On a frictionless horizontal surface, assumed to be the x-y plane, a small trolley A is moving along

Q: Why must the relative velocity be exactly along the line OA?

Imagine you are the trolley. In your frame of reference, you are completely stationary. For a ball to hit a stationary object, it must travel in a straight line directly towards it. That straight line connecting the ball's starting point to you is the line of sight, $OA$.

Q: Why can't we just equate the positions of the ball and the trolley at some time $t$?

You absolutely can! If you write the kinematic equations $x_B(t) = x_A(t)$ and $y_B(t) = y_A(t)$, you will get the exact same result. However, using the relative velocity vector is a powerful shortcut that bypasses the need to introduce and solve for a time variable $t$.

Q: What if the line OA was at 30 degrees instead of 45 degrees?

Then the relative velocity vector would need to be at $30^\circ$. You would set $\tan 30^\circ = \frac{v_{BA,y}}{v_{BA,x}}$, which means $\frac{1}{\sqrt{3}} = \frac{v_{BA,y}}{v_{BA,x}}$. The core logic remains exactly the same!

Q: Why did we subtract $\mathbf{v}_A$ from $\mathbf{v}_B$ and not the other way around?

We needed the velocity of the ball *with respect to* the trolley, denoted as $\mathbf{v}_{BA}$. By definition, $\mathbf{v}_{BA} = \mathbf{v}_B - \mathbf{v}_A$. If we did $\mathbf{v}_A - \mathbf{v}_B$, we would be finding the velocity of the trolley as seen by the ball, which would point in the exact opposite direction.

Q: Is it possible for the ball to never hit the trolley, regardless of its speed $v$?

Yes! If the ball is thrown at an angle $\phi$ that is too small or too large, the required relative velocity angle might be impossible to achieve. The math would yield a negative or undefined speed $v$, indicating that a collision is physically impossible under those conditions.

Question

On a frictionless horizontal surface, assumed to be the $x$-$y$ plane, a small trolley $A$ is moving along a straight line parallel to the $y$-axis (see figure) with a constant velocity of $(\sqrt{3}-1)	ext{ m/s}$. At a particular instant when the line $OA$ makes an angle of $45^\circ$ with the $x$-axis, a ball is thrown along the surface from the origin $O$. Its velocity makes an angle $\phi$ with the $x$-axis and it hits the trolley. (2002)
(a) The motion of the ball is observed from the frame of the trolley. Calculate the angle $	heta$ made by the velocity vector of the ball with the $x$-axis in this frame.
(b) Find the speed of the ball with respect to the surface, if $\phi = 4	heta/3$.

Sigma Percentile · Accepted Answer

### The Thrill of the Intercept

Intercepting a moving target is one of the most thrilling and classic problems in physics. Whether it is a missile defense system calculating the trajectory to intercept an incoming threat, or a quarterback throwing a football to a running receiver, the underlying physics is identical. In this problem, we are tasked with throwing a ball across a frictionless surface to perfectly strike a moving trolley.

Solving this problem in the ground frame can be a nightmare. You have two objects moving simultaneously, and you have to find a specific time $t$ where their $x$ and $y$ coordinates perfectly match. But physics offers us a beautiful shortcut: **Relative Motion**.

### Shifting Perspectives: The Magic of Relative Motion

Instead of watching the chaos from the ground, imagine you are sitting on the moving trolley. From your perspective, you are completely stationary. The entire world, including the ground, is moving backwards.

Now, if you are stationary, what must the ball do to hit you? It must travel in a perfectly straight line directly towards you. This straight line is called the **Line of Sight**. By shifting our frame of reference to the trolley, we have reduced a complex interception problem into a simple geometry problem.

### Decoding the Geometry: The Line of Sight

At the instant the ball is thrown, the line connecting the origin (where the ball is) to the trolley makes an angle of $45^\circ$ with the $x$-axis. This is our line of sight, $OA$.

For the ball to travel along this line in the trolley's frame, its relative velocity vector, $\mathbf{v}_{BA}$, must also point exactly at $45^\circ$. This immediately gives us the answer to the first part of the problem: the angle $	heta$ made by the velocity vector of the ball in the trolley's frame is simply $45^\circ$.

### Translating to the Ground Frame

Now that we know $	heta = 45^\circ$, we can use the mathematical relationship provided in the problem to find the launch angle in the ground frame, denoted as $\phi$.

The problem states that $\phi = \frac{4	heta}{3}$. Substituting our known value:

$$ \phi = \frac{4}{3} 	imes 45^\circ = 60^\circ $$

This tells us that the ball must be thrown at a $60^\circ$ angle relative to the $x$-axis from the ground.

### The Vector Arsenal

To proceed, we need to express the velocities of both the trolley and the ball as mathematical vectors.

The trolley $A$ is moving purely parallel to the $y$-axis with a given speed. Therefore, its velocity vector is:

$$ \mathbf{v}_A = (\sqrt{3}-1)\hat{j} $$

The ball $B$ is thrown with an unknown speed $v$ at an angle of $60^\circ$. We can resolve this into its $x$ and $y$ components using basic trigonometry:

$$ \mathbf{v}_B = (v \cos 60^\circ)\hat{i} + (v \sin 60^\circ)\hat{j} $$

Substituting the values of $\cos 60^\circ = \frac{1}{2}$ and $\sin 60^\circ = \frac{\sqrt{3}}{2}$, we get:

$$ \mathbf{v}_B = \frac{v}{2}\hat{i} + \frac{\sqrt{3}v}{2}\hat{j} $$

### The Master Equation

The core of relative motion lies in the relative velocity equation. The velocity of the ball with respect to the trolley is the vector difference of their absolute velocities:

$$ \mathbf{v}_{BA} = \mathbf{v}_B - \mathbf{v}_A $$

Substituting our vector expressions:

$$ \mathbf{v}_{BA} = \left(\frac{v}{2}\hat{i} + \frac{\sqrt{3}v}{2}\hat{j}ight) - (\sqrt{3}-1)\hat{j} $$

Grouping the $\hat{i}$ and $\hat{j}$ components carefully, we obtain the master vector for our relative velocity:

$$ \mathbf{v}_{BA} = \frac{v}{2}\hat{i} + \left(\frac{\sqrt{3}v}{2} - \sqrt{3} + 1ight)\hat{j} $$

### The Final Strike

We established earlier that this relative velocity vector $\mathbf{v}_{BA}$ must point at an angle of $45^\circ$. The tangent of a vector's angle is the ratio of its $y$-component to its $x$-component.

Since $	an 45^\circ = 1$, it dictates a beautiful physical constraint: **the $x$ and $y$ components of the relative velocity must be perfectly equal**.

Let's equate them:

$$ \frac{v}{2} = \frac{\sqrt{3}v}{2} - \sqrt{3} + 1 $$

Now, we just need to solve this linear equation for $v$. Let's bring all the constant terms to the left side and the terms containing $v$ to the right side:

$$ \sqrt{3} - 1 = \frac{\sqrt{3}v}{2} - \frac{v}{2} $$

Factoring out $\frac{v}{2}$ on the right side reveals a stunning algebraic symmetry:

$$ \sqrt{3} - 1 = \frac{v}{2}(\sqrt{3} - 1) $$

Since $(\sqrt{3} - 1)$ is a non-zero constant, we can safely divide both sides by it. The terms cancel out perfectly, leaving us with:

$$ 1 = \frac{v}{2} $$

Multiplying by 2, we arrive at our final, elegant answer:

$$ v = 2 	ext{ m/s} $$

By shifting our perspective and trusting the geometry of relative motion, a complex 2D interception problem collapses into a simple, satisfying algebraic cancellation.

Visualized Solution

$Visualizing the Setup$

$Condition for Collision$

$Finding θ$

$Finding ϕ$

$Velocity Vectors$

$Relative Velocity v_{B A}$

$Applying the Direction Constraint$

$Equating Components$

$Solving for v$

$Final Calculation$

The Sigma Insight: Relative Velocity

The Thrill of the Intercept

Shifting Perspectives

Decoding the Geometry

Translating to the Ground Frame

The Vector Arsenal

The Master Equation

The Final Strike

Similar Questions

List I describes four systems, each with two particles A and B in relative motion as shown in figure. List II gives possible magnitudes of their relative velocities (in $ms^{- 1}$ ) at time $t = \frac{π}{3} s$ .

List-I

List-II

Comprehension Passage

When and for how long can the particle C fulfil the given condition?

What is speed of the particle C at the instant $t = 5 s$ ?

What is modulus of acceleration of the particle C at the instant $t = 5 s$ ?

At the instant, when the line joining locations of A and B is perpendicular to the line joining locations of B and C, what are the magnitudes of velocities of C relative to A and B respectively?

Visualized Solution

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

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

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

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

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

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

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

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

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

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