A cart is moving along x-direction with a velocity of 4 m/s. A person on the cart throws a stone wit

Q: Why doesn't the cart's velocity affect the $y$ and $z$ components of the stone?

Velocity is a vector, and its components in orthogonal directions ($x$, $y$, $z$) are completely independent of each other. Since the cart only moves along the $x$-axis, it only adds to the stone's $x$-velocity.

Q: Why is the vertical velocity zero at the highest point?

Gravity constantly accelerates the stone downwards. As it rises, its vertical velocity decreases until it momentarily stops rising (velocity becomes zero) before it starts falling back down. This is the definition of the highest point.

Q: Why do we use conservation of momentum and not energy during the collision?

In a completely inelastic collision, kinetic energy is lost to heat and deformation (the stone embedding itself). However, because there are no external horizontal forces during the split-second impact, linear momentum is strictly conserved.

Q: Why does tension becoming zero mean velocity is zero at the horizontal position?

At the horizontal position, gravity acts purely downwards (tangential to the circular path), so it doesn't contribute to tension. The only thing providing tension would be the centripetal force $\frac{mv^2}{L}$. If tension is zero, then $\frac{mv^2}{L}$ must be zero, which means $v = 0$.

Q: Does the mass of the stone matter for the final length $L$?

Surprisingly, no! When we equate kinetic energy to potential energy ($\frac{1}{2}(2m)v_0^2 = (2m)gL$), the mass term $(2m)$ cancels out completely. The height a pendulum swings depends only on its initial speed and gravity.

Question

A cart is moving along $x$-direction with a velocity of $4	ext{ m/s}$. A person on the cart throws a stone with a velocity of $6	ext{ m/s}$ relative to himself. In the frame of reference of the cart, the stone is thrown in $y$-$z$ plane making an angle of $30^\circ$ with vertical $z$-axis. At the highest point of its trajectory, the stone hits an object of equal mass hung vertically from branch of a tree by means of a string of length $L$.
A completely inelastic collision occurs, in which the stone gets embedded in the object. Determine ($g = 9.8	ext{ m/s}^2$)
(a) the speed of the combined mass immediately after the collision with respect to an observer on the ground.
(b) the length $L$ of the string such that tension in the string becomes zero when the string becomes horizontal during the subsequent motion of the combined mass.

Sigma Percentile · Accepted Answer

Imagine standing on the ground, watching a cart roll by at a steady pace. Suddenly, a person on the cart throws a stone. But they don't just throw it forward; they throw it at an angle, into the $y$-$z$ plane. This creates a beautiful, complex 3D trajectory. Our goal is to track this stone, find out what happens when it crashes into a target at its highest point, and then analyze the swinging motion that follows.

### Setting the Stage: The Moving Cart and the 3D Throw

Let's break this down systematically. We start by defining our coordinate system. The cart is moving purely along the $x$-axis. So, its velocity is simply $\mathbf{v}_{	ext{cart}} = 4\hat{i} 	ext{ m/s}$.

Now, consider the stone. The person on the cart throws it in the $y$-$z$ plane, making a $30^\circ$ angle with the vertical $z$-axis. This velocity is **relative** to the cart. Using basic trigonometry, we can resolve this relative velocity into its components. The vertical component is adjacent to the angle, so it uses cosine, while the horizontal $y$-component uses sine.

$$\mathbf{v}_{	ext{rel}} = 6 \sin 30^\circ \hat{j} + 6 \cos 30^\circ \hat{k}$$

Since $\sin 30^\circ = 0.5$ and $\cos 30^\circ = \frac{\sqrt{3}}{2}$, this simplifies to:

$$\mathbf{v}_{	ext{rel}} = 3\hat{j} + 3\sqrt{3}\hat{k} 	ext{ m/s}$$

### The Absolute Reality: Vector Addition

To understand the stone's actual path through the air as seen by an observer on the ground, we need its absolute velocity. This is a classic application of relative motion. The absolute velocity of the stone is the vector sum of its velocity relative to the cart and the cart's velocity.

$$\mathbf{v}_{	ext{stone}} = \mathbf{v}_{	ext{cart}} + \mathbf{v}_{	ext{rel}}$$

Substituting our vectors, we get a full 3D velocity vector:

$$\mathbf{v}_{	ext{stone}} = 4\hat{i} + 3\hat{j} + 3\sqrt{3}\hat{k} 	ext{ m/s}$$

### The Peak of the Arc: 2D Motion

The stone is now a projectile in 3D space. Gravity acts purely in the negative $z$-direction, constantly pulling it down. This means gravity only affects the $z$-component of the velocity. The $x$ and $y$ components remain completely unaffected throughout the flight.

At the very highest point of its trajectory, the stone stops rising. For a brief, infinitesimal moment, its vertical velocity becomes zero. However, it is still moving horizontally! The velocity at the highest point is simply the initial velocity minus the $z$-component.

$$\mathbf{v} = 4\hat{i} + 3\hat{j} 	ext{ m/s}$$

To find the speed at this point, we calculate the magnitude of this 2D vector using the Pythagorean theorem:

$$v = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = 5 	ext{ m/s}$$

### The Inelastic Impact: Momentum Conservation

Right at this peak, the stone encounters an object of equal mass hanging from a tree. The problem states that a completely inelastic collision occurs, meaning the stone gets embedded in the object. They become a single, combined mass of $2m$.

Because the collision happens in mid-air and there are no external horizontal forces acting on the system during the brief moment of impact, linear momentum is strictly conserved.

$$m \mathbf{v} = (m + m) \mathbf{v}_0$$

Here, $\mathbf{v}_0$ is the velocity of the combined mass immediately after the collision. Since the masses are equal, the equation simplifies beautifully:

$$m(5) = 2m(v_0)$$

$$v_0 = \frac{5}{2} = 2.5 	ext{ m/s}$$

This is the answer to the first part of our problem. The combined mass moves off at $2.5 	ext{ m/s}$.

### The Pendulum's Swing: Energy Conservation

Now the scenario shifts. The combined mass is attached to a string of length $L$, turning it into a pendulum. With its newly acquired kinetic energy, it begins to swing upwards.

The problem gives us a crucial clue: the tension in the string becomes zero exactly when the string becomes horizontal. For a pendulum, the tension $T$ at the horizontal position is provided entirely by the centripetal force required to keep it moving in a circle. If $T = 0$, it means the velocity at that exact moment must also be zero. The pendulum has reached its maximum height.

We can now apply the principle of conservation of mechanical energy. The kinetic energy at the bottom of the swing is completely converted into gravitational potential energy at the horizontal position.

$$K_i + U_i = K_f + U_f$$

$$\frac{1}{2}(2m)v_0^2 + 0 = 0 + (2m)gL$$

Notice how the mass $(2m)$ cancels out from both sides. This is a beautiful feature of gravity; the mass doesn't matter!

$$v_0^2 = 2gL$$

Now, we simply rearrange the equation to solve for the length of the string, $L$:

$$L = \frac{v_0^2}{2g}$$

Substituting the values we found ($v_0 = 2.5 	ext{ m/s}$ and $g = 9.8 	ext{ m/s}^2$):

$$L = \frac{(2.5)^2}{2 	imes 9.8} = \frac{6.25}{19.6}$$

$$L \approx 0.3188 	ext{ m}$$

Rounding to two decimal places, we get **$L = 0.32 	ext{ m}$**. And with that, we have completely unraveled the mystery of the 3D throw and the swinging pendulum!

Visualized Solution

Coordinate System & Cart Velocity

Relative Velocity of the Stone

Absolute Velocity of the Stone

Velocity at the Highest Point

Inelastic Collision

Speed of the Combined Mass

Motion of the Pendulum

Conservation of Mechanical Energy

Calculating the Length of the String

The Sigma Insight: Conservation of Linear Momentum

Setting the Stage

The Absolute Reality

The Peak of the Arc: 2D Motion

The Inelastic Impact

The Pendulum's Swing

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