WEBVTT
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So we've now drawn
pictures of the interaction
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of a person jumping off
a cart in both the ground
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frame and a reference frame
moving with velocity, vc, which
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is the final speed of the cart.
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These were pictures in the
reference frame moving with vc,
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and these are the
pictures-- momentum
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diagrams-- of the person and
the cart in the ground frame.
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Now, we would like to apply
the momentum principle
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to solve for vp and vc.
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We are given that the velocity
of the person in the moving
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frame, u.
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We'll treat this as
a given quantity.
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And we'll express
the velocity, vc,
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and so we want to find vc and
vp in terms of u and p and mc.
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Now, we'll use the
ground frame first,
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and our assumption here is that
the person jumped horizontally
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and that there are
no external forces
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in the horizontal direction.
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So all of the vectors
we've drawn are horizontal,
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and there's no external forces
in the horizontal direction.
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Then we know that the initial
momentum of the system
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is equal to the final
momentum of the system.
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In our initial picture,
nothing is moving.
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And in our final
picture, we have
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m cart v cart plus m
person v person is 0.
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Now, recall that we also showed
that the velocity of the person
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in the ground frame is related
to the velocity of the person
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in the moving frame by
adding the relative velocity
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of the two frames.
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The relative velocity
of the two frames
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is the velocity of the cart, vc,
so what we have is vc plus u.
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So this is how vp
is related to u.
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And now, I can use
this start equation
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to write it as 0 equals mc
vc plus mp times vc plus u.
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And by adding terms, I get that
mc plus mp vc equals minus mpu.
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Or the velocity of the cart
is equal to minus mpu divided
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by mc plus mp.
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Now that I have the
velocity of the cart,
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I can just substitute
that in here
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to find the velocity of the
person, which, remember,
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was the velocity
of the cart plus u.
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So now, I've solved this
problem in the ground frame.
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Next, I'll do the same
analysis in the moving frame.