Problem 55 - College Physics

Question

A stone with a mass of \(0.100 \mathrm{~kg}\) rests on a frictionless, horizontal surface. A bullet of mass \(2.50 \mathrm{~g}\) traveling horizontally at \(500 \mathrm{~m} / \mathrm{s}\) strikes the stone and rebounds horizontally at right angles to its original direction with a speed of \(300 \mathrm{~m} / \mathrm{s}\). (a) Compute the magnitude and direction of the velocity of the stone after it is struck. (b) Is the collision perfectly elastic?

Step-by-Step Solution

Answer

(a) The stone's velocity is 14.5 m/s at 31°; (b) The collision is not perfectly elastic.

Step 1: Understand the Problem

We have a bullet that strikes a stone, and we need to determine the stone's velocity after the collision. The problem involves the conservation of momentum and possibly the coefficient of restitution for determining elasticity.

Step 2: Convert Units

Convert the bullet's mass to kilograms for consistency. \( 2.50 \text{ g} = 0.00250 \text{ kg} \).

Step 3: Set Up Conservation of Momentum Equations

For the initial state, the bullet moves, and the stone is stationary. After collision, momentum is conserved:The initial momentum of the bullet: \( p_{b,i} = m_{b} \times v_{b,i} = 0.00250 \times 500 \). The initial momentum of the stone: \( p_{s,i} = m_{s} \times v_{s,i} = 0 \times 0 \).The final momentum of the bullet in the y-direction: \( p_{b,f_y} = m_{b} \times v_{b,f_y} = 0.00250 \times (-300) \). The final momentum of the bullet in the x-direction is zero because the bullet rebounds and moves at a right angle: \( p_{b,f_x} = 0 \).Let \( v_{s_x} \) and \( v_{s_y} \) be the components of the stone's velocity. The conservation of momentum in x and y directions can be expressed as:In the x-direction: \( 0.00250 \times 500 = 0.100 \times v_{s_x} \)In the y-direction: \( 0 = 0.100 \times v_{s_y} + 0.00250 \times (-300) \)

Step 4: Solve the Equations Simultaneously

From the x-direction equation:\[v_{s_x} = \frac{0.00250 \times 500}{0.100} = 12.5 \text{ m/s} \]From the y-direction equation:\[v_{s_y} = \frac{-0.00250 \times (-300)}{0.100} = 7.5 \text{ m/s} \]

Step 5: Calculate Magnitude and Direction

The magnitude of the velocity is given by:\[ v_s = \sqrt{v_{s_x}^2 + v_{s_y}^2} = \sqrt{12.5^2 + 7.5^2} = 14.5 \text{ m/s} \] The direction (angle θ with the x-axis) can be found using:\[ \theta = \tan^{-1}\left( \frac{v_{s_y}}{v_{s_x}} \right) = \tan^{-1}\left( \frac{7.5}{12.5} \right) \approx 31.0^\circ \]

Step 6: Check for Elastic Collision

An elastic collision conserves kinetic energy. Calculate the initial and final kinetic energies.Initial kinetic energy of the bullet:\( KE_{i} = \frac{1}{2} \times 0.00250 \times 500^2 = 312.5 \text{ J} \)Final kinetic energies:For the bullet:\( KE_{b,f} = \frac{1}{2} \times 0.00250 \times 300^2 = 112.5 \text{ J} \)For the stone:\( KE_{s,f} = \frac{1}{2} \times 0.100 \times 14.5^2 = 10.5 \text{ J} \)The total final kinetic energy: \( KE_{total,f} = 112.5 + 10.5 = 123.0 \text{ J} \)Since initial kinetic energy \( KE_{i} = 312.5 \text{ J} \) is not equal to the total final kinetic energy \( KE_{total,f} = 123.0 \text{ J} \), the collision is not perfectly elastic.

Key Concepts

Inelastic Collision

In the world of physics, collisions can be broadly categorized into two types: elastic and inelastic. An inelastic collision is a type of collision where the kinetic energy is not conserved. This means that some of the kinetic energy is converted into other forms of energy, such as sound or thermal energy.

In the provided exercise, the bullet and the stone experience an inelastic collision. Although the momentum is conserved during the collision process, there is a significant loss of kinetic energy. The exercise demonstrates this by comparing the total kinetic energy before and after the collision. With a drop from an initial value of 312.5 J to a final total of 123.0 J, it is evident that this is an inelastic collision.

Remember, the key feature of inelastic collisions is the lack of kinetic energy conservation, even though momentum remains conserved. This concept is crucial to distinguishing between different types of collisions in physics.

Kinetic Energy

Kinetic energy is the energy an object possesses due to its motion. It is given by the formula \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass and \( v \) is the velocity of the object. In collision problems, we often examine the kinetic energy before and after the event to understand energy dynamics.

In this particular scenario, we start with calculating the initial kinetic energy of the bullet before the collision. The bullet's kinetic energy was 312.5 J, which results from its high velocity of 500 m/s. After the collision, the bullet and stone have separate kinetic energies: 112.5 J and 10.5 J respectively. The significant difference between the initial and final total kinetic energies confirms that kinetic energy is not conserved here, pointing to an inelastic collision.

Understanding kinetic energy and its transformation during collisions is crucial in physics, as it helps describe how energy is redistributed in various processes.

Problem-Solving Steps

Solving physics problems systematically can greatly enhance understanding and accuracy. Here, the problem-solving process for the collision problem is broken down into clear steps.

First, it is vital to comprehend the problem thoroughly by identifying the known variables and the question asked. Then, converting units consistently helps avoid mistakes during calculations. For instance, converting the bullet's mass from grams to kilograms ensures uniformity with the stone's mass.

Next, applying fundamental physical principles, such as conservation of momentum, is necessary. Set up the equations based on the calculated momentum before and after the collision for each direction. After equations are established from these principles, solve them systematically for the required unknowns.

Finally, calculating the resulting values and interpreting them in the context of the problem will give you the answers. Following these structured steps allows one to approach complex physics problems with confidence and clarity.

Velocity Calculation

Velocity is a vector quantity, involving both magnitude and direction. In this exercise, after the collision between the bullet and the stone, it was necessary to calculate the final velocity of the stone. This involves tackling both the magnitude of this velocity and its direction.

The velocity components of the stone, calculated using the conservation of momentum, reveal \( v_{s_x} = 12.5 \text{ m/s} \) and \( v_{s_y} = 7.5 \text{ m/s} \). These components describe its motion in the x and y directions post-collision. To determine the overall velocity magnitude and direction of the stone, we employ Pythagoras’ theorem and trigonometry.

The magnitude is calculated as \( v_s = \sqrt{v_{s_x}^2 + v_{s_y}^2} = 14.5 \text{ m/s} \). Additionally, using the arctangent function, we determine the direction: \( \theta = \tan^{-1}\left( \frac{7.5}{12.5} \right) \approx 31.0^\circ \), relative to the x-axis. Accurate velocity calculations are pivotal in analyzing motion in physics.