Question

A 77.0-kg fullback running east with a speed of 5.60 m/s is tackled by a 95.0-kg opponent running north with a speed of 3.00 m/s. Calculate the velocity of the players immediately after the tackle.

Answer 1

Step-by-step explanation:

This is a Law of Momentum Conservation problem involving an inelastic collision (where the 2 objects are stuck together and move as one mass after they collide). The Law of Momentum conservation says that the momentum after the collision has to be the same as it was before. The only thing that changes is the kinetic energy (because the velocity is changed). The player running east has a momentum of

p = (77.0)(5.60) so

p =
`431(kg*m)/(s)` and the player running north has a momentum of

p = (95.0)(3.00) so

p =
`285(kg*m)/(s)`

If you learned anything about vector addition to get to this point in physics, you know that in order to add vectors you have to take use a somewhat glorified version of Pythagorean's Theorem to find the resultant vector, which, in our case will be the final momentum. Plugging in:

`p_f=√((431)^2+(285)^2)` so

`p_f=517(kg*m)/(s)` What we are looking for, though, is not the final momentum but the final velocity. The equation for momentum is

p = mv, but don't forget that our 2 players are stuck together after the collision, so they move as one mass, altering our equation to look like this:

`p=(m_1+m_2)v` and filling in:

`517=172.0v` and

v = 3.00 m/s. We can find the angle, too (might as well, since it's usually part of this type of problem).

`\theta=tan^(-1)((285)/(431))` so

`\theta=` 33.5° N of E

`517=172.0v`