Question

Two ice skaters are holding hands at the center of a frozen pond when an argument ensues. Skater A shoves skater B\mathrm{B} along a horizontal direction. Identify (e) If AA has a mass of 0.9000.900 times that of BB, and BB begins to move away with a speed of 2.00 m/s2.00 \mathrm{~m} / \mathrm{s}, find the speed of AA.

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The change in the momentum of an isolated system is zero where the initial momentum equals the final momentum. The particles interact in the system without interact with the environment. So, the momentum is conserved and we use the law of conservation of momentum as given by equation (6.7) in the form

pf=pi\begin{equation} p_{\mathrm{f}}=p_{\mathrm{i}} \end{equation}

The ice is frictionless, so the system is isolated. When an object that has mass mm and moves with speed vv has momentum pp, this momentum is a vector and it is the product of the object's mass and its velocity. The momentum is given by

p=mv\begin{equation*} p=mv \end{equation*}

To keep the balance, the momentum of skater A should equal the momentum of skater B. We use the expression of pp into equation (1) to be in the form

mAvA=mBvB\begin{equation} m_A v_A=-m_Bv_B \end{equation}

The mass of skater A mA=0.900mBm_A= 0.900 m_B and the velocity of skater B is vBv_B = 2 m/s. So, we solve equation (2) for vAv_A

vA=mBvBmA\begin{equation} v_A =-\frac{m_B v_B}{m_A} \end{equation}

Now, we plug the values for mAm_A and vBv_B into equation (3) to get vAv_A

vA=mBvBmA=(mB)(2 m/s)0.900mB=2.22 m/s\begin{align*} v_A &=-\frac{m_B v_B}{m_A}\\ &=-\frac{(m_B)(2 \mathrm{~m/s} )}{0.900 m_B} \\ &=-2.22 \mathrm{~m/s} \end{align*}

The velocity of skater A is 2.22 m/s\boxed{2.22 \mathrm{~m/s}} in the opposite direction of skater B.

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