Answer:
Total momentum of the system is 378 kg-m/s
Step-by-step explanation:
It is given that,
Mass of first bumper car, m₁ = 222 kg
Velocity of first bumper car, v₁ = 3.10 m/s (in right)
Mass of other bumper car, m₂ = 165 kg
Velocity of second bumper car, v₂ = -1.88 m/s (in left)
Momentum of the system is given by the product of its mass and velocity. So, the total momentum of this system is given by :
![p=m_1v_1+m_2v_2](https://img.qammunity.org/2020/formulas/physics/college/rzvqeinphyfwrr9gqensxwyth3weo472i5.png)
![p=222\ kg* 3.10\ m/s+165\ kg* (-1.88\ m/s)](https://img.qammunity.org/2020/formulas/physics/college/quy6n0loj70v99pu380j45643eabhzu6n0.png)
p = 378 kg-m/s
Hence, the total momentum of the system is 378 kg-m/s