Question

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

Answer 1

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.

`\begin{gathered} p_(before)=p_(after) \\ m_1u_1+m_2u_2=m_1v_1+m_2v_2 \end{gathered}`

The initial velocities of both skaters are 0 m/s

`m_1\cdot0_{}+m_2\cdot0=m_1v_1+m_2v_2`

Also, m = W/g

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

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.