Answers:
a) -2.54 m/s
b) -2351.25 J
Explanation:
This problem can be solved by the Conservation of Momentum principle, which establishes that the initial momentum [tex]p_{o}[/tex] must be equal to the final momentum [tex]p_{f}[/tex]:
[tex]p_{o}=p_{f}[/tex] (1)
Where:
[tex]p_{o}=m_{1} V_{o} + m_{2} U_{o}[/tex] (2)
[tex]p_{f}=(m_{1} + m_{2}) V_{f}[/tex] (3)
[tex]m_{1}=110 kg[/tex] is the mass of the first football player
[tex]V{o}=-7 m/s[/tex] is the velocity of the first football player (to the south)
[tex]m_{2}=75 kg[/tex] is the mass of the second football player
[tex]U_{o}=4 m/s[/tex] is the velocity of the second football player (to the north)
[tex]V_{f}[/tex] is the final velocity of both football players
With this in mind, let's begin with the answers:
a) Velocity of the players just after the tackle
Substituting (2) and (3) in (1):
[tex]m_{1} V_{o} + m_{2} U_{o}=(m_{1} + m_{2}) V_{f}[/tex] (4)
Isolating [tex]V_{f}[/tex]:
[tex]V_{f}=\frac{m_{1} V_{o} + m_{2} U_{o}}{m_{1} + m_{2}}[/tex] (5)
[tex]V_{f}=\frac{(110 kg)(-7 m/s) + (75 kg) (4 m/s)}{110 kg + 75 kg}[/tex] (6)
[tex]V_{f}=-2.54 m/s[/tex] (7) The negative sign indicates the direction of the final velocity, to the south
b) Decrease in kinetic energy of the 110kg player
The change in Kinetic energy [tex]\Delta K[/tex] is defined as:
[tex]\Delta K=\frac{1}{2} m_{1}V_{f}^{2} - \frac{1}{2} m_{1}V_{o}^{2}[/tex] (8)
Simplifying:
[tex]\Delta K=\frac{1}{2} m_{1}(V_{f}^{2} - V_{o}^{2})[/tex] (9)
[tex]\Delta K=\frac{1}{2} 110 kg((-2.5 m/s)^{2} - (-7 m/s)^{2})[/tex] (10)
Finally:
[tex]\Delta K=-2351.25 J[/tex] (10) Where the minus sign indicates the player's kinetic energy has decreased due to the perfectly inelastic collision
