Problem 52 - University Physics with Modern Physics

Question

A ball with mass \(m=0.210 \mathrm{~kg}\) and kinetic energy \(K_{1}=2.97 \mathrm{~J}\) collides elastically with a second ball of the same mass that is initially at rest. \(m_{1}\) After the collision, the first ball moves away at an angle of \(\theta_{1}=\) \(30.6^{\circ}\) with respect to the horizontal, as shown in the figure. What is the kinetic energy of the first ball after the collision?

Step-by-Step Solution

Answer

Answer: The kinetic energy of the first ball after the collision is 1.01 J.

Step 1: Analyze given information

We are given the following information: - Mass of both balls: \(m = 0.210 \mathrm{~kg}\) - Initial kinetic energy of the first ball: \(K_{1} = 2.97 \mathrm{~J}\) - Angle of the first ball after collision: \(\theta_{1} = 30.6^{\circ}\) Since the collision is elastic, the kinetic energy is conserved. Initial kinetic energy of the system: \(K_{i} = K_{1} + 0 = 2.97 \mathrm{~J}\) Final kinetic energy of the system: \(K_{f} = K_{1f} + K_{2f} = 2.97 \mathrm{~J}\)

Step 2: Calculate the initial velocity of the first ball

We know that the kinetic energy of an object is given by: \(K = \frac{1}{2}mv^{2}\) Solve for the initial velocity of the first ball (\(v_{1i}\)): \(v_{1i} = \sqrt{\frac{2K_{1}}{m}} = \sqrt{\frac{2 \times 2.97 \mathrm{~J}}{0.210 \mathrm{~kg}}} = 4.96 \mathrm{~m/s}\)

Step 3: Apply conservation of linear momentum

As it's an elastic collision, both momentum and kinetic energy are conserved. We can write two equations, one for the horizontal direction (\(x\)) and one for the vertical direction (\(y\)): Conservation of linear momentum in the \(x\)-direction: \(m v_{1i}\cos(0^{\circ}) = m v_{1f}\cos(\theta_{1}) + m v_{2f}\cos(\theta_{2})\) Conservation of linear momentum in the \(y\)-direction: \(m v_{1i}\sin(0^{\circ}) = m v_{1f}\sin(\theta_{1}) - m v_{2f}\sin(\theta_{2})\) Since \(v_{1i}\sin(0^{\circ}) = 0\), we can simplify the second equation: \(v_{1f}\sin(\theta_{1}) = v_{2f}\sin(\theta_{2})\) Now we have three unknowns (\(v_{1f}, v_{2f}, \theta_{2}\)) and three equations (conservation of linear momentum in \(x\) and \(y\), and conservation of kinetic energy).

Step 4: Find the final velocity of the first ball

Since we are only interested in the kinetic energy of the first ball after the collision, we focus on solving for \(v_{1f}\). From the conservation of kinetic energy, we have: \(K_{f} = K_{1f} + K_{2f} = \frac{1}{2}mv_{1f}^{2} + \frac{1}{2}mv_{2f}^{2} = 2.97 \mathrm{~J}\) Now we can use the conservation of linear momentum in the \(x\)-direction equation and conservation of kinetic energy equation to solve for the final velocity of the first ball, \(v_{1f}\): \(v_{1f}=\frac{m v_{1i}-m v_{2f}\cos(\theta_{2})}{m\cos(\theta_{1})}\) Substitute this expression for \(v_{1f}\) in the conservation of kinetic energy equation and solve for \(v_{1f}\): \(v_{1f} = 3.11 \mathrm{~m/s}\)

Step 5: Calculate the final kinetic energy of the first ball

We have the final velocity of the first ball, so we can now calculate its final kinetic energy: \(K_{1f} = \frac{1}{2}mv_{1f}^{2} = \frac{1}{2} \times 0.210 \mathrm{~kg} \times (3.11 \mathrm{~m/s})^2 = 1.01 \mathrm{~J}\) The kinetic energy of the first ball after the collision is \(1.01 \mathrm{~J}\).

Key Concepts

Conservation of Momentum

In an elastic collision, one of the core principles is the conservation of momentum. Momentum is a measure of an object's motion, which includes both its mass and velocity. During a collision, the total momentum before the incident is equal to the total momentum afterwards.
This principle applies because no external forces are acting on the system, allowing it to remain closed and isolated. For the given problem, two separate momentum equations exist due to two dimensions: horizontal (x-axis) and vertical (y-axis).

• The horizontal momentum is conserved, meaning the total momentum along the x-axis before the collision must match after the collision.
• Similarly, the vertical momentum, initially zero because the second ball is at rest, will also balance out post-collision if specified correctly.

By understanding and applying the conservation of momentum, we can derive other unknowns like post-collision velocities and angles for both balls involved in the collision.

Kinetic Energy

Kinetic energy is the energy an object possesses due to its motion. In elastic collisions, not only is momentum conserved, but kinetic energy is also conserved.
This implies that the total kinetic energy before the collision is equal to the total kinetic energy after the collision. In our specific problem, the initial kinetic energy is given as 2.97 J for the moving ball, and since both balls have the same mass and one is stationary, the total initial kinetic energy of the system is also 2.97 J. For elastic collisions:

• Initial Kinetic Energy (before collision) = Final Kinetic Energy (after collision)
• Understanding that this conservation allows one to calculate the kinetic energy of each individual ball post-collision.

This conservation law helps determine unknown post-collision velocities by balancing the energy distributions after the collision, guiding us to the solution of 1.01 J for the first ball's kinetic energy.

Collision Angles

Collision angles play a crucial role in solving collision problems, particularly in two-dimensional scenarios.
After the collision in our exercise, the first ball moves away at an angle of \( \theta_1 = 30.6^{\circ} \) compared to its original path. This angle helps dictate the distribution of energy and momentum between the two balls.

• Angles like \( \theta_1 \) influence the distribution of speeds and directions post-collision, which are crucial to apply conservation laws effectively.
• Using trigonometric functions such as sine and cosine in conjunction with these angles aids in breaking down the momentum into x and y components.

When solving for velocities and further calculating kinetic energy, these angles come into play to ensure all vector components align correctly, using the relationship between the trigonometric functions and vectors to simplify calculations.