To solve this problem, we can use the principle of conservation of momentum. The momentum before the collision equals the momentum after the collision.
Let's define:
- \( m_1 = 20 \) kg (mass of the eastbound cart)
- \( v_1 = 6 \) m/s (velocity of the eastbound cart)
- \( m_2 = 30 \) kg (mass of the westbound cart)
- \( v_2 \) = unknown speed of the westbound cart, which we will consider as negative since it is moving in the opposite direction (west).
Before the collision, the total momentum of the system is given by the sum of the momenta of both carts:
\[ \text{Total momentum before} = m_1 v_1 + m_2 v_2 \]
Substituting the known values:
\[ \text{Total momentum before} = (20 , \text{kg})(6 , \text{m/s}) + (30 , \text{kg})(v_2) \]
After the collision, both carts come to rest, which means:
\[ \text{Total momentum after} = 0 \]
Setting the total momentum before equal to the total momentum after, we have:
\[ (20 , \text{kg})(6 , \text{m/s}) + (30 , \text{kg})(v_2) = 0 \]
Calculating the momentum of the eastbound cart:
\[ 120 , \text{kg m/s} + (30 , \text{kg})(v_2) = 0 \]
Now, we can solve for \( v_2 \):
\[ 30 , \text{kg}(v_2) = -120 , \text{kg m/s} \]
\[ v_2 = \frac{-120 , \text{kg m/s}}{30 , \text{kg}} = -4 , \text{m/s} \]
Thus, the speed of the westbound cart before the collision was 4 m/s (in the westward direction).
The final answer is:
4 meters per second (westbound).