Problem 6 - Applied Physics

Question

Ball A with a mass of \(0.500 \mathrm{~kg}\) is moving east at a velocity of \(0.800 \mathrm{~m} / \mathrm{s}\). It strikes ball \(\mathrm{B}\), also of mass \(0.500 \mathrm{~kg}\), which is stationary. Ball A glances off \(\mathrm{B}\) at an angle of \(40.0^{\circ}\) north of its original path. Ball \(\mathrm{B}\) is pushed along a path perpendicular to the final path of ball A. (a) What is the momentum of ball A after the collision? (b) What is the momentum of ball B after the collision? (c) What is the velocity of ball A after the collision? (d) What is the velocity of ball B after the collision?

Step-by-Step Solution

Answer

Ball A's velocity and momentum are 1.044 m/s and 0.522 kg⋅m/s; Ball B's velocity and momentum are 0.671 m/s and 0.3355 kg⋅m/s, respectively.

Step 1: Initial Momentum of Ball A

Calculate the initial momentum of Ball A using the formula: \( p = mv \). The mass of Ball A is \(0.500~\text{kg}\) and velocity is \(0.800~\text{m/s}\).\[ p_{A_i} = 0.500~\text{kg} \times 0.800~\text{m/s} = 0.400~\text{kg} \cdot \text{m/s}\]The initial momentum of Ball A is \(0.400~\text{kg} \cdot \text{m/s}\) to the east.

Step 2: Conservation of Momentum in the X-direction

The x-component of momentum is conserved in the collision. Let's denote the final velocity of Ball A as \( v_{A_f} \) and the velocity of Ball B as \( v_B \). The angle \( \theta = 40.0^\circ \).Using conservation of momentum:\[ 0.400 = 0.500 \cdot v_{A_f} \cos(40.0^\circ) \]Since Ball B moves perpendicularly to Ball A's final path, its contribution in x-direction is 0.

Step 3: Conservation of Momentum in the Y-direction

In the y-direction, the initial momentum is 0 since both balls move initially east. In the final state:\[ 0 = 0.500 \cdot v_{A_f} \sin(40.0^\circ) - 0.500 \cdot v_B\]This equation represents the conservation of momentum in the y-axis.

Step 4: Solving for Final Velocity of Ball A

Substitute from Step 2: \[ v_{A_f} = \frac{0.400}{0.500 \cos(40.0^\circ)}\]Compute the value of \( v_{A_f} \):\[ v_{A_f} = \frac{0.400}{0.500 \times 0.766} \approx 1.044~\text{m/s}\]The final velocity of Ball A is approximately \(1.044~\text{m/s}\).

Step 5: Solving for Velocity of Ball B

Using the equation from Step 3:\[ 0 = 0.500 \cdot 1.044 \sin(40.0^\circ) - 0.500 \cdot v_B\]Solving for \( v_B \):\[ v_B = 1.044 \times 0.643 \]\[ v_B \approx 0.671~\text{m/s}\]The final velocity of Ball B is approximately \(0.671~\text{m/s}\).

Step 6: Final Momentum of Ball A

Calculate the momentum of Ball A after the collision using its final velocity and mass:\[ p_{A_f} = 0.500 \cdot 1.044~\text{m/s}\]\[ p_{A_f} = 0.522~\text{kg} \cdot \text{m/s}\]The momentum of Ball A after the collision is \(0.522~\text{kg} \cdot \text{m/s}\).

Step 7: Final Momentum of Ball B

Calculate the momentum of Ball B after the collision using its final velocity and mass:\[ p_{B} = 0.500 \cdot 0.671~\text{m/s}\]\[ p_{B} = 0.3355~\text{kg} \cdot \text{m/s}\]The momentum of Ball B after the collision is approximately \(0.3355~\text{kg} \cdot \text{m/s}\).

Key Concepts

Collision Analysis

Understanding collision analysis is crucial in comprehending how objects interact in reality. In physics, a collision is any event where two or more bodies exert forces on each other for a relatively short time. When examining the collision, we focus on the forces and the effects on the moving objects. In our scenario, two equal-mass balls collide, providing us an opportunity to explore an elastic collision, where kinetic energy and momentum are conserved.

The direction of movement changes post-collision. Ball A initially travels east while Ball B is stationary. After the collision, Ball A changes direction, moving at an angle, and Ball B begins moving perpendicularly. Collision analysis involves evaluating how these direction changes impact velocity and momentum. Understanding angles and directions is essential. For Ball A, an angle of 40 degrees north of its original path affects how we calculate its momentum components.

• Types of collisions: elastic and inelastic
• Direction and angles are key in analyzing collisions
• Momentum components in different directions (x and y axes)

Momentum Calculation

Momentum is a measure of the motion of an object, calculated using the formula: \( p = mv \), where \( p \) is the momentum, \( m \) is the mass, and \( v \) is the velocity. Calculating momentum accurately helps predict post-collision velocities and directions. The law of conservation of momentum states that the total momentum before a collision equals the total momentum after a collision.

In this exercise, we initially calculate the momentum of Ball A, which has a mass of 0.500 kg and moves east at 0.800 m/s. Its initial momentum is \( 0.400 \; \text{kg} \cdot \text{m/s} \). After the collision, we find Ball A's new direction and final momentum. For Ball B, which starts from rest, calculation after collision becomes more complex as it involves momentum components in different directions. This method is key in determining both the magnitude and direction of each ball's movement post-collision.

• Initial momentum calculated using \( p = mv \)
• Conservation of momentum ensures total momentum is unchanged post-collision
• Complex calculations involve components (x and y directions)

Physics Problem Solving

Physics problem solving requires logical thinking and a systematic approach to analyze situations. In solving momentum-related problems like this one, it is crucial to work through the problem step-by-step. Start with a clear understanding of initial conditions and known values. Then apply the conservation laws and solve for unknowns one at a time.

In the given exercise, begin by calculating initial momentum values. Use conservation of momentum laws in both the x and y directions to understand how each component is conserved independently. Physics problem solving often involves breaking down complex concepts into simpler, solvable steps through equations. Remember to double-check each step for unit consistency and logical progression. Result validation through back-checking ensures the solution is both accurate and sensible.

• Breakdown of problems into manageable steps
• Application of conservation laws sequentially
• Regular validation of results and logical consistency