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Two figure skaters, one weighing 625 N and the other 725 N, push off against each other onfrictionless ice. If the heavier skater travels at 1.5 m/s, how fast will the lighter one travel?A 1.7 m/sB 2.8 m/sC -1.7 m/sD -2.8 m/s

Sagot :

We are given the following information:

Weight of skater 1 = 625 N

Weight of skater 2 = 725 N

Final velocity of skater 2 = 1.5 m/s

Final velocity of skater 1 = ?

Recall from the law of conservation of momentum, the total momentum before the collision and after the collision must be equal.

[tex]\begin{gathered} p_{before}=p_{after} \\ m_1u_1+m_2u_2=m_1v_1+m_2v_2 \end{gathered}[/tex]

The initial velocities of both skaters are 0 m/s

[tex]m_1\cdot0_{}+m_2\cdot0=m_1v_1+m_2v_2[/tex]

Also, m = W/g

[tex]\begin{gathered} 0=m_1v_1+m_2v_2 \\ 0=(\frac{625}{9.8})\cdot_{}v_1+(\frac{725}{9.8})\cdot1.5 \\ (\frac{625}{9.8})\cdot_{}v_1=-(\frac{725}{9.8})\cdot1.5 \\ (63.78)\cdot_{}v_1=-110.97 \\ _{}v_1=-\frac{110.97}{63.78} \\ _{}v_1=-1.7\: \frac{m}{s} \end{gathered}[/tex]

So, the lighter skater will travel with a velocity of 1.7 m/s

The negative sign means that the lighter skater will be traveling oppositely to the heavier skater.

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