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

Answer:

The post-collision velocity of the two players is $v= 2 ms^{-1}$.

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 $v= 2 ms^{-1}$.