Two hockey players of opposite teams, while trying to hit a hockey ball on the ground collide and immediately become entangled. Once has a mass of `60kg`, and was moving with a velocity `5*0m//s`, while the other has a mass of `55kg` and was moving faster with a velocity of `6*0m//s` towards the first player. In which direction and with what velocity will they move after they become entangled? Assume that the frictional force acting between the feet of the two players and ground is negligible.

Explanation

If v is the velocity of the two entangled players after the collision, the total players after the collision, the total momentum then
`=(m_(1)+m_(2))xxv`
`=(60+55) kg xx v ms^(-1)`
`=155xx v kg ms^(-1)`.
Equating the momenta of the system before and after collision, in accordance with the law of consevation of momentum , we get
`v= - 30//115`
`= - 0.26 ms^(-1)`
Thus, the two entangled players would move with velocity `0.26 ms^(-1)` from right to left, that is , in the direction the second player was moving before the collision.

Let the first player be moving from left to right. By concention left to right is taken as the positive direction and thus right to left negative direction ( Fig. 9.19). If symbols m and u represent the mass and initial velocity of the two players, respectively. Subscripts 1 and 2 in these physical quantities refer to the two hockey players. Thus,
`m_(1)=60 kg ,u_(1)=+5 s^(-1) ,` and
`m_(2)=55 kg , u_(2)= -6 kg s^(-1)`.
The total momentum of the two players before the collision
`=60 kg xx(+5 ms^(-1))+ 55 kg xx(-6 ms^(-1))`
`=-30 kg ms^(-1)`