Problem 76 - College Physics

Question

\(\bullet\) A blue puck with mass 0.0400 \(\mathrm{kg}\) , sliding with a velocity of magnitude 0.200 \(\mathrm{m} / \mathrm{s}\) on a frictionless, horizontal air table, makes a perfectly elastic, head-on collision with a red puck with mass \(m,\) initially at rest. After the collision, the velocity of the blue puck is 0.050 \(\mathrm{m} / \mathrm{s}\) in the same direction as its initial velocity. Find (a) the velocity (magnitude and direction) of the red puck after the collision; and (b) the mass \(m\) of the red puck.

Step-by-Step Solution

Answer

The red puck moves at 0.125 m/s, and its mass is 0.048 kg.

Step 1: Conservation of Momentum

Use the law of conservation of linear momentum, which states that the total momentum before the collision is equal to the total momentum after the collision. Let \( m_b = 0.0400 \, \text{kg} \) be the mass of the blue puck and \( v_b = 0.200 \, \text{m/s} \) be its initial velocity. The red puck is initially at rest, so its initial velocity is 0. The final velocity of the blue puck is \( v_{b_f} = 0.050 \, \text{m/s} \). Let \( v_{r_f} \) be the final velocity of the red puck and \( m \) be the mass of the red puck. The conservation of momentum equation is: \ m_b \cdot v_b + 0 = m_b \cdot v_{b_f} + m \cdot v_{r_f}.

Step 2: Solve for Red Puck's Velocity

Rearrange the momentum equation from Step 1 to solve for \( v_{r_f} \): \ v_{r_f} = \frac{m_b \cdot (v_b - v_{b_f})}{m}. \ Substitute the known values into the equation: \ v_{r_f} = \frac{0.0400 \, \text{kg} \times (0.200 \, \text{m/s} - 0.050 \, \text{m/s})}{m} \ Simplifying, we get \ v_{r_f} = \frac{0.0400 \, \text{kg} \times 0.150 \, \text{m/s}}{m} \ v_{r_f} = \frac{0.0060 \, \text{kg} \cdot \text{m/s}}{m}.

Step 3: Conservation of Kinetic Energy

Use the conservation of kinetic energy because the collision is perfectly elastic. Initial kinetic energy \( KE_i \) of the system is: \ KE_i = \frac{1}{2} m_b v_b^2. \ Final kinetic energy \( KE_f \) of the system is given by the sum of kinetic energy of both pucks after collision: \ KE_f = \frac{1}{2} m_b v_{b_f}^2 + \frac{1}{2} m v_{r_f}^2. \ Equating \( KE_i \) to \( KE_f \), we have: \ \frac{1}{2} m_b v_b^2 = \frac{1}{2} m_b v_{b_f}^2 + \frac{1}{2} m v_{r_f}^2.

Step 4: Simplify and Solve for Mass of Red Puck

Substituting the values and simplifying the kinetic energy equation, \ 0.5 \cdot 0.0400 \cdot (0.200)^2 = 0.5 \cdot 0.0400 \cdot (0.050)^2 + 0.5 \cdot m \cdot \left(\frac{0.0060}{m}\right)^2. \ Solving for \( m \), \ 0.5 \cdot 0.0400 \times 0.04 = 0.5 \cdot 0.0400 \times 0.0025 + 0.5 \cdot \frac{0.0060^2}{m}. \ Simplifying the equation, we get: \ 0.0008 = 0.00005 + \frac{0.000036}{m}, \ 0.00075 = \frac{0.000036}{m}, \ m = \frac{0.000036}{0.00075} = 0.048 \text{ kg}.

Step 5: Calculate Red Puck's Velocity with Known Mass

Now that we know \( m = 0.048 \text{ kg} \), substitute it into the equation from Step 2 to find \( v_{r_f} \): \ v_{r_f} = \frac{0.0060}{0.048} \ Thus, \( v_{r_f} = 0.125 \, \text{m/s}\). The velocity is in the direction of the initial blue puck's motion.

Key Concepts

Conservation of Momentum

In every collision, whether elastic or inelastic, the principle of conservation of momentum is a key concept. This principle states that the total momentum of a system of particles remains constant if no external forces act on the system. Momentum, a vector quantity, is the product of an object's mass and velocity. In our problem, the blue puck initially possesses all the momentum since the red puck is at rest.

To calculate the conservation of momentum during the collision, we use the formula:

• Initial Total Momentum = Final Total Momentum.
• \[ m_b imes v_b = m_b imes v_{b_f} + m imes v_{r_f} \]

where:

• \( m_b \) is the mass of the blue puck.
• \( v_b \) and \( v_{b_f} \) are the initial and final velocities of the blue puck, respectively.
• \( m \) is the mass of the red puck, initially at rest, so its initial velocity is zero.
• \( v_{r_f} \) is the velocity of the red puck after the collision.

This allows us to solve for the unknown variables in the problem, like the mass and velocity of the red puck.

Kinetic Energy

Kinetic energy, the energy an object possesses due to its motion, also plays an integral role in analyzing elastic collisions. For an elastic collision, such as in the given problem, not only is momentum conserved but kinetic energy is too.

The formula for kinetic energy is:

• \( KE = \frac{1}{2} mv^2 \)

In an elastic collision scenario:

• Initial Kinetic Energy = Final Kinetic Energy.
• \[ \frac{1}{2} m_b v_b^2 = \frac{1}{2} m_b v_{b_f}^2 + \frac{1}{2} m v_{r_f}^2 \]

This states that the sum of the particles' kinetic energy before impact is equal to the sum after impact.

In applying this to the exercise, knowing the initial kinetic energies of both pucks and using the equations, one can solve for unknowns such as the red puck's mass and its velocity after collision. This part of the theoretical framework ensures the integrity of the final solution.

Collision Mechanics

The mechanics of collision involve understanding how two objects interact during the contact phase. In physics, an elastic collision is one where both momentum and kinetic energy are conserved. This type of collision is idealized because it assumes no energy is lost to sound, heat, or deformation.

During the collision process:

• The objects exert equal but opposite forces on each other, as per Newton's third law.
• The duration of contact is typically very short, making calculations focus mostly on initial and final states attributes of the objects involved.

Since our problem features a head-on collision on a frictionless surface, it simplifies the analysis greatly:

• No frictional force affects the system.
• The velocity after collision becomes directly calculable.

Understanding collision mechanics helps explain the resultant movements and can be applied to predict outcomes in various physical scenarios where studies of collisions are relevant.

Physics Problems

Physics problems can seem daunting because of the required understanding of multiple principles and the mathematics involved. However, breaking them down into smaller, manageable parts can simplify the process.

The exercise involves concepts such as:

• Identifying known values from the problem statement.
• Applying appropriate physical laws, like the conservation of momentum and kinetic energy.
• Using algebra to solve for unknowns like mass or velocity.

Common steps to approach physics problems include:

• Reading the problem carefully.
• Listing out known and unknown quantities.
• Writing down relevant equations.
• Solving the equations step by step.
• Checking the solution against physical intuition and ensuring the units make sense.

Solid grasp of fundamentals combined with practice equips students to tackle similar problems efficiently in real-world physics applications and exams. This systematic approach ensures clarity and understanding of the problem-solving process.