ s}$, the block is $0.1	ext{ m}$ to the right of its equilibrium position.

Therefore, the total distance of the block from the origin $O$ is:

$$d = L_0 + x(1) = 4.9 + 0.1 = 5.0	ext{ m}$$

### Tracking the Pebble's Flight

Now, let us turn our attention to the pebble projected from point $P$.

Point $P$ is located at a horizontal distance of $10	ext{ m}$ from the origin $O$.

Since the pebble strikes the block at a distance of $5.0	ext{ m}$ from $O$, the horizontal distance traveled by the pebble in $1	ext{ s}$ must be:

$$s_x = 10 - 5.0 = 5.0	ext{ m}$$

Because there is no horizontal acceleration acting on the pebble, its horizontal motion is completely uniform.

We can express the horizontal distance as:

$$s_x = v_x t = v \cos(45^\circ) t$$

Substituting the known values $s_x = 5.0	ext{ m}$ and $t = 1	ext{ s}$:

$$5.0 = v \cos(45^\circ) 	imes 1$$

$$v \left(\frac{1}{\sqrt{2}}ight) = 5.0 \implies v = 5\sqrt{2}	ext{ m/s}$$

To match the options, we can write this as:

$$v = \sqrt{50}	ext{ m/s}$$

### Verifying the Physics: A Sanity Check

But wait! Does the pebble actually land on the horizontal surface at $t = 1	ext{ s}$?

If the pebble were still high in the air or had already hit the ground earlier, the collision would not occur.

Let us verify the vertical displacement of the pebble at $t = 1	ext{ s}$:

$$y(t) = v \sin(45^\circ) t - \frac{1}{2} g t^2$$

Substituting $v = 5\sqrt{2}	ext{ m/s}$, $	heta = 45^\circ$, and $t = 1	ext{ s}$:

$$y(1) = (5\sqrt{2}) \left(\frac{1}{\sqrt{2}}ight) (1) - \frac{1}{2} (10) (1)^2 = 5 - 5 = 0	ext{ m}$$

This is absolutely brilliant! The vertical displacement is exactly zero, which means the pebble lands perfectly on the horizontal surface at the precise instant it reaches the block.

Thus, the launch speed of the pebble must be **$\sqrt{50}	ext{ m/s}$**, which corresponds to option **(a)**.

Select Answer:

Visualized Solution

Understanding the Physical Setup

Formulating the Block's Motion

Displacement of the Block at $t = 1 s$

Total Distance from Origin

Analyzing the Pebble's Horizontal Displacement

Relating Launch Speed to Horizontal Distance

Calculating the Value of $v$

Verifying the Vertical Position at Impact

The Sigma Insight: Simple Harmonic Motion (SHM)

Introduction

Deconstructing the Block's SHM

Tracking the Pebble's Flight

Verifying the Physics

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