While a family is taking a walk on a frozen pond, the two small children (a boy and a girl) manage to run into each other. They become entangled, resulting in a totally inelastic collision. The boy has a mass of 15 kg and was initially running at 1.0 m/s and the girl has a mass of 20 kg and was initially running at 2.0 m/s. Before their collision, the relative angle between their velocities was 45°. Assuming that the coefficient of friction between the resulting boy+girl combination and the ice is 0.15, how far do they slide after the collision before coming to rest?
System: Boy and girl as point particles. External influences are neglected during the collision under the assumption that collision forces dominate. After the collision, the system experiences external influences from the earth (gravity) and the ice (normal force and friction).
Models: [Momentum and Impulse] followed by the [Work-Energy Theorem].
Approach: We begin by analyzing the collision using momentum. When a problem gives a relative angle, it is important to develop a coordinate system to orient ourselves as we solve. We therefore begin with a picture. We have arbitrarily assigned the boy to move along the x-axis, and the girl to have positive x- and y-velocity components.
With our picture developed, we can write the equations of constant momentum, since we are assuming that during the collision external forces are negligible compared to the collision forces.
\begin
[ p^
_
+ p^
_
= p^
_
]
[p^
_
+ p^
_
= p^
_
]\end
Rewriting in terms of the masses and velocities, and substituting the appropriate zeros gives:
\begin
[ m^
v^
_
+ m^
v^
_
= (m^
+m^
)v_
]
[ m^
v^
_
= (m^
+m^
)v_
]\end
We have all the givens we need to solve these equations directly for the final velocity of the system:
\begin
[ v_
= \frac{m^
v^
_
+m^
v^
_{x,i}}{m^
+m^{G}} = \mbox
]
[ v_
= \frac{m^
v^
_{y,i}}{m^
+m^{G}} = \mbox
]\end
so that the magnitude of the final velocity is:
\begin
[ v_
= \sqrt{(\mbox
)^2+(\mbox
)^{2}} = \mbox
]\end
This final velocity for the collision of the boy and girl is the initial velocity for their post-collision slide. We can now consider a free body diagram for the boy+girl system as they slide across the ice. Assuming the pond is flat, gravity and the normal force will do no work. Friction will do negative work, as it is directed at 180° from the direction of motion of the system. Since only one force can do work, the question of how far the system slides after the collision is easily addressed using the work-energy theorem. The relevant form for the Law of Change is:
\begin
[ K_
= K_
+ \int F_
\cos\theta \:ds = K_
+\int \mu_
N\cos\theta \:ds]\end
Considering the free body diagram plus the fact that the boy and girl are not accelerating in the z-directtion, it is clear that the normal force and the weight of the system must balance. Further, we want to know how far the boy and girl slide before they stop, indicating that vf should be zero for this part of the problem. Thus, we can write:
\begin
[ 0 = \frac
(m^
+m^
)v_
^
+ \mu (m^
+m^
)gs ]\end
Again, recognizing that the initial velocity for the slide is the final velocity of the collision, we can solve to find:
\begin
[ s = \frac{v_
^{2}}
= \mbox
]\end