Question

A catapult on a level field tosses a 3 -kg stone a horizontal distance of \(100 \mathrm{~m}\). A second 3 -kg stone tossed in an identical fashion breaks apart in the air into 2 pieces, one with a mass of \(1 \mathrm{~kg}\) and one with a mass of \(2 \mathrm{~kg} .\) Both of the pieces hit the ground at the same time. If the 1 -kg piece lands a distance of \(180 \mathrm{~m}\) away from the catapult, how far away from the catapult does the 2 -kg piece land? Ignore air resistance. a) \(20 \mathrm{~m}\) c) \(100 \mathrm{~m}\) e) \(180 \mathrm{~m}\) b) \(60 \mathrm{~m}\) d) \(120 \mathrm{~m}\)

Short answer

Based on the information given and by applying the principles of physics and the conservation of momentum, the 2 kg piece lands at a distance of 60 meters from the catapult.

Step-by-step solution

Identifying the known values

The given values are: 1. Mass of the first stone: 3 kg 2. Mass of the broken pieces: 1 kg and 2 kg 3. Landing distance for the unbroken stone: 100 m 4. Landing distance for the 1 kg piece: 180 m Note that the final velocity of the 2 kg piece is the same in the x and y components as the velocity of the 1 kg piece, since they both hit the ground at the same time.

Using the conservation of momentum

Since momentum is conserved, we can write: Initial momentum (before breaking) = Final momentum (after breaking) Let's denote the velocities of the two broken pieces as: \(v_1\) - velocity of 1 kg piece \(v_2\) - velocity of 2 kg piece The initial momentum (3 kg stone) can be found by using the range equation for projectile motion: \(v_{initial} = \frac{100 \mathrm{~m} \cdot g}{v_{vertical}}\) The vertical component of velocity will cancel out due to the stone's parabolic nature. Therefore: \(M_initial \cdot v_{initial}=M_1 \cdot v_1 + M_2 \cdot v_2\)

Finding the velocity of 2 kg piece

Since both broken pieces hit the ground at the same time, they have the same horizontal velocity. Let's denote the horizontal velocity as \(v_{horizontal}\). We can find \(v_1\) by using the range equation again: \(v_1 = \frac{180 \mathrm{~m} \cdot g}{v_{vertical}}\) Now we can plug this back into the momentum conservation equation: \(3 \cdot v_{initial} = 1 \cdot v_1 + 2 \cdot v_2\) Using the range equations: \(3 \cdot \frac{100 \mathrm{~m} \cdot g}{v_{vertical}} = 1 \cdot \frac{180 \mathrm{~m} \cdot g}{v_{vertical}} + 2 \cdot v_2\) Cancel out the \(g\) and \(v_{vertical}\), and we get: \(3 \cdot 100 = 180 + 2 \cdot v_2\)

Calculating the distance for the 2 kg piece

After calculating the velocity of the 2 kg piece from the above equation, we can use the range equation again to find the distance it travels: \(v_2 = \frac{3\cdot100 - 180}{2} = 60\) Finally, using the range equation: \(distance_{2kg} = \frac{v_2 \cdot v_{vertical}}{g} = \frac{60 \mathrm{~m} \cdot v_{vertical}}{g}\) Since the proportion is consistent, we can set this equal to: \(100 \mathrm{~m} = \frac{v_{initial} \cdot v_{vertical}}{g}\) Solve for \(distance_{2kg}\): \(\frac{60 \mathrm{~m} \cdot v_{vertical}}{g} = \frac{v_{initial} \cdot v_{vertical}}{100} \Rightarrow distance_{2kg} = 60 \mathrm{~m}\) The 2 kg piece lands at a distance of 60 meters from the catapult. The correct answer is (b) \(60 \mathrm{~m}\).

Key concepts

Conservation of Momentum

Conservation of momentum is a fundamental principle in physics that states the momentum of a closed system is constant, provided no external forces act on it. In this scenario, when the 3 kg stone breaks into a 1 kg and a 2 kg piece in midair, the system is conserved because there are no external forces like air resistance affecting the stone besides gravity, which acts equally on both pieces.

Momentum is defined as the product of an object's mass and velocity. Therefore, before the breakup, the total momentum is simply the mass of the stone multiplied by its initial velocity. After breaking, each piece carries a part of the stone's momentum. By applying the conservation of momentum formula, \( M_{initial} \cdot v_{initial} = M_1 \cdot v_1 + M_2 \cdot v_2 \), we equate the initial and final momentum to find unknown velocities.

This principle helps us solve problems where particles split or collide because the total momentum before and after the event remains unchanged, which is crucial in determining the subsequent paths and velocities of the particles.

Range Equation

The range equation in projectile motion calculates the horizontal distance traveled by an object when launched. This equation is particularly helpful when analyzing how far projectiles, like the stone from the catapult, travel when the conditions are optimal.

The standard range formula is given by: \[ R = \frac{v^2 \cdot \sin(2\theta)}{g} \]where \( R \) is the horizontal range, \( v \) is the initial velocity, \( \theta \) is the launch angle, and \( g \) is the acceleration due to gravity. In situations with no air resistance and a level launch level, as with the catapult, this equation provides accurate predictions of where the projectile will land.

In the exercise, both the 1 kg piece and 2 kg piece are affected equally by gravity but launched horizontally, so the vertical component of their velocity cancels in the problem setup, making their horizontal distances a direct function of their horizontal velocities. Thus, by knowing how one part behaves under these conditions, we can solve for the other using the conservation principles.

Horizontal Velocity

Horizontal velocity refers to the constant speed of an object moving parallel to the Earth's surface in projectile motion. Unlike vertical velocity, which changes due to gravity, horizontal velocity remains constant if air resistance is ignored, as in this exercise.

For objects in projectile motion that start with an initial horizontal velocity, this component plays a crucial role in determining how far an object will travel. Here, when the stone splits into two pieces, both maintain equal horizontal velocity because they remain unaffected by vertical forces until they hit the ground.

Knowing the horizontal velocity of one piece allows us to calculate the distance the other piece travels using the principles of conservation of momentum. For the 1 kg and 2 kg pieces in this exercise, the horizontal velocity was key in calculating how far each piece would land from the launch site.

Parabolic Trajectory

Parabolic trajectory describes the path followed by an object in projectile motion. Under ideal conditions, like absent air resistance, this path is a perfect parabola due to the influence of gravity acting downwards while the projectile moves horizontally.

The nature of this trajectory can be understood by considering that the object's horizontal velocity remains constant, while its vertical velocity changes uniformly under gravity. This combination creates the curved path of a parabola.

In our exercise, both pieces of the stone follow a parabolic trajectory after the stone breaks, with the point where they hit the ground dictating the horizontal range they've covered. An understanding of this concept helps in predicting and explaining motion patterns of projectiles in various physics problems, demonstrating the physics behind motion through equations like the range equation.