To get off a frozen, frictionless lake, a 70.0 kg person takes off a 0.150 kg shoe and throws it horizontally, directly away from the shore with a speed of 2.00 m/s. If the person is 5.00 m from the shore, how long does it take for him to reach it?
To get off a frozen, frictionless lake, a 70.0 kg person takes off a 0.150 kg shoe and throws it horizontally, directly away from the shore with a speed of 2.00 m/s. If the person is 5.00 m from the shore, how long does it take for him to reach it?
This is actually two problems.
The first is a conservation of momentum problem. Before the person throws the shoe, what is the momentum? 0 kgm/s, because nothing is moving. Now, we know the shoe is thrown and we know how fast it is being thrown. The net momentum of the system (person + shoe) before the throw has to equal the momentum of the system after the throw (because there are no net external forces acting on the system.) So:
$\displaystyle P_{tot,0} = P_{tot}$
$\displaystyle 0 = m_sv_s + m_pv_p$
where $\displaystyle m_s$ is the mass of the shoe, etc.
$\displaystyle v_p = -\frac{m_s}{m_p} \cdot v_s$
(The negative sign merely indicates that the person is moving in the opposite direction as the shoe.)
The second part of the problem is easy. It's a frictionless surface, so if the person starts moving with a velocity they will continue with that velocity. So the person is moving with a speed $\displaystyle v_p$ over a distance of 5.00 m. How long does it take?
-Dan