Two carts with masses of 4.24 kg and 3.44 kg move toward each other on a frictionless track with speeds of 4.90 m/s and 3.51 m/s respectively. The carts stick together after colliding head-on. Find the final speed.
To find the final speed of the two carts after collision, we can apply the principle of conservation of momentum.
The momentum of an object is given by the product of its mass and velocity:
\[ momentum = mass \times velocity \]
According to the principle of conservation of momentum, the total momentum before the collision should be equal to the total momentum after the collision.
Let's denote the initial velocities of the two carts as \(v_1\) and \(v_2\), where \(v_1 = 4.90 \, \text{m/s}\) is the velocity of the first cart and \(v_2 = -3.51 \, \text{m/s}\) is the velocity of the second cart. The negative sign for \(v_2\) indicates that the second cart is moving in the opposite direction.
We can calculate the initial momentum (\(p\)) of each cart using the formula:
\[ p = m \times v \]
where \(m\) is the mass of the cart and \(v\) is its velocity.
For the first cart:
Initial momentum of the first cart (\(p_1\)) = mass of the first cart (\(m_1\)) x velocity of the first cart (\(v_1\))
\[ p_1 = m_1 \times v_1 = 4.24 \, \text{kg} \times 4.90 \, \text{m/s} \]
For the second cart:
Initial momentum of the second cart (\(p_2\)) = mass of the second cart (\(m_2\)) x velocity of the second cart (\(v_2\))
\[ p_2 = m_2 \times v_2 = 3.44 \, \text{kg} \times (-3.51) \, \text{m/s} \]
Since the carts stick together after collision, the final velocity (\(v_f\)) of the combined carts is the same.
The total momentum before the collision is given by:
\[ p_{\text{total\_initial}} = p_1 + p_2 \]
The total momentum after the collision is given by:
\[ p_{\text{total\_final}} = (m_1 + m_2) \times v_f \]
Since momentum is conserved, we can equate the total initial momentum to the total final momentum:
\[ p_{\text{total\_initial}} = p_{\text{total\_final}} \]
\[ p_1 + p_2 = (m_1 + m_2) \times v_f \]
Substituting the values we have:
\[ 4.24 \, \text{kg} \times 4.90 \, \text{m/s} + 3.44 \, \text{kg} \times (-3.51) \, \text {m/s} = (4.24 \, \text{kg} + 3.44 \, \text{kg}) \times v_f \]
\[ 20.776 \, \text{kg m/s} + (-11.9944) \, \text{kg m/s} = 7.68 \, \text{kg} \times v_f \]
\[ 8.7816 \, \text{kg m/s} = 7.68 \, \text{kg} \times v_f \]
Simplifying:
\[ v_f = \frac{8.7816 \, \text{kg m/s}}{7.68 \, \text{kg}} \]
\[ v_f = 1.142 \, \text{m/s} \]
Therefore, the final speed of the two carts after collision is \(1.142 \, \text{m/s}\).