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 second cart after the collision?

Answer 1

To solve the problem of the elastic collision between the two carts, we can use the conservation of momentum and the conservation of kinetic energy.

Let:
- \( m_1 = 5 \, \text{kg} \) (mass of the first cart)
- \( m_2 = 1 \, \text{kg} \) (mass of the second cart)
- \( u_1 = 0 \, \text{m/s} \) (initial velocity of the first cart)
- \( u_2 = 3 \, \text{m/s} \) (initial velocity of the second cart)

Let:
- \( v_1 \) be the final velocity of the first cart
- \( v_2 \) be the final velocity of the second cart

### Step 1: Conservation of Momentum
The total momentum before the collision must equal the total momentum after the collision.
\[
m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2
\]
Substituting the known values:
\[
5 \cdot 0 + 1 \cdot 3 = 5 v_1 + 1 v_2
\]
This simplifies to:
\[
3 = 5 v_1 + v_2 \quad \text{(1)}
\]

### Step 2: Conservation of Kinetic Energy
The total kinetic energy before the collision must equal the total kinetic energy after the collision.
\[
\frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2
\]
Substituting the known values:
\[
\frac{1}{2} \cdot 5 \cdot 0^2 + \frac{1}{2} \cdot 1 \cdot 3^2 = \frac{1}{2} \cdot 5 v_1^2 + \frac{1}{2} \cdot 1 v_2^2
\]
This simplifies to:
\[
\frac{1}{2} \cdot 1 \cdot 9 = \frac{1}{2} \cdot 5 v_1^2 + \frac{1}{2} \cdot v_2^2
\]
\[
4.5 = \frac{5}{2} v_1^2 + \frac{1}{2} v_2^2
\]
Multiply everything by 2 to eliminate the fraction:
\[
9 = 5 v_1^2 + v_2^2 \quad \text{(2)}
\]

### Step 3: Solving the System of Equations
Now we have a system of two equations:
1. \( 3 = 5 v_1 + v_2 \)
2. \( 9 = 5 v_1^2 + v_2^2 \)

From equation (1), we can express \( v_2 \) in terms of \( v_1 \):
\[
v_2 = 3 - 5 v_1
\]

Substituting \( v_2 \) into equation (2):
\[
9 = 5 v_1^2 + (3 - 5 v_1)^2
\]
Expanding the square:
\[
9 = 5 v_1^2 + (9 - 30 v_1 + 25 v_1^2)
\]
Combining the terms:
\[
9 = 30 v_1^2 - 30 v_1 + 9
\]
Subtracting 9 from both sides:
\[
0 = 30 v_1^2 - 30 v_1
\]
Factoring out \( 30 v_1 \):
\[
0 = 30 v_1(v_1 - 1)
\]
This gives two solutions:
\[
v_1 = 0 \quad \text{or} \quad v_1 = 1
\]

### Step 4: Finding \( v_2 \)
If \( v_1 = 0 \), then substituting into (1):
\[
v_2 = 3 - 5(0) = 3
\]
This is the initial condition, so we discard this solution.

If \( v_1 = 1 \):
\[
v_2 = 3 - 5(1) = 3 - 5 = -2
\]

### Final Answer
Thus, the velocity of the second cart after the collision is:
\[
\boxed{-2 \, \text{m/s}}
\]