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Sagot :
Use the Law of Conservation of Linear Momentum to analyze this situation, according to which the total linear momentum of the system formed by the two skaters must be the same before and after the collision:
[tex]p_i=p_f[/tex]The linear momentum p of a body with mass m and velocity v is:
[tex]\vec{p}=m\vec{v}[/tex]If we consider the right direction as positive and the left direction as negative, then, the linear momentum of the 65kg ice skater is:
[tex]p_{65\operatorname{kg}}=(65\operatorname{kg})(6.0\frac{m}{s})=390\operatorname{kg}\cdot\frac{m}{s}[/tex]The linear momentum of the 85kg ice skater is:
[tex]p_{85\operatorname{kg}}=(85\operatorname{kg})(-4.5\frac{m}{s})=-382.5\operatorname{kg}\cdot\frac{m}{s}[/tex]Then, the total linear momentum before the collision, is:
[tex]\begin{gathered} p_i=p_{65\operatorname{kg}}+p_{85\operatorname{kg}} \\ =390\operatorname{kg}\cdot\frac{m}{s}-382.5\operatorname{kg}\cdot\frac{m}{s} \\ =7.5\operatorname{kg}\cdot\frac{m}{s} \end{gathered}[/tex]The linear momentum of both ice skaters together after the collision, is equal to the product of their combined masses and their unknown speed v. Then:
[tex]\begin{gathered} p_f=(65\operatorname{kg}+85\operatorname{kg})\cdot v \\ =(150\operatorname{kg})\cdot v \end{gathered}[/tex]Using the Law of Conservation of Linear Momentum:
[tex](150\operatorname{kg})v=7.5\operatorname{kg}\cdot\frac{m}{s}[/tex]Divide by 150kg to isolate the speed v:
[tex]v=\frac{7.5\operatorname{kg}\cdot\frac{m}{s}}{150\operatorname{kg}}=0.05\frac{m}{s}[/tex]Since the velocity is positive, then the ice skaters are moving to the right.
Therefore, the speed of the ice skaters after the collision is 0.05 m/s, and they are moving to the right.
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