Answer:
Velocity of the players afterwards = 2.88 m/s towards east.
Explanation:
Mass of football player A [tex](m_1)[/tex]= 91.5 kg
Velocity of player A [tex](v_1)[/tex]= 2.73 m/s
Mass of football player B [tex](m_2)[/tex] =63.5 kg
Velocity of player B [tex](v_2)[/tex]= 3.09 m/s
Since both players move in same direction east, so their velocity afterwards will also be in same direction east.
By law of conservation of momentum we have:
Momentum before collision = Momentum after collision
This can be written as:
[tex]p_i=p_f[/tex]
[tex]m_1v_1+m_2v_2=(m_1+m_2)\ v[/tex]
where [tex]v[/tex] is the velocity of the players together afterwards.
We can plugin the given value to find [tex]v[/tex]
[tex](91.5)(2.73)+(63.5)(3.09)=(91.5+63.5)\ v[/tex]
[tex]249.795+196.215=155\ v[/tex]
[tex]446.01=155\ v[/tex]
Dividing both sides by 155.
[tex]\frac{446.01}{155}=\frac{155\ v}{155}[/tex]
[tex]2.88=v[/tex]
∴ [tex]v=2.88[/tex] m/s
Velocity of the players afterwards = 2.88 m/s towards east.
