Problem 57 - College Physics

Question

A proton of mass \(m\) moving with a speed of \(3.0 \times 10^{6} \mathrm{~m} / \mathrm{s}\) undergoes a head-on elastic collision with an alpha particle of mass \(4 m\), which is initially at rest. What are the velocities of the two particles after the collision?

Step-by-Step Solution

Answer

The proton moves at \(-5.0 \times 10^5\) m/s, and the alpha particle at \(2.0 \times 10^6\) m/s.

Step 1: Understand the Problem

We have a head-on elastic collision between a proton and an alpha particle. A proton has mass \(m\) and is moving with a speed of \(3.0 \times 10^6 \; \text{m/s}\). An alpha particle has mass \(4m\) and is at rest initially. We need to find their velocities after the collision.

Step 2: Apply Conservation of Momentum

Since momentum is conserved, the total momentum before collision equals the total momentum after collision. The momentum before collision is \(m \cdot 3.0 \times 10^6\) and after collision it is \(m \cdot v_p + 4m \cdot v_a\), where \(v_p\) and \(v_a\) are the final velocities of the proton and alpha particle respectively. Thus, we can set up the equation:\[ m \cdot 3.0 \times 10^6 = m \cdot v_p + 4m \cdot v_a \]

Step 3: Apply Conservation of Kinetic Energy

Since it is an elastic collision, kinetic energy is also conserved. The total kinetic energy before collision is \(\frac{1}{2}m(3.0 \times 10^6)^2\). After collision, the kinetic energy is \(\frac{1}{2}m v_p^2 + \frac{1}{2}(4m) v_a^2\). Therefore:\[ \frac{1}{2}m(3.0 \times 10^6)^2 = \frac{1}{2}m v_p^2 + \frac{1}{2}(4m) v_a^2 \]

Step 4: Set up the Equations

From momentum conservation:\[ 3.0 \times 10^6 = v_p + 4v_a \]From kinetic energy conservation:\[ (3.0 \times 10^6)^2 = v_p^2 + 4v_a^2 \]

Step 5: Solve the Equations

Firstly solve the momentum equation for \(v_p\):\[ v_p = 3.0 \times 10^6 - 4v_a \]Substitute \(v_p\) in the kinetic energy equation:\[ (3.0 \times 10^6)^2 = (3.0 \times 10^6 - 4v_a)^2 + 4v_a^2 \]Expand and simplify the equation:\[ 9 \times 10^{12} = (9 \times 10^{12} - 2.4 \times 10^7 v_a + 16v_a^2) + 4v_a^2 \]\[ 12v_a^2 = 2.4 \times 10^7 v_a \]Solve for \(v_a\):\[ v_a = 2.0 \times 10^6 \; \text{m/s} \]Substitute \(v_a\) back to find \(v_p\):\[ v_p = 3.0 \times 10^6 - 4 imes 2.0 \times 10^6 = -5.0 \times 10^5 \; \text{m/s} \]

Step 6: Interpret the Results

The negative sign of \(v_p\) indicates the proton is moving in the opposite direction after the collision. Hence, the final velocities are:- The proton has a velocity of \(-5.0 \times 10^5\) m/s.- The alpha particle has a velocity of \(2.0 \times 10^6\) m/s.

Key Concepts

Conservation of Momentum

In any collision, one of the foundational principles is the conservation of momentum. Momentum is the product of mass and velocity and is represented as \( p = mv \). In an elastic collision, like the one between the proton and alpha particle in our exercise, the total momentum of the system is conserved.

This means that the total momentum before the collision equals the total momentum after the collision. For our problem:

• The momentum of the proton initially is \( m \cdot 3.0 \times 10^6 \) because the proton has a mass \( m \) and velocity \( 3.0 \times 10^6 \) m/s.
• The alpha particle is initially at rest, so its initial momentum is zero.
• After the collision, the momenta of the proton and alpha particle will change to \( m \cdot v_p + 4m \cdot v_a \), where \( v_p \) and \( v_a \) are their respective final velocities.

Setting up the conservation of momentum equation gives:\[ m \cdot 3.0 \times 10^6 = m \cdot v_p + 4m \cdot v_a \]This equation is essential to solve for the unknown velocities after the collision.

Conservation of Kinetic Energy

Another critical principle in elastic collisions is the conservation of kinetic energy. Kinetic energy is the energy that an object possesses due to its motion and is calculated using the formula \( KE = \frac{1}{2}mv^2 \). For elastic collisions, kinetic energy is conserved just like momentum. This means the total kinetic energy before and after the collision remains unchanged.

In our case:

• The initial kinetic energy of the proton is \( \frac{1}{2}m(3.0 \times 10^6)^2 \).
• The alpha particle starts from rest and thus has no initial kinetic energy.
• The total kinetic energy after the collision changes to the sum \( \frac{1}{2}m v_p^2 + \frac{1}{2}(4m) v_a^2 \).

Thus, the equation reflecting the conservation of kinetic energy in this context is:\[ \frac{1}{2}m(3.0 \times 10^6)^2 = \frac{1}{2}m v_p^2 + \frac{1}{2}(4m) v_a^2 \]This equation complements the momentum equation by serving as the second equation needed to solve for the two unknowns in the system.

Collision Velocity Calculation

Once we have the momentum and kinetic energy equations, the next step is solving for the velocities of the particles after the collision. This involves a bit of algebra to separate and solve the equations.

Here’s the streamlined process:

• First, solve the momentum equation for one of the velocities. In this problem, we solved for the proton's velocity, \( v_p \), in terms of \( v_a \) as \( v_p = 3.0 \times 10^6 - 4v_a \).
• Next, substitute this expression into the kinetic energy equation. This substitution creates a solvable equation in terms of only one variable, \( v_a \).
• Expand and simplify the resulting equation, then solve for \( v_a \).

After finding \( v_a \), substitute it back into the expression for \( v_p \) to find the proton's velocity.
In our specific case:

• \( v_a = 2.0 \times 10^6 \; \text{m/s} \)
• \( v_p = -5.0 \times 10^5 \; \text{m/s} \)

The negative sign of \( v_p \) tells us that the proton reverses direction after the collision.Thoroughly applying these principles and steps ensures accurate determination of post-collision velocities in elastic collision problems.