Problem 96 - University Physics with Modern Physics

Question

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

Red puck's velocity: 2.64 m/s; mass: 0.00227 kg.

Step 1: Understand the Problem

The exercise involves a collision between two pucks on a frictionless surface where momentum and energy are conserved, as it is a perfectly elastic collision. The initial conditions for the blue puck are given, and we need to find the post-collision velocity of the red puck and its mass.

Step 2: Write down the conservation of momentum equation

For the system, use conservation of momentum: \( m_1 \cdot v_{1i} + m_2 \cdot v_{2i} = m_1 \cdot v_{1f} + m_2 \cdot v_{2f} \). Here, \(m_1 = 0.0400 \text{ kg}, v_{1i} = 0.200 \text{ m/s}\) (initial blue puck velocity), \(v_{2i} = 0\) (red puck initially at rest), \(v_{1f} = 0.050 \text{ m/s}\) (blue puck final velocity). Substitute these into the equation to get \(0.0400 \times 0.200 = 0.0400 \times 0.050 + m \cdot v_{2f}\).

Step 3: Solve for final velocity of red puck

Rearrange the equation from Step 2 to solve for \(v_{2f}\): \(v_{2f} = \frac{0.0400 \times 0.150}{m} = \frac{0.0060}{m}\). This gives the expression for the final velocity of the red puck.

Step 4: Write down the conservation of kinetic energy equation

In an elastic collision, kinetic energy is conserved: \( \frac{1}{2} m_1 v_{1i}^2 = \frac{1}{2} m_1 v_{1f}^2 + \frac{1}{2} m_2 v_{2f}^2 \). Substitute known values \( \frac{1}{2} \times 0.0400 \times 0.200^2 = \frac{1}{2} \times 0.0400 \times 0.050^2 + \frac{1}{2} \times m \times \left(\frac{0.0060}{m}\right)^2 \). Simplify to \(0.0080 - 0.0001 = \frac{0.0060^2}{2m}\).

Step 5: Solve for mass of the red puck

Rearrange the equation from Step 4 to solve for \(m\): \( 0.0079 = \frac{0.000036}{2m} \), which simplifies to \( m = \frac{0.000036}{2 \times 0.0079} = 0.00227 \text{ kg}\).

Step 6: Substitute mass to find the velocity of red puck

Now that we have the mass \(m = 0.00227\text{ kg}\), substitute it back into the equation for \(v_{2f}\): \(v_{2f} = \frac{0.0060}{0.00227} \approx 2.643\text{ m/s}\). This is the velocity of the red puck post-collision in the same direction as the initial velocity of the blue puck.

Key Concepts

Conservation of Momentum

Momentum is like the fairy godmother of physics, ensuring that all is balanced in motion. In a frictionless and closed system, such as the collision scenario described, the total momentum before the collision equals the total momentum after the collision.

This concept is crucial for solving problems related to elastic collisions. Momentum, defined as the product of mass and velocity, is symbolized by the equation: \( p = mv \). During our puck collision, the principle can be expressed using the formula:

• Initial Momentum = Final Momentum
• \( m_1 \cdot v_{1i} + m_2 \cdot v_{2i} = m_1 \cdot v_{1f} + m_2 \cdot v_{2f} \)

Here, \( m_1 \) and \( m_2 \) represent the masses of the blue and red pucks, respectively. The given velocities represent the initial and final states of the pucks.

By plugging in the known values, you can solve for unknown variables such as the final velocity of the red puck after the collision, ensuring the balance of momentum remains intact.

Conservation of Kinetic Energy

Kinetic energy is the energy of motion, and in elastic collisions like ours, it is carefully preserved. This means that the total kinetic energy before the collision is equal to the total kinetic energy after the collision.

The equation for kinetic energy (KE) is expressed as:

• \( KE = \frac{1}{2} mv^2 \)

For a collision between two objects, the conservation of kinetic energy is described by:

• \( \frac{1}{2} m_1 v_{1i}^2 + \frac{1}{2} m_2 v_{2i}^2 = \frac{1}{2} m_1 v_{1f}^2 + \frac{1}{2} m_2 v_{2f}^2 \)

In our exercise, this principle helps us determine unknowns like the mass of the red puck.
After expressing the equivalence and substituting known values, you can solve for missing quantities such as the mass \( m \) of the red puck. This equation accounts for every watt of energy, illustrating how it's conserved in elastic collisions.

Head-On Collision

A head-on collision occurs when two objects move directly towards each other along the same line. In our discussion, imagine two hockey pucks on a perfectly smooth ice rink.

Such a collision involves two objects colliding along a single, straight path. These collisions are fundamental to understanding momentum and energy distribution.
Understanding head-on collisions:

• The total momentum of the system is conserved.
• Kinetic energy remains constant through the collision.
• The direction(s) of the velocities after the collision are determined by initial conditions. In perfectly elastic collisions, objects may bounce back or one might gain speed depending on mass and initial velocity profiles.

For the exercise at hand, the head-on nature simplifies complexity because all movements are linear and direct, making calculations straightforward based on known principles of momentum and energy conservation.