Question

A physics teacher (mass = 80.4 kg) is jogging through the woods and runs straight into a large oak tree at 4.9 m/s. Rebound speed is measured at 4.0 m/s in the opposite direction. If the time of contact with the tree is 56 milliseconds, what is the magnitude of the force that the tree exerts on the teacher?

Answer 1

12777.86N

Step-by-step explanation:

This problem is approached by understanding that the impulse felt by the physics teacher is equal to his change in momentum. This is mathematically represented as follows;

`Ft=mv-(-mu)................(1)`

where m is his mass, v is his final velocity, u is his initial velocity, t is the time for which the force F acts on him.

It should also be noted that while he rebounds there is a change in his direction and that is why the mu carries a negative sign.

Equation (1) can therefore be written as,

`Ft=mv+mu\\F=(m(v+u))/(t).............(2)`

equation (2) is a statement of Newton's second law of motion.

Given;

m = 80.4kg

u = 4.9m/s

v = 4.0m/s

t = 56 milliseconds = 0.056s

Therefore;

`F=(80.4(4.9+4.0))/(0.056)\\F=(80.4*8.9)/(0.056)\\F=12777.86N`