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

Q: Wait, why is the absolute velocity the sum of the cart's velocity and the relative velocity?

According to Galilean relativity, the velocity of an object with respect to the ground is the vector sum of its velocity relative to a moving frame (the cart) and the velocity of that frame itself. It's like walking on a moving walkway; your total speed is your walking speed plus the walkway's speed.

Q: Why does the $z$-component of velocity become zero at the highest point?

Gravity acts downwards (in the negative $z$-direction), constantly decelerating the vertical motion. At the peak of the trajectory, the vertical velocity is momentarily reduced to zero before the object starts falling back down. It's the turning point of the vertical motion.

Q: Why do the $x$ and $y$ components of velocity remain constant throughout the flight?

In standard projectile motion, we neglect air resistance. This means there are absolutely no forces acting in the horizontal $x$ and $y$ directions. According to Newton's First Law, without a net force, the velocity components in these directions must remain perfectly constant.

Q: Why is linear momentum conserved during the collision, even though gravity is acting?

The collision happens in a very short, almost instantaneous time interval. During this brief moment, the internal impulsive forces between the stone and the block are infinitely larger than any external forces like gravity. Therefore, we can safely conserve momentum in the direction of motion just before and just after the impact.

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

At the horizontal position, the string is horizontal, and gravity acts vertically. The tension $T$ is the only force acting along the radial direction towards the pivot, so it must provide the entire centripetal force: $T = \frac{mv^2}{L}$. If the tension is zero, the required centripetal force is zero, which mathematically forces the velocity $v$ to be zero.

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

### The 3D Setup and Relative Velocity

The beauty of physics lies in its ability to break down complex, multi-dimensional motions into simple, independent components. Imagine standing on a cart moving steadily along the $x$-axis. From this moving platform, you throw a stone into the $y$-$z$ plane. To an observer on the ground, the stone's motion is a fascinating 3D curve, but we can decode it step by step.

First, we establish our coordinate system with unit vectors $\hat{i}$, $\hat{j}$, and $\hat{k}$. The cart's velocity is purely in the $x$-direction:
$$ \mathbf{v}_{	ext{cart}} = 4\hat{i} 	ext{ m/s} $$

The stone is thrown relative to the cart in the $y$-$z$ plane, making an angle of $30^\circ$ with the vertical $z$-axis. We resolve this relative velocity into its components:
$$ \mathbf{v}_{	ext{rel}} = (6 \sin 30^\circ)\hat{j} + (6 \cos 30^\circ)\hat{k} = 3\hat{j} + 3\sqrt{3}\hat{k} 	ext{ m/s} $$

### The Absolute Trajectory

To understand the stone's true path as seen from the ground, we must find its absolute velocity. According to Galilean relativity, this is the vector sum of the cart's velocity and the stone's relative velocity:
$$ \mathbf{v}_{	ext{abs}} = \mathbf{v}_{	ext{cart}} + \mathbf{v}_{	ext{rel}} = 4\hat{i} + 3\hat{j} + 3\sqrt{3}\hat{k} 	ext{ m/s} $$

This vector dictates the stone's parabolic flight. As it arcs through the air, gravity acts only in the negative $z$-direction, constantly decelerating its vertical ascent.

### The Zenith and The Inelastic Collision

At the very peak of its trajectory, the stone experiences a moment of pure horizontal motion. Its vertical velocity component, $v_z$, becomes exactly zero. However, because there are no horizontal forces (neglecting air resistance), the $x$ and $y$ components remain unchanged. The velocity at this highest point is:
$$ \mathbf{v}_{	ext{top}} = 4\hat{i} + 3\hat{j} 	ext{ m/s} $$

The magnitude of this velocity is the speed just before impact:
$$ v = \sqrt{4^2 + 3^2} = 5 	ext{ m/s} $$

At this exact zenith, the stone collides with a stationary object of equal mass $m$ hanging from a tree. The collision is completely inelastic, meaning they stick together to form a combined mass of $2m$. Because the collision happens instantaneously, impulsive forces dominate, and we can safely conserve linear momentum in the horizontal plane:
$$ m v = (2m) v_0 $$

Solving for the new velocity $v_0$, we find:
$$ v_0 = \frac{v}{2} = 2.5 	ext{ m/s} $$

### The Pendulum's Ascent

Now, the combined mass acts as a pendulum of length $L$, swinging upwards. We are given a crucial constraint: the tension in the string becomes zero exactly when the string reaches the horizontal position.

At the horizontal position, gravity acts vertically downwards, perpendicular to the string. The only force acting along the string is the tension $T$, which must provide the necessary centripetal force:
$$ T = \frac{(2m)v_{	ext{horiz}}^2}{L} $$

If $T = 0$, it strictly implies that $v_{	ext{horiz}} = 0$. The mass has just enough energy to reach the horizontal line before stopping and falling back. This is a perfect setup for the conservation of mechanical energy. The kinetic energy immediately after the collision is entirely converted into gravitational potential energy at the horizontal position:
$$ \frac{1}{2}(2m)v_0^2 = (2m)gL $$

We can now solve for the length of the string $L$:
$$ L = \frac{v_0^2}{2g} = \frac{(2.5)^2}{2(9.8)} = \frac{6.25}{19.6} \approx 0.32 	ext{ m} $$

This problem is a magnificent symphony of kinematics, momentum, and energy, demonstrating how fundamental laws seamlessly hand off to one another to describe complex physical realities.

Visualized Solution

$Visualizing the 3D Setup$

$v_{cart}$

$v_{rel} Setup$

$v_{rel} Calculation$

$v_{abs} Formula$

$v_{abs} Calculation$

$Highest Point Logic$

$Velocity at Highest Point$

$The Collision Event$

$Conservation of Momentum$

$Combined Velocity v_{0}$

$Pendulum Motion$

$Tension Condition$

$Energy Conservation$

$Calculating Length L$

The Sigma Insight: Conservation of Linear Momentum

The 3D Setup and Relative Velocity

The Absolute Trajectory

The Zenith and The Inelastic Collision

The Pendulum's Ascent

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