Question
A 5 kg metal cart is sitting at rest on a frictionless ice surface. Another metal cart whose mass is 1 kg is propelled at the cart and strikes it in a one-dimensional elastic collision. If the original velocity of the second cart was 3 m/s, what is the velocity of the first cart after the collision?
Answers
GPT-4o mini
In an elastic collision, both momentum and kinetic energy are conserved. We can use these conservation laws to solve for the velocities after the collision.
Let's denote:
- \( m_1 = 5 \, \text{kg} \) (mass of the first cart)
- \( v_1 = 0 \, \text{m/s} \) (initial velocity of the first cart)
- \( m_2 = 1 \, \text{kg} \) (mass of the second cart)
- \( v_2 = 3 \, \text{m/s} \) (initial velocity of the second cart)
- \( v_1' \) (final velocity of the first cart)
- \( v_2' \) (final velocity of the second cart)
### Step 1: Conservation of Momentum
The equation for conservation of momentum is:
\[
m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2'
\]
Substituting the values we know:
\[
5 \cdot 0 + 1 \cdot 3 = 5 v_1' + 1 v_2'
\]
This simplifies to:
\[
3 = 5 v_1' + v_2' \quad \text{(1)}
\]
### Step 2: Conservation of Kinetic Energy
The equation for conservation of kinetic energy is:
\[
\frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 = \frac{1}{2} m_1 {v_1'}^2 + \frac{1}{2} m_2 {v_2'}^2
\]
Substituting the known values:
\[
\frac{1}{2} \cdot 5 \cdot 0^2 + \frac{1}{2} \cdot 1 \cdot 3^2 = \frac{1}{2} \cdot 5 \cdot {v_1'}^2 + \frac{1}{2} \cdot 1 \cdot {v_2'}^2
\]
This simplifies to:
\[
\frac{1}{2} \cdot 1 \cdot 9 = \frac{1}{2} \cdot 5 \cdot {v_1'}^2 + \frac{1}{2} \cdot 1 \cdot {v_2'}^2
\]
\[
4.5 = \frac{5}{2} {v_1'}^2 + \frac{1}{2} {v_2'}^2 \quad \text{(2)}
\]
### Step 3: Solve the system of equations
From equation (1):
\[
v_2' = 3 - 5 v_1' \quad \text{(3)}
\]
Substituting equation (3) into equation (2):
\[
4.5 = \frac{5}{2} {v_1'}^2 + \frac{1}{2} (3 - 5 v_1')^2
\]
Expanding this:
\[
4.5 = \frac{5}{2} {v_1'}^2 + \frac{1}{2} (9 - 30 v_1' + 25 {v_1'}^2)
\]
\[
4.5 = \frac{5}{2} {v_1'}^2 + \frac{1}{2} \cdot 9 - 15 v_1' + \frac{25}{2} {v_1'}^2
\]
Combining like terms:
\[
4.5 = \left( \frac{5}{2} + \frac{25}{2} \right) {v_1'}^2 - 15 v_1' + \frac{9}{2}
\]
\[
4.5 = 15 {v_1'}^2 - 15 v_1' + 4.5
\]
Subtract \( 4.5 \) from both sides:
\[
0 = 15 {v_1'}^2 - 15 v_1'
\]
Factoring out \( 15 v_1' \):
\[
0 = 15 v_1' (v_1' - 1)
\]
This gives us the solutions:
\[
v_1' = 0 \quad \text{or} \quad v_1' = 1 \, \text{m/s}
\]
### Conclusion
The first cart (5 kg) has a final velocity of \( \mathbf{1 \, m/s} \) after the collision.
Let's denote:
- \( m_1 = 5 \, \text{kg} \) (mass of the first cart)
- \( v_1 = 0 \, \text{m/s} \) (initial velocity of the first cart)
- \( m_2 = 1 \, \text{kg} \) (mass of the second cart)
- \( v_2 = 3 \, \text{m/s} \) (initial velocity of the second cart)
- \( v_1' \) (final velocity of the first cart)
- \( v_2' \) (final velocity of the second cart)
### Step 1: Conservation of Momentum
The equation for conservation of momentum is:
\[
m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2'
\]
Substituting the values we know:
\[
5 \cdot 0 + 1 \cdot 3 = 5 v_1' + 1 v_2'
\]
This simplifies to:
\[
3 = 5 v_1' + v_2' \quad \text{(1)}
\]
### Step 2: Conservation of Kinetic Energy
The equation for conservation of kinetic energy is:
\[
\frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 = \frac{1}{2} m_1 {v_1'}^2 + \frac{1}{2} m_2 {v_2'}^2
\]
Substituting the known values:
\[
\frac{1}{2} \cdot 5 \cdot 0^2 + \frac{1}{2} \cdot 1 \cdot 3^2 = \frac{1}{2} \cdot 5 \cdot {v_1'}^2 + \frac{1}{2} \cdot 1 \cdot {v_2'}^2
\]
This simplifies to:
\[
\frac{1}{2} \cdot 1 \cdot 9 = \frac{1}{2} \cdot 5 \cdot {v_1'}^2 + \frac{1}{2} \cdot 1 \cdot {v_2'}^2
\]
\[
4.5 = \frac{5}{2} {v_1'}^2 + \frac{1}{2} {v_2'}^2 \quad \text{(2)}
\]
### Step 3: Solve the system of equations
From equation (1):
\[
v_2' = 3 - 5 v_1' \quad \text{(3)}
\]
Substituting equation (3) into equation (2):
\[
4.5 = \frac{5}{2} {v_1'}^2 + \frac{1}{2} (3 - 5 v_1')^2
\]
Expanding this:
\[
4.5 = \frac{5}{2} {v_1'}^2 + \frac{1}{2} (9 - 30 v_1' + 25 {v_1'}^2)
\]
\[
4.5 = \frac{5}{2} {v_1'}^2 + \frac{1}{2} \cdot 9 - 15 v_1' + \frac{25}{2} {v_1'}^2
\]
Combining like terms:
\[
4.5 = \left( \frac{5}{2} + \frac{25}{2} \right) {v_1'}^2 - 15 v_1' + \frac{9}{2}
\]
\[
4.5 = 15 {v_1'}^2 - 15 v_1' + 4.5
\]
Subtract \( 4.5 \) from both sides:
\[
0 = 15 {v_1'}^2 - 15 v_1'
\]
Factoring out \( 15 v_1' \):
\[
0 = 15 v_1' (v_1' - 1)
\]
This gives us the solutions:
\[
v_1' = 0 \quad \text{or} \quad v_1' = 1 \, \text{m/s}
\]
### Conclusion
The first cart (5 kg) has a final velocity of \( \mathbf{1 \, m/s} \) after the collision.