4. A time traveling brontosaurus with a mass of \( 22,226 \mathrm{~kg} \) is running with a velocity of \( 75 \mathrm{~m} / \mathrm{s} \) East. It collides head-on with a semi-truck with a mass of \( 9,100 \mathrm{~kg} \) moving in the opposite direction of \( 120 \mathrm{~m} / \mathrm{s} \) (west), which results in an elastic collision, sending the semi-truck backward with a final velocity of \( 5 \mathrm{~m} / \) s(east). What is the final velocity of the brontosaurus?

In an elastic collision, both momentum and kinetic energy are conserved. First, we can calculate the initial momentum of both objects. The momentum of the brontosaurus (\(p_{B}\)) is \(22,226 \, \text{kg} \times 75 \, \text{m/s} = 1,666,950 \, \text{kg m/s}\) (east). The momentum of the semi-truck (\(p_{T}\)) is \(9,100 \, \text{kg} \times (-120 \, \text{m/s}) = -1,092,000 \, \text{kg m/s}\) (west). The total initial momentum is: \[ p_{initial} = p_{B} + p_{T} = 1,666,950 + (-1,092,000) = 574,950 \, \text{kg m/s}. \] Next, after the collision, we know the final momentum of the truck is: \[ p'_{T} = 9,100 \, \text{kg} \times 5 \, \text{m/s} = 45,500 \, \text{kg m/s}. \] We set the total momentum after the collision equal to the total momentum before the collision: \[ 574,950 = p'_{B} + 45,500. \] Where \(p'_{B}\) is the final momentum of the brontosaurus. Thus, \[ p'_{B} = 574,950 - 45,500 = 529,450 \, \text{kg m/s}. \] To find the final velocity of the brontosaurus \(v'_{B}\), we divide its final momentum by its mass: \[ v'_{B} = \frac{p'_{B}}{m_{B}} = \frac{529,450 \, \text{kg m/s}}{22,226 \, \text{kg}} \approx 23.83 \, \text{m/s}. \] Since momentum is a vector quantity, we must also point out that the collision sends the brontosaurus in the same direction as it originally was (east). Therefore, the final velocity of the brontosaurus is approximately \(23.83 \, \text{m/s}\) east.