Question

A 0.100-kg stone rests on a frictionless, horizontal surface. A bullet of mass 6.00 \(g\), traveling horizontally at 350 m/s, strikes the stone and rebounds horizontally at right angles to its original direction with a speed of 250 m/s. (a) Compute the magnitude and direction of the velocity of the stone after it is struck. (b) Is the collision perfectly elastic?

Short answer

The stone's velocity is 25 m/s at \( -35.54^\circ \) from the x-axis. The collision is not perfectly elastic.

Step-by-step solution

Convert units

First, convert the mass of the bullet from grams to kilograms.Mass of bullet \( m_b = 6.00 \text{ g} = 0.006 \text{ kg} \).

Apply Conservation of Momentum in the x-direction

The initial velocity of the bullet is \( v_{b_x,i} = 350 \text{ m/s} \), and the final velocity in the y-direction is \( v_{b,y} = 250 \text{ m/s} \). There is no initial momentum in the y-direction for the bullet. Using conservation of momentum in the x direction: \[ m_b \cdot v_{b_x,i} = m_b \cdot v_{b_x,f} + m_s \cdot v_{s_x} \].Since the bullet rebounds perpendicularly, \( v_{b_x, f} = 0 \). Thus, \( m_b \cdot v_{b_x,i} = m_s \cdot v_{s_x} \).\( 0.006 \cdot 350 = 0.100 \cdot v_{s_x} \).Solve for \( v_{s_x} \): \( v_{s_x} = \frac{0.006 \times 350}{0.100} \).

Solve for Velocity in the X-Direction

Calculate the x-component of the stone's velocity.\( v_{s_x} = \frac{2.1 \text{ kg}\cdot\text{m/s}}{0.100 \text{ kg}} = 21 \text{ m/s} \).

Apply Conservation of Momentum in the y-direction

Initially, the bullet has no y-component velocity. After collision, its velocity in the y-direction is \( v_{b,y} = 250 \text{ m/s} \) and therefore: \[ 0 = m_s \cdot v_{s_y} + m_b \cdot v_{b_y} \].This can be rearranged to find \( v_{s_y} \):\( m_s \cdot v_{s_y} = -m_b \cdot v_{b,y} \).Solve for \( v_{s_y} \):\( 0.100 \cdot v_{s_y} = -0.006 \times 250 \).

Solve for Velocity in the Y-Direction

Calculate the y-component of the stone's velocity.\( v_{s_y} = \frac{-1.5 \text{ kg}\cdot\text{m/s}}{0.100 \text{ kg}} = -15 \text{ m/s} \).The negative sign indicates the stone moves in the opposite direction to the bullet's y-component velocity.

Calculate Magnitude and Direction of the Stone's Velocity

The stone's velocity has components \( v_{s_x} = 21 \text{ m/s} \) and \( v_{s_y} = -15 \text{ m/s} \).Use the Pythagorean theorem to find the magnitude: \[ v_s = \sqrt{v_{s_x}^2 + v_{s_y}^2} = \sqrt{21^2 + (-15)^2} \].Calculate \( v_s \) to find its magnitude.Direction (angle \( \theta \)) can be found using: \[ \tan \theta = \frac{v_{s_y}}{v_{s_x}} = \frac{-15}{21} \].Determine the angle \( \theta \).

Determine If the Collision is Perfectly Elastic

Check the kinetic energy before and after collision. Initial kinetic energy \( KE_{i} = \frac{1}{2} m_b (v_{b_x,i})^2 \).After collision: \( KE_{f} = \frac{1}{2} m_b (v_{b,y})^2 + \frac{1}{2} m_s v_s^2 \).Compare \( KE_i \) and \( KE_f \). If they are equal, the collision is perfectly elastic.

Key concepts

Elastic Collision

An elastic collision is a type of collision where total kinetic energy is conserved. This means that the sum of kinetic energy before the collision equals the sum of kinetic energy after the collision. In mechanics, these collisions are ideal because they imply no energy is lost to sound, heat, or deformation.

In the provided exercise, a bullet hits a stone and rebounds at right angles. For this to be a perfectly elastic collision, the kinetic energy before the impact needs to be equal to the kinetic energy after the impact. Elastic collisions also conserve momentum, which is why both conservation of momentum and kinetic energy are used to evaluate the results.

Given this, when analyzing whether the collision between the bullet and the stone is elastic, one must calculate the total kinetic energy before and after the collision. If they match, the collision can indeed be classified as perfectly elastic.

Kinetic Energy

Kinetic energy is the energy possessed by an object due to its motion. The kinetic energy (KE) can be calculated using the formula: \[KE = \frac{1}{2} m v^2\]where \( m \) is the mass and \( v \) is the velocity of the object. In any collision, examining the kinetic energy helps to understand whether some energy was lost during the impact.

For the bullet hitting the stone, the initial kinetic energy is calculated with the bullet's mass and velocity before the collision. After the collision, you calculate the kinetic energies of both the bullet and stone using their respective masses and velocities. Why is this important? Because if the total kinetic energy remains constant, it indicates no energy was lost and helps confirm if the collision was elastic.

• Initial kinetic energy depends on the bullet only, as the stone is initially at rest.
• After the collision, both the bullet's new path and the stone's movement contribute to the kinetic energy.

By comparing these kinetic energy values, you can determine if there’s any loss or if the energy remained the same, thus checking for elasticity.

Momentum in Two Dimensions

Momentum is the product of an object's mass and velocity and is a fundamental concept in physics because it is always conserved in a closed system. In two dimensions, momentum conservation involves both the x and y directions, adding complexity but also providing a comprehensive assessment of the collision effects.

In this particular exercise, the bullet initially travels in the x-direction and rebounds in the y-direction. This necessitates evaluating momentum in both dimensions separately. In the x-direction, conservation of momentum considers the initial velocity of the bullet and its lack of movement post-collision. In the y-direction, the calculation starts with zero initial momentum as the bullet was not moving vertically before impact.

• Momentum in x-direction: Start with the bullet's initial velocity and solve for the stone's motion after the bullet rebound.
• Momentum in y-direction: Explore post-collision dynamics where only the stone moves in response to the bullet's new direction.

By solving these equations, the stone's velocity components (both x and y) are discovered, allowing us to understand the total change in motion resulting from the collision.