Question

During a goal-line stand, a 75-kg fullback moving eastward with a speed of 10m/s collides head-on with a 100-kg lineman moving westward with a speed of 4 m/s. The two players collide and stick together, moving at the same velocity after the collision. Determine the post collision velocity of the two players.

Answer 1

The post-collision velocity of the two players is
`v= 2 ms^(-1)`.

Step-by-step explanation:

The expression for the conservation of momentum is as follows as;

`m_(1)u_(1)+m_(2)u_(2)=m_(1)v_(1)+m_(2)v_(2)`

Here,
`m_(1)` is the mass of the first object,
`m_(2)` is the mass of the second object,
`u_(1)` and
`u_(2)` are the initial velocities of the first and second objects and
`v_(1)` and
`v_(2)` are the final velocities of the objects.

It is given in the problem that during a goal-line stand, a 75-kg fullback moving eastward with a speed of 10m/s collides head-on with a 100-kg lineman moving westward with a speed of 4 m/s. The two players collide and stick together, moving at the same velocity after the collision,
`v_(1)=v_(2)`.

Calculate the post collision velocity of the two players.

`m_(1)u_(1)+m_(2)u_(2)=m_(1)v_(1)+m_(2)v_(2)`

Put
`m_(1)=75 kg`,
`m_(2)=100 kg`,
`u_(1)=10 ms^(-1)`,
`u_(2)=-4 ms^(-1)`,
`v_(1)=v` and
`v_(2)=v`.

`(75)(10)+(100)(-4)=(75)v+(100)v`

`750-400=175v`

`v=2 ms^(-1)`

Therefore, the post-collision velocity of the two players is
`v= 2 ms^(-1)`.