A car P is moving with a uniform speed of 5√3 m/s towards a carriage of mass 9 kg at rest kept on th

Q: Why do we ignore the car's velocity when calculating the ball's initial velocity?

The problem states the ball is fired with an initial velocity of $100 \text{ m/s}$. In physics problems of this type, unless specified as 'relative to the car', such velocities are taken relative to the ground.

Q: Why doesn't the vertical velocity of the ball affect the carriage's horizontal speed?

The vertical and horizontal dimensions are independent. The vertical momentum is absorbed by the ground (the normal force from the rails), so only the horizontal momentum is conserved in the direction of the carriage's motion.

Q: How do we know the second ball will definitely hit the carriage?

After the first collision, the carriage moves at $5\sqrt{3} \text{ m/s}$, which is exactly the speed of the car. Since they are moving at the same horizontal speed, their relative horizontal distance remains constant. Thus, a ball fired with the same parameters will perfectly hit the carriage again.

Q: Why is the collision considered perfectly inelastic?

The problem explicitly states that the cannon ball 'hits the stationary carriage... and sticks to it.' When two bodies stick together and move with a common velocity after impact, the collision is perfectly inelastic.

Q: What happens to the kinetic energy during these collisions?

Since the collisions are perfectly inelastic, a significant amount of kinetic energy is lost (converted into heat, sound, and deformation energy). Only linear momentum is conserved.

Question

A car $P$ is moving with a uniform speed of $5\sqrt{3} 	ext{ m/s}$ towards a carriage of mass $9 	ext{ kg}$ at rest kept on the rails at a point $B$ as shown in figure. The height $AC$ is $120 	ext{ m}$. Cannon balls of $1 	ext{ kg}$ are fired from the car with an initial velocity $100 	ext{ m/s}$ at an angle $30^\circ$ with the horizontal. The first cannon ball hits the stationary carriage after a time $t_0$ and sticks to it. Determine $t_0$. At $t_0$, the second cannon ball is fired. Assume that the resistive force between the rails and the carriage is constant and ignore the vertical motion of the carriage throughout.
If the second ball also hits and sticks to the carriage, what will be the horizontal velocity of the carriage just after the second impact?

Sigma Percentile · Accepted Answer

### The Setup: A Cinematic Physics Problem

Imagine an action movie sequence: a car is speeding along the edge of a $120 	ext{ m}$ high cliff at $5\sqrt{3} 	ext{ m/s}$. Suddenly, it fires a $1 	ext{ kg}$ cannon ball at $100 	ext{ m/s}$ at an angle of $30^\circ$ to the horizontal. Down below, a $9 	ext{ kg}$ carriage sits peacefully on the tracks.

Our mission is to find out exactly when the first ball hits the carriage, and what happens when a second ball is fired immediately after the first impact. This problem is a beautiful blend of 2D projectile motion and the conservation of linear momentum.

### Analyzing the First Cannon Ball's Flight

To determine the time of flight $t_0$, we must isolate the vertical motion of the cannon ball. In physics, horizontal and vertical motions are completely independent.

The initial vertical velocity is given by $u_y = 100 \sin 30^\circ = 50 	ext{ m/s}$. The ball must travel from the top of the $120 	ext{ m}$ cliff to the ground, meaning its vertical displacement is $s_y = -120 	ext{ m}$.

Using the second equation of motion, we can set up our quadratic equation:

$$s_y = u_y t + \frac{1}{2} a_y t^2$$

$$-120 = 50t - \frac{1}{2}(10)t^2$$

Simplifying this, we get $t^2 - 10t - 24 = 0$. Factoring this quadratic yields $(t-12)(t+2) = 0$. Since time cannot be negative, we find that the time of flight is exactly $t_0 = 12 	ext{ s}$.

### The First Impact: Momentum Takes the Wheel

While the ball was flying through the air, it was also moving horizontally. Its horizontal velocity is $u_x = 100 \cos 30^\circ = 50\sqrt{3} 	ext{ m/s}$. Since there is no air resistance, this velocity remains constant right up until the moment of impact.

At $t = 12 	ext{ s}$, the $1 	ext{ kg}$ ball strikes the $9 	ext{ kg}$ carriage and sticks to it. This is a perfectly inelastic collision. We apply the principle of conservation of linear momentum in the horizontal direction.

$$m_{	ext{ball}} u_x + m_{	ext{carriage}} (0) = (m_{	ext{ball}} + m_{	ext{carriage}}) v_1$$

$$1 	imes 50\sqrt{3} = (1 + 9) v_1$$

Solving for $v_1$, we get $v_1 = 5\sqrt{3} 	ext{ m/s}$. **This is a magical result!** The carriage is now moving at the exact same horizontal speed as the car on the cliff.

### The Synchronization: Why the Second Ball Hits

At the exact moment of the first impact ($t = 12 	ext{ s}$), the car fires a second cannon ball. Will it hit the carriage? Let's think about the relative motion.

The second ball will also take $12 	ext{ s}$ to reach the ground. During this time, the car (which fired the ball) and the carriage are both moving horizontally at $5\sqrt{3} 	ext{ m/s}$.

Because their horizontal speeds are identical, the relative horizontal distance between the car and the carriage remains perfectly constant. This synchronization guarantees that the second ball will land exactly where the carriage is $12 	ext{ s}$ later!

### The Final Collision: Calculating the Ultimate Speed

Now we analyze the second collision. The carriage system now has a mass of $10 	ext{ kg}$ (original carriage plus the first ball) and is moving at $5\sqrt{3} 	ext{ m/s}$.

The second ball arrives with a mass of $1 	ext{ kg}$ and a horizontal velocity of $50\sqrt{3} 	ext{ m/s}$. We once again apply the conservation of linear momentum.

$$p_{	ext{initial}} = 10 	imes 5\sqrt{3} + 1 	imes 50\sqrt{3} = 100\sqrt{3} 	ext{ kg m/s}$$

After the second ball sticks, the total mass becomes $11 	ext{ kg}$. Let the final velocity be $v_2$.

$$11 v_2 = 100\sqrt{3}$$

$$v_2 = \frac{100\sqrt{3}}{11} \approx 15.75 	ext{ m/s}$$

**And there we have it!** The final horizontal velocity of the carriage just after the second impact is approximately $15.75 	ext{ m/s}$. This problem beautifully demonstrates how complex physical events can be broken down into simple, logical steps.

Visualized Solution

Visualizing the Setup

Vertical Motion of the First Ball

Calculating Time of Flight $t_{0}$

Horizontal Velocity of the Ball

First Collision: Momentum Conservation

Velocity of Carriage After First Impact

The Second Cannon Ball

Second Collision: Momentum Conservation

Final Velocity of the Carriage

The Sigma Insight: Conservation of Linear Momentum

The Setup

Analyzing the First Cannon Ball's Flight

The First Impact

The Synchronization

The Final Collision

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