Question

# Two ice skaters are holding hands at the center of a frozen pond when an argument ensues. Skater A shoves skater $\mathrm{B}$ along a horizontal direction. Identify (e) If $A$ has a mass of $0.900$ times that of $B$, and $B$ begins to move away with a speed of $2.00 \mathrm{~m} / \mathrm{s}$, find the speed of $A$.

Solution

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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

$$$p_{\mathrm{f}}=p_{\mathrm{i}}$$$

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

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

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

$$$m_A v_A=-m_Bv_B$$$

The mass of skater A $m_A= 0.900 m_B$ and the velocity of skater B is $v_B$ = 2 m/s. So, we solve equation (2) for $v_A$

$$$v_A =-\frac{m_B v_B}{m_A}$$$

Now, we plug the values for $m_A$ and $v_B$ into equation (3) to get $v_A$

\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 $\boxed{2.22 \mathrm{~m/s}}$ in the opposite direction of skater B.