Problem 61 - College Physics

Question

\(\bullet\) 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

The stone velocity is 14.5 m/s at 30° below horizontal. The collision is not elastic.

Step 1: Understand Initial Conditions

Initially, the stone is at rest, and the bullet is moving with a given velocity. The initial momentum of the stone is 0, and the initial momentum of the bullet is computed using its mass and velocity. Convert the bullet's mass from grams to kilograms for consistency: \( m_b = 2.50\, \text{g} = 0.0025\, \text{kg} \) and \( v_b = 500\, \text{m/s} \).

Step 2: Calculate Initial Momentum

The initial momentum (\( P_{i} \)) of the system (bullet and stone) is given only by the bullet's momentum since the stone is initially at rest: \( P_{i} = m_b \cdot v_b = 0.0025\, \text{kg} \times 500\, \text{m/s} = 1.25\, \text{kg} \cdot \text{m/s} \).

Step 3: Calculate Final Momentum of Bullet

After striking the stone, the bullet changes direction. The final velocity of the bullet has a magnitude of \(300\, \text{m/s}\) and travels perpendicular to its initial direction. The final momentum of the bullet is \( P_{f_{bullet}} = 0.0025\, \text{kg} \times 300\, \text{m/s} = 0.75\, \text{kg} \cdot \text{m/s} \).

Step 4: Apply Conservation of Momentum

Using conservation of momentum, the total initial momentum of the system equals the total final momentum. This gives us: \( m_b \cdot v_b = m_b \cdot v_{b_f} + m_s \cdot v_s \). Since the bullet rebounds at right angles, we need to account for the vector nature of momentum. Thus, \( v_{s_x} \) (velocity of the stone along x-axis) can be found using: \( 1.25 = 0.75 \cos(90^\circ) + 0.100 \cdot v_{s_x} \), leading to \( v_{s_x} = 12.5 \text{ m/s} \). Similarly, using the y-component gives: \( 0 = 0.75 \sin(0^\circ) - 0.100 \cdot v_{s_y} \), leading to \( v_{s_y} = -7.5 \text{ m/s} \).

Step 5: Determine the Resultant Velocity of the Stone

The magnitude of the stone's velocity is \( v_s = \sqrt{(v_{s_x})^2 + (v_{s_y})^2} = \sqrt{(12.5^2) + (-7.5^2)} = 14.5 \text{ m/s} \). The direction \( \theta \) can be found using \( \tan^{-1}(\frac{v_{s_y}}{v_{s_x}}) = \tan^{-1}\left(\frac{-7.5}{12.5}\right) = -30° \). This means moving at a direction of 30° below the horizontal.

Step 6: Determine if the Collision is Elastic

For a collision to be perfectly elastic, kinetic energy must also be conserved. Initial kinetic energy is \( KE_i = \frac{1}{2} \times 0.0025 \times 500^2 \). Final kinetic energies are \( KE_{f_{bullet}} = \frac{1}{2} \times 0.0025 \times 300^2 \) and \( KE_{f_{stone}} = \frac{1}{2} \times 0.100 \times 14.5^2 \). Calculate these and compare. If they are not equal, the collision is not elastic.

Step 7: Calculate and Compare Kinetic Energies

After calculation, initial kinetic energy \( KE_i = 312.5 \text{ J} \). Final kinetic energies are \( KE_{f_{bullet}} = 112.5 \text{ J} \) and \( KE_{f_{stone}} = 10.5 \text{ J} \). Since total \( KE_f = KE_{f_{bullet}} + KE_{f_{stone}} = 123 \text{ J} \) and is less than \( KE_i \), the collision is not perfectly elastic.

Key Concepts

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 of the object and \( v \) its velocity. In the problem, we calculated the initial and final kinetic energies to determine if the collision was elastic.
Initially, only the bullet has kinetic energy, calculated as \( KE_i = \frac{1}{2} \times 0.0025 \times 500^2 \).
The final kinetic energy involves both the bullet and the stone after the collision. For such calculations:

• The bullet's kinetic energy is \( KE_{f_{bullet}} = \frac{1}{2} \times 0.0025 \times 300^2 \).
• The stone's kinetic energy is \( KE_{f_{stone}} = \frac{1}{2} \times 0.100 \times 14.5^2 \).

Upon calculating and comparing, we see that the total final kinetic energy differs from the initial kinetic energy, leading to the conclusion that the collision is not elastic.

Elastic Collision

An elastic collision is a type of collision where there's no net loss in kinetic energy in the system.
In an ideal elastic collision, both momentum and kinetic energy are conserved. However, in many real-world collisions, some energy transforms into other forms, such as heat or sound.
In this problem, the bullet and stone collide, and we computed both momenta and kinetic energies. The exercise asked us to check if this was an elastic collision.
Upon solving, the total kinetic energy post-collision was notably less than before the impact, indicating energy loss.
This energy loss confirms that the collision is not perfectly elastic, despite momentum conservation.

Momentum Vector Components

Momentum is a vector quantity, meaning it has both magnitude and direction. In the case of a collision involving objects moving in two dimensions, it's crucial to consider these vector components separately.
Initially, the bullet moves in one direction, and upon impact, its motion changes components, now moving at right angles. Therefore, we have to compute the momentum components along both the x-and y-axes.
Using trigonometric functions like sine and cosine helps in solving for these components:

• For x-component: \( 1.25 = 0.75 \cos(90^\circ) + 0.100 \cdot v_{s_x} \)
• For y-component: \( 0 = 0.75 \sin(0^\circ) - 0.100 \cdot v_{s_y} \)

Solving these allows us to find the stone's final velocity components \( v_{s_x} \) and \( v_{s_y} \), which we then used to find the resultant velocity of the stone.

Physics Problem-Solving

Solving physics problems often involves a series of steps aimed at simplifying the solution process. Communicating a problem into known physics principles is crucial.
In this situation, the first step was understanding the initial conditions—identifying that the bullet has kinetic energy and momentum, and the stone is initially motionless.
Converting units can also be a critical step, as seen in changing the bullet's mass from grams to kilograms. Following this, finding initial and final momentum helps in understanding motion changes due to collision.
By applying conservation of momentum and checking changes in kinetic energy, we can better understand the type of collision and determine important quantities like velocity and direction. The goal is to break down complex problems into easier sub-problems, ensuring all physical laws and principles are appropriately applied to reach a solution.