The question is,

A $45.5~kg$ girl is standing on a $140~kg$ plank. Both originally at rest on a frozen lake that constitutes a friction-less, flat surface. The girl begins to walk along the plank at a constant velocity of $1.47~m/s$ relative to the plank.

(a) What is the velocity of the plank relative to the ice surface?

(b) What is the girl's velocity relative to the ice surface?

In a commentary given on this problem, they are essentially that, when the velocity of the girl relative to the plank is $\vec{v}_{gp}$, and the velocity of the plank relative to the ice is $\vec{v}_{gp}$, the velocity of the girl relative to the ice is simply $\vec{v}_{gi}= \vec{v}_{gp} + \vec{v}_{gp}$.

I just can't conceive why finding the velocity of the girl relative to the earth is a simply sum. Could someone please elucidate this for me?

Also, they treat this situation as if momentum is conserved, that is, $\Delta \vec{P}=0$; but isn't it necessary that friction be present, between the plank and the girl's shoes, for there to be any motion? Wouldn't this force cause a change in momentum?


I will try to attempt an answer here. You have two questions: One on relative velocity and the other on momentum.

For the first, I believe you're confused on the notion of relative velocity. So let me intuitively explain the idea: When we say what is the velocity of B with respect to A, what we mean is that if A were a person, with which velocity would he see B moving? Please note that A thinks that he is stationary. So if A is riding a bicycle, he would realize that he is moving towards the street lamps in front of him. Of course, this is because he knows that he is the one who is actually moving. If he instead thought that he is the one who is stationary then he would think that the street lamps are instead moving towards him, and the velocity with which they move towards A is the relative velocity of the lamps with respect to A.

Okay now let's say there are two people, A and B and they start moving towards each other with velocities $5$ and $7~m/s$ respectively. What is the relative velocity of B with respect to A? Every second B moves $7~m$ towards A but A in the same amount of time covers $5~m$ towards B. So if A thinks he is stationary then he would think that B just moved $7~m + 5~m = 12~m$ toward him in one second, hence the answer is $12~m/s$. I hope you can now relate this to the case where both of them are moving in the same direction.

Now for your example. Let's say the magnitude of the velocity of the plank and the girl is $v_p$ and $v_g$. By magnitude of the velocity I mean the magnitude of the velocity of the plank with respect to the ice, so don't get confused. Also these two variables are simply the magnitudes, so both of them are positive numbers with units $m/s$.

Now if you think about this in terms of the analogy given above, it should make sense that relative to the plank, the girl is moving away from it with a speed of $v_p + v_g$, the value of which is given to be $1.47~m/s$. So if $v_{gp}$ is the relative velocity of the girl with respect to the plank then, $v_{gp} = v_p + v_g$. You may find the equation that I have given here is different to the one given in the commentary in your book, that's because my variables do not include the negative sign of the direction in the variables themselves, it is written explicitly in the equation.

The normal force from the ice and the weight of the girl and the plank combined cancel each other, so there is no external force on the girl-plank system. Hence the momentum is conserved. Since the initial momentum was zero, the final momentum must also be zero. Therefore, $m_gv_g + (-m_pv_p) = 0$ where $m_g$ and $m_p$ is the mass of the girl and the plank respectively. Note that the momentum of the plank is negative since the velocity is in the opposite direction and I haven't included the negative sign in the velocity variable of the plank. You can now solve the two equations to give you the solution.

The frictional force between the girl and the plank is within the system i.e. it is an internal force. This force is what causes the change in momentum of the girl and the plank from zero to whatever they have now. But the momentum of them combined is still the same i.e. zero.

I think I may have not explained the frictional force part clearly, so if you have doubts, feel free to ask.

  • $\begingroup$ I have a small query. I understand that the frictional forces are internal forces, and that there is no net force when the entire system is considered. But, what if we were to consider the girl or the plank alone? There would be an unbalanced force on both of them. Then, shouldn't they accelerate individually? Then, how can we talk about a constant velocity? Is this velocity ever changing then? But, in many practice problems Like this, I'm told to calculate the time taken to reach the end of the plank( referring to equivalent scenarios) so, I have to take the velocity to be constant. But how? $\endgroup$ – SNB Jan 17 '17 at 17:32
  • $\begingroup$ Just one more question. The girl is on the plank and hence the plank feels her weight through the normal force. As the plank moves, the girl obtains it's velocity, along with her own velocity. So, when applying conservation of momentum why don't we add the girl's mass to the boat's mass calculate the corresponding velocity? Any help is appreciated $\endgroup$ – SNB Jan 17 '17 at 17:38

Try this method for solving the majority of Momentum/Impulse Problems with these two simple equations. Michel Van Biezen on Youtube teaches this method. Sure beats m1v1 + m2v2 (initial) = m1v1 + m2v2 (final).


Girl- Mass of 45.5kg; Velocity +1.47m/s

Plank- Mass of 140kg


QA Find the VELOCITY of Plank (that is girl + plank) on ice.

QB Find the VELOCITY of Girl on ice.


[I = mΔV] [I = ΔP = mΔV] [Vf = Vi + ΔV]

Ans A:

I = mΔV --> I = (45.5kg)(1.47m/s) = 66.9 kg m/s

I = mΔV --> ΔV = i/m --> ΔV = (-66.9 kg m/s) / (45.5kg + 140kg) = -0.361m/s

Ans B:

Vf = Vi + ΔV --> Vf = (+1.47m/s) + (-0.361 m/s) = 1.11 m/s


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