in this video we're looking at an inelastic collision between two identical blocks one of
them is initially stationary but this collision is complicated by the fact that we have a rough
surface and we have a kinetic friction coefficient given to us throughout the entire process that's
going to be 0.15 so in our initial state one of the blocks is moving with a given initial velocity
of three meters per second and it slides through one meter before the impact occurs and during that
one meter it's slowing down so in part a we have to figure out what's the speed just before the
impact in part b we're going to apply conservation of momentum to figure out the speed of the stuck
together blocks after the impact and then in part c after those blocks are stuck together
they're going to slow down as friction continues to work on them and eventually they're going to
come to a stop and we're trying to figure out how far do they slide after they're stuck together
so to get started with part a we want the speed of that initially moving block after it slides one
meter we've got to get the acceleration of it to get that done that means we need to figure out the
size of the friction force on it so we're going to pop in all the vectors for a force analysis
and there are three forces acting on this block of course we have gravity pulling
straight down with a magnitude of mg since m is 1 and we're using 9.8 as an approximation for g
that's 9.8 newtons pointing down then we have a normal force pushing up
and because there are no other forces acting vertically in this problem the normal force must
exactly balance the force of gravity pointing down so that's 9.8 newtons
and then pointing to the left we have the friction force i'll call that fk that's given by the
coefficient of kinetic friction mu k multiplied by the normal force little n multiplying 0.15
times 9.8 we get a kinetic friction force of 1.47 newtons now that we have the horizontal
force acting on this block we can apply newton's second law to figure out the magnitude of the
acceleration i'm going to write newton's second law as a equals f over m and we just plug in
our force of 1.47 newtons plug in our mass of one kilogram dividing by one is easy enough
and i end up with a magnitude for my acceleration of 1.47 meters per second squared now when we take
this acceleration and plug it into a kinematics equation you have to remember to put a minus
sign on it because we're going to use a coordinate system where rightward means positive and leftward
means negative so this is negative 1.47 when we plug it in so we want the speed of this
block after it slides for one meter and there's a kinematics equation that gets this done in one
step it's the time independent equation v squared is equal to v naught squared plus two a times
the quantity x minus x naught x naught is our initial position which we're going to call zero
and to find the final speed i just have to plug everything in on the right hand side and take the
square root so v naught squared that's 3 squared now our acceleration is going to take a minus
sign again because it points to the left so i have minus 2 times 1.47 and then our sliding distance
is 1 meter so it's kind of optional to even show that since multiplying by one doesn't change
anything when we run the numbers on this and round to three significant digits we get 2.46 meters per
second just before the collision occurs now in part b we want the speed of the blocks just after
the collision occurs so we're just looking at this instant in time where an inelastic collision
occurs meaning the blocks get stuck together there is no motion during that instant for friction to
do any work so we don't have to think about that we're just doing pure momentum conservation here
so p initial is equal to p final for that instantaneous collision in our initial momentum
just before the collision occurs that requires the speed we computed in part a the leftmost
block was moving at 2.46 meters per second so my initial momentum will be the mass of that
block times its speed coming into the collision 2.46 meters per second my final momentum now i
have two blocks stuck together both moving for a total of two kilograms and i have an unknown final
velocity so to solve for v f well the kilograms cancel there and we're just dividing both sides
by two and we get that the final velocity that's just after the impact occurs is 1.23 meters
per second finally in part c we want to get the sliding distance for these stuck together blocks
after the collision occurs and this is basically just the reverse of what happened in part a we
know the initial velocity of the stuck together blocks is going to be 1.23 meters per second
we know the final velocity of the stuck together blocks is zero because we're trying to figure out
how far they slide before they stop well at the end of the process v is equal to zero and the
acceleration of these blocks as they slide to a stop is actually going to be equal to the value
that we computed in part a it might not be obvious at first why this is true so we'll go ahead and
analyze it as a two kilogram mass and show that it's true so just to be thorough here here's
a two kilogram mass that's the stuck together blocks and then the force analysis is familiar
we've got gravity pulling down on this mass where the magnitude of mg this time m is 2 so our weight
is 2 times 9.8 or 19.6 newtons again no forces are interfering vertically so our normal force is
equal to the weight 19.6 newtons and our friction force again is the coefficient of kinetic friction
multiplying the normal force which is now twice as much as it was before so our friction force has
doubled simply because our mass has doubled it's now 2.94 newtons but notice from newton's second
law a equals f over m if i double the force on a mass and i also double the mass the factors of 2
are going to cancel out and my acceleration is going to be the same so just to see that
this works my friction force is 2.94 newtons my mass is now 2 kilograms and i arrive at the same
number for the acceleration 1.47 meters per second squared and again that counts as negative once we
get it into a kinematics equation so we're going to apply the time independent kinematics equation
again v squared equals v naught squared plus 2a times x minus x naught we're going to take our
initial position as zero just after the impact occurs our final speed is also zero because
these blocks are sliding until they come to rest and what we're after here is that little x and we
can solve for that very quickly by subtracting v naught squared from both sides and dividing by 2a
so our sliding distance is going to be a negative v naught squared divided by twice the acceleration
and the initial speed well that's the speed after the collision occurs that's 1.23 and i
need to square that and remember that a takes a minus sign because it points to the left
and this is good because the minus signs cancel and i get a positive value for x in other words
the blocks end up to the right of where they began and our sliding distance to three significant
figures is 0.515 meters and we're done if you enjoyed this video or at least found it useful
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