Problem 84 - University Physics with Modern Physics

Question

A Ricocheting Bullet. 0.100 -kg stone rests on a frictionless, horizontal surface. A bullet of mass 6.00 g, traveling horizontally at 350 \(\mathrm{m} / \mathrm{s}\) , strikes the stone and rebounds horizontally at right angles to its original direction with a speed of 250 \(\mathrm{m} / \mathrm{s}\) . (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 approximately 25.8 m/s at a 35.54° angle. (b) The collision is not perfectly elastic.

Step 1: Convert Units

Convert the mass of the bullet from grams to kilograms. Given the mass of the bullet is 6.00 g, convert it to kilograms: \[6.00 \text{ g} = 0.006 \text{ kg}\]

Step 2: Calculate Momentum in x-direction

Use the conservation of momentum in the x-direction. Initially, only the bullet has momentum. Initial momentum in x-direction = momentum of bullet = \[0.006 \times 350 = 2.1 \text{ kg}dot \text{m/s} \]After collision, bullet travels perpendicular, thus stone's x-momentum = 2.1 kg·m/s.

Step 3: Calculate Stone's Velocity in x-direction

The stone in the x-direction must also have a momentum of 2.1 kg·m/s after being struck. Given the stone's mass is 0.100 kg, its velocity in the x-direction is:\[v_{x, stone} = \frac{2.1}{0.1} = 21 \text{ m/s}\]

Step 4: Calculate Momentum in y-direction

Since the bullet travels in the y-direction at 250 m/s after the collision, its momentum is:\[0.006 \times 250 = 1.5 \text{ kg}dot \text{m/s} \]This is the stone's momentum in the y-direction.

Step 5: Calculate Stone's Velocity in y-direction

Now calculate the velocity of the stone in the y-direction using momentum:\[v_{y, stone} = \frac{1.5}{0.1} = 15 \text{ m/s}\]

Step 6: Calculate Resultant Velocity and Direction of Stone

The resultant velocity of the stone is found by combining the x and y components:\[v = \sqrt{v_{x, stone}^2 + v_{y, stone}^2} = \sqrt{21^2 + 15^2} = \sqrt{441 + 225} = \sqrt{666} \approx 25.8 \text{ m/s} \]The direction \( \theta \) relative to the original direction of the bullet (x-axis) is:\( \theta = \tan^{-1} \left( \frac{15}{21} \right) \approx 35.54^\circ \).

Step 7: Check for Perfect Elasticity

To determine if a collision is perfectly elastic, the total kinetic energy before and after must be equal. Compute initial kinetic energy (KE) and final KE (separately for bullet and stone). Initial KE:\[\frac{1}{2} \times 0.006 \times 350^2 \approx 367.5 \text{ J}\].Final KE (Bullet) + Final KE (Stone):\[\frac{1}{2} \times 0.006 \times 250^2 + \frac{1}{2} \times 0.1 \times 25.8^2 \approx 187.5 + 33.249 = 220.749 \text{ J}\]Since the kinetic energies do not match, the collision is not perfectly elastic.

Key Concepts

Perfectly Elastic Collision

A perfectly elastic collision is an ideal scenario in physics. In such collisions, both momentum and kinetic energy are conserved. This means that after the collision, the total kinetic energy of the objects involved is the same as before. In reality, perfectly elastic collisions are rare, especially at macroscopic levels. They are most commonly seen in particle physics. When examining a collision to check if it is perfectly elastic, you need to compare the total kinetic energy before and after the event. If there is any difference, the collision isn't perfectly elastic. In the case of the ricocheting bullet and the stone, the initial kinetic energy doesn't match the final kinetic energy post-collision. Thus, it cannot be perfectly elastic.

Kinetic Energy

Kinetic energy is the energy an object possesses due to its motion. It's an important concept in physics, formulated as: \[ KE = \frac{1}{2} mv^2 \] where \( m \) is the object's mass and \( v \) its velocity. The faster an object moves, or the heavier it is, the greater its kinetic energy.
In the ricocheting bullet scenario, you calculate the kinetic energy of both the bullet and the stone before and after the collision. The results help determine whether the collision was elastic. If the total kinetic energies before and after the collision differ, this indicates energy was lost (often due to heat or deformation), thus showing the collision isn't perfectly elastic.
Remember, kinetic energy can change dramatically in collisions based on the masses and velocities involved. It's a key player in understanding many types of physical interactions.

Ricocheting Bullet

When a bullet ricochets, it bounces off a surface at an angle after hitting it, sometimes with altered speed or direction. Several factors affect a bullet's ricochet behavior such as the angle of incidence, the surface material, and the bullet's design.
In physics problems, like our exercise, a ricocheting bullet often involves analyzing how energy and momentum are conserved and transferred during the collision. For the bullet and stone exercise, once the bullet hits the stone and bounces at a right angle, calculations for the stone's new path and velocity include components of motion in both the x and y directions.
Understanding the dynamics of a ricocheting bullet highlights how momentum conservation plays out in real-world scenarios. It shows that while direction and speed may alter, total system momentum remains constant, allowing us to solve for unknowns like velocity and impact angles after such interactions.