Question

What is the magnitude of the momentum of a basketball player weighing 72 kg running towards the back court for a fast break play with a speed of 7m/s?

Answer 1

Solution. Using the definition of the momentum

$$p=mv$$

where m=72kg is the mass of the body, v=7m/s is the velocity of the body.

Therefore, the magnitude of the momentum is equal to

$$p=72kg\times7\frac{m}{s}=504\frac{kg\times m}{s}$$

Answer.

$$p=504\frac{kg\times m}{s}$$