Answer:
[tex]v_f=0.825m/s[/tex]
Explanation:
We must use conservation of linear momentum before and after the collision, [tex]p_i=p_f[/tex]
Before the collision we have:
[tex]p_i=p_1+p_2=m_1v_1+m_2v_2[/tex]
where these are the masses are initial velocities of both players.
After the collision we have:
[tex]p_f=(m_1+m_2)v_f[/tex]
since they clong together, acting as one body.
This means we have:
[tex]m_1v_1+m_2v_2=(m_1+m_2)v_f[/tex]
Or:
[tex]v_f=\frac{m_1v_1+m_2v_2}{m_1+m_2}[/tex]
Which for our values is:
[tex]v_f=\frac{(98.5kg)(6.05m/s)+(119kg)(-3.5m/s)}{(98.5kg)+(119kg)}=0.825m/s[/tex]
The final velocity of the two players when they cling together is
+4.65 m/s in the direction of player one, this is because player one possess more momentum than player two.
Given Data
Mass of player one M1= 98.5kg
Initial Velocity of player one U1= 6.05m/s
Mass of player two M2= 119kg
Initial velocity of plater two U2= -3.50m/s
Final Velocity of the two players after impact V= ??
Applying the principle of conservation of linear momentum for inelastic collision we have
M1U1 + M2U2 = (M1+M2)V
Substitute our given data
98.5*6.05 + 119*3.50 = (98.5+119)*V
Solving for V
595.925 + 416.5 = 217.5V
1012.425 = 217.5V
Divide both sides by 217.5
V = 1012.425/217.5
V = 4.65 m/s
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