Problem 77 - College Physics

Question

A 0.30-kg puck, initially at rest on a frictionless horizontal surface, is struck by a 0.20-kg puck that is initially moving along the x-axis with a velocity of 2.0 m/s. After the collision, the 0.20-kg puck has a speed of 1.0 m/s at an angle of u 5 53° to the positive x-axis. (a) Determine the velocity of the 0.30-kg puck after the collision. (b) Find the fraction of kinetic energy lost in the collision.

Step-by-Step Solution

Answer

(a) The velocity of the 0.30-kg puck after the collision is 0.93 m/s in the positive x direction and -0.53 m/s in the y direction with the speed of 1.07 m/s.(b) The fraction of kinetic energy lost in the collision is 0.325 or 32.5%.

Step 1: Initial Total Momentum

First, calculate the total initial momentum before the collision which would only be due to the second puck (0.20-kg), because the first is at rest. The momentum can be found by multiplying mass and velocity, i.e., momentum = mass \times velocity:\( P_{initial} = m_{20} \times v_{20} = 0.20kg \times 2.0m/s = 0.40 kg.m/s \). So, the total initial momentum is 0.40 kg.m/s directed along the positive x-axis.

Step 2: Calculate momentum of 0.20-kg puck after collision

The momentum after the collision is along an angle of 53 degrees. So, we need to find its x and y-components. The x-component of momentum after collision, \( P_{20x, final} = m_{20} \times v_{20_{final}} \times cos(53) = 0.20kg \times 1.0m/s \times cos(53) = 0.12 kg.m/s \). The y-component of momentum after collision, \( P_{20y, final} = m_{20} \times v_{20_{final}} \times sin(53) = 0.20kg \times 1.0m/s \times sin(53) = 0.16 kg.m/s \).

Step 3: Determine the velocity of the 0.30-kg puck after the collision

Using the law of conservation of momentum, the total initial momentum must be equal to total final momentum. Therefore, the momentum of the 0.30-kg puck could be found by subtracting the final x and y-component of momentum of the 0.20-kg puck from the total momentum in the respective directions.Lets define the velocity of 0.30-kg puck in the x and y directions as \( v_{30x} \) and \( v_{30y} \) respectively.Then, \( P_{30x, final} = P_{initial} - P_{20x, final} = m_{30} \times v_{30x} \) => \( v_{30x} = (P_{initial} - P_{20x, final}) / m_{30} = (0.4 - 0.12) / 0.30 = 0.93 m/s \) in positive x direction. Similarly, \( P_{30y, final} = - P_{20y, final} = m_{30} \times v_{30y} \) => \( v_{30y} = - P_{20y, final} / m_{30} = - 0.16 / 0.30 = -0.53 m/s \) in negative y direction.The speed of the 0.30-kg puck could then be calculated using the Pythagorean Theorem: \( v_{30} = sqrt(v_{30x}^2 + v_{30y}^2) = sqrt((0.93)^2 + (-0.53)^2) = 1.07 m/s \).

Step 4: Calculate the initial kinetic energy

Now calculate the initial kinetic energy (KE) before the collision. Since the 0.30-kg puck is stationary, it has no kinetic energy. The kinetic energy for the 0.20-kg puck is KE=1/2*m*v^2. Hence \( KE_{initial} = 0.5 \times 0.20 \times (2.0)^2 = 0.40 J \).

Step 5: Calculate the final kinetic energy

We find the final kinetic energy by adding up the individual final kinetic energies of the two pucks. For the 0.20-kg puck: \( KE_{20final} = 0.5 \times 0.20 \times (1.0)^2 = 0.10 J \)For the 0.30-kg puck: \( KE_{30final} =0.5 \times 0.30 \times (1.07)^2 = 0.17 J \)Then add these two to get the total: \( KE_{final} = KE_{20final} + KE_{30final} = 0.10 J + 0.17 J = 0.27 J \)

Step 6: Calculate the fraction of kinetic energy lost

The fraction of the kinetic energy lost is equal to the difference between initial and final kinetic energies divided by the initial kinetic energy.Hence, \(KE_{lost} = (KE_{initial} - KE_{final}) / KE_{initial} = (0.40J - 0.27J) / 0.40J = 0.325 \) which gives us that 32.5% of the kinetic energy was lost during the collision.

Key Concepts

Kinetic Energy

Kinetic energy is a fundamental concept in physics that refers to the energy an object possesses due to its motion. You may think of it as the energy needed to accelerate an object from rest to a certain speed. In this problem, we calculate the kinetic energy associated with the motion of two pucks.
Initially, only the moving puck (0.20 kg) has kinetic energy because the other puck is at rest. The formula to calculate kinetic energy is:

• \( KE = \frac{1}{2} m v^2 \), where \( m \) is mass and \( v \) is velocity.

When the velocity changes due to a collision, so does the kinetic energy of the objects involved. However, because energy is often lost to factors like heat or sound during a collision, the total final kinetic energy is usually less than the initial kinetic energy. This phenomenon is clearly observed in inelastic collisions like the one in this exercise, where kinetic energy accounts for changes in the moving pucks' speeds.

Collision Physics

Collision physics deals with the interaction between two or more bodies when they collide. The nature of collisions can vary greatly, from perfectly elastic, where kinetic energy is conserved, to completely inelastic, where the objects stick together afterward, resulting in maximum energy loss.

• In elastic collisions, both momentum and kinetic energy are conserved.
• In inelastic collisions, only momentum is conserved, while kinetic energy is not.

In this scenario, the collision between the pucks is inelastic. We determine this by comparing the kinetic energy before and after the collision. The significant loss in kinetic energy (32.5% in this case) indicates that not all of the initial kinetic energy remains in the system after the collision. Understanding how collisions affect kinetic energy and momentum is crucial for grasping energy transformation and how forces interact in physics.

Momentum Components

Momentum, a key concept in physics, describes the quantity of motion an object has, linked directly to both its mass and velocity. It is a vector, meaning it has both magnitude and direction. This exercise helps understand its components, which are vital when examining objects moving in multiple directions.
The total momentum of a system remains constant in the absence of external forces, following the conservation of momentum principle. This exercise specifically uses the momentum conservation law to find components:

• For the x-component, momentum is disrupted due to the motion change, calculated as \( P_{20x} = 0.20 \times 1.0 \times \cos(53) = 0.12 \; \text{kg}\cdot\text{m/s} \).
• The y-component involves calculating \( P_{20y} = 0.20 \times 1.0 \times \sin(53) = 0.16 \; \text{kg}\cdot\text{m/s} \).

By determining both the x and y components separately, we can resolve complex motions into simpler 1-dimensional problems. Understanding these components allows us to successfully find the new velocity of the second puck using momentum conservation.