Main content

### Course: Class 11 Physics (India) > Unit 10

Lesson 9: Elastic and inelastic collisions- Elastic and inelastic collisions
- Properties of inelastic and elastic collisions
- Solving elastic collision problems the hard way
- Deriving the shortcut to solve elastic collision problems
- How to use the shortcut for solving elastic collisions
- What are elastic and inelastic collisions?
- Elastic collisions review
- Inelastic collision review

© 2024 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# How to use the shortcut for solving elastic collisions

In this video, David solves an example elastic collision problem to find the final velocities using the easier/shortcut approach. Created by David SantoPietro.

## Want to join the conversation?

- so V1i + v1f = V2i + V2f only used for Elastic Collision?(10 votes)
- Yes, it is correct as it was derived from conservation of momentum and conservation of kinetic energy which will happen only in elastic collisions(4 votes)

- What if there is elastic collision in two dimension?

Will we take components and then apply the formula?(5 votes)- Yes, apply conservation of momentum for each dimension separately. That means breaking up the momentum vectors into components parallel to each dimension.

If the collision is elastic, kinetic energy will also remain unchanged before and after the collision, but for inelastic collisions, this is not true. Momentum however will always be conserved for both elastic and inelastic collisions.(6 votes)

- Dear Tutors:

Please explain how the golf ball gained speed after the collision? Initial velocity of the golf ball was 50 m/s. Final velocity of golf ball was calculated to be 51.36 m/s. Is this an error?(4 votes)- @S26Patel, as explained here by Andrew M here above, it means in other words that the energy of the objects is transferred to one another. The law of conservation of energy says that energy is never really 'lost', it only is transferred to another object or another form of energy. Either within the system of the problem or to outside of the problem, but that is another story.

Basicly this means that a huge mass(M1) with high velocity (and thus high energy) when colliding with a way smaller mass(M2) with a certain velocity, M1 will transfer some of it's energy to M2. And because E_kin = 1/2 * mass * velocity^2; this means that M2 will gain more velocity because it has less mass. If the exactly same amount of velocity lost by M1 would be gained by M2, there must have been some energy transferred into heat or something else.(2 votes)

- can you solve of either velocity originally or should you pick a certain one to make it easier? Say if one object was at rest already would it be easier to solve for it's velocity since it's initial KE is 0?(2 votes)
- If one of the objects was initially at rest, then its initial momentum would cancel out because the initial velocity of that object would be zero. So, the momentum equation would be just V1f = V2i + V2f.

Hope this helps!(3 votes)

- So am I correct in assuming that V1ix + V1fx = V2ix + V2fx and the same for any other dimension if I needed to decompose velocity vectors in a higher dimensional problem?(2 votes)
- The best way to start a collision problem in 2 or 3 dimensions (3 dimensional collisions problems are very rare) is to select an appropriate set of coordinates. If you have two dots or spheres colliding, which is almost always the case, then the best coordinates will have an axis connecting their centres at the moment of collision, and an axis perpendicular to that. This system will give you the easiest equations. In one of the axis, the one connecting the centres, the solution is exactly like a one dimensional collision, and the other 2 axis show no change in each of the objects' velocity (if there's no friction).(3 votes)

- Is there a way of roughly estimating how much final velocity is lost, for example, by heat and sound? How would I deduct this from my final answer?(2 votes)
- When elastic collisions occur, the momentum is conserved (or constant), so there is no velocity lost to heat and sound, in this specific problem, with particular factors assumed.(1 vote)

- is there any video explaining coeffeicient of restitution(1 vote)
- what happened to the kilograms when he multiplied it(1 vote)
- Canceled out(2 votes)

- Hello! I had an exam today and the first question was asking to find the INITIAL and FINAL velocities of the second ball, when the first ball's initial and final velocity values were given. How would one go about solving this problem?

Ball 1: m=3.0kg

vi= 6.0m/s

vf= 0.5m/s

ELASTIC COLLISION

Ball 2: m= 2.0kg

Vi= ?

Vf= ?(1 vote)- >Let Ball 1(A) have a mass(m) of 3kg velocity of vi m/s initially and vf m/s finally in the +x direction.

>Let Ball 2(B) have a mass(M) of 2kg velocity of Vi m/s initially and Vf m/s finally in the +x direction.

So, applying the conservation of linear momentum:

=>*m(vi)+M(Vi)=m(vf)+M(Vf)*

OR:

=> 3(6)+2Vi=3(1/2)+2Vf

=> 18+2Vi=(3+4Vf)/2

=> 36+4Vi=3+4Vf

=> 4Vf-4Vi=33 ...(i)

Now, using the formula for the coefficient of restitution:*e=(v2-v1)/(u1-u2)*

OR:

(Vf-0.5)/(6-Vi)=1 [Elastic Collision]

=> Vf-(1/2)=6-Vi

=> 2Vf-1=12-2Vi

=> 2Vf+2Vi=13

=> 4Vf+4Vi=26 ...(ii)

Adding equations (i) and (ii):

=>8Vf=59 =>**Vf=7.375m/s**

Putting Vf in (ii):

=>4*7.375+4Vi=26

=>**Vi=-3.5m/s**

I hope it helped!(2 votes)

- How would you solve a problem that involves external forces (like gravity and such during points of contact) in cases where it is not negligible?(1 vote)

## Video transcript

- [Narrator] Alright, here's
pretty much the fastest way you can solve one of these
elastic collision problems, when you don't know two of the velocities. In this case we don't
know the final velocities. We know the initial velocity of the tennis ball and its mass. We know the initial
velocity of the golf ball and its mass, but we don't
know the final velocities of either ball, and the trick
to make these calculations go faster for an elastic
collision is to use this equation, which says the initial
velocity of one of the objects before the collision,
plus the final velocity of that same object after
the collision should equal, if it's an elastic collision, it'll equal the initial velocity of the second object before the collision
plus the final velocity of that second object after the collision. If you want to see where this comes from, we derived it in the previous video. But now that we know
it, we can just use it for any elastic collision. So this expression right here only holds for elastic collisions. But you should love
this formula right here, because what this does
is it allows you to avoid having to use conservation
of energy, and it avoids having to square terms and create horrible messes of algebra. This formula's gonna be
much cleaner, much nicer. Let's see how to use it. Let's just say object
one is the tennis ball and object two here is the golf ball. So in that case, the initial
velocity of the tennis ball, that'd be 40, and it'd be positive 40. I'm gonna write positive just so I know that's to the right, and
it should be positive. Plus the final velocity
of the tennis ball, I'll write that as Vt. So instead of writing one,
I'm gonna get confused which one was number
one, so I'll write a Vt. That way I get to know which object's velocity I'm talking about. So Vt final, that's what
Vt final's gonna mean, final velocity of the tennis ball. And that should equal the initial velocity of the second object, our
second object is the golf ball. The initial velocity of the second object, of our golf ball, is not 50. It's negative 50, you've gotta be careful. These are velocities in this formula. So if you've got a
velocity, that's directed in the negative direction,
you better make it negative. And if you solve in here,
you might get a negative, so these are vector values up here. You gotta plug them in
with the proper sign. So this initial velocity of the golf ball would be negative 50 meters per second, 'cause we're gonna assume
leftward is negative, and rightward is positive. Plus the final velocity
of the second object, second object is our golf ball. I'll call this Vg instead of V2. Vg will be V of the golf
ball and then f for final, i.e. after the collision. So we can solve this for Vt final. I can subtract 40 meters
per second from both sides and I get that the final
velocity of the tennis ball is gonna equal, I'll have
this Vg final just sittin' over here, final velocity of the golf ball after the collision. And then negative 50
minus a 40 is gonna be negative 90 meters per second. So this formula alone was not enough, 'cause I've still got two unknowns. I can't solve for either. I've gotta use another equation, and the other equation we're gonna use is conservation of momentum. 'Cause during this collision, the momentum should be conserved, assuming
the collision happens so fast, any net external impulse is negligible. So we can say that the
total initial momentum is equal to the total final momentum. We basically do this for
every single collision because we make that
assumption, that the net external impulse during this
collision is gonna be small. That means the momentum
should be conserved. So the formula for momentum
is mass times velocity. So the momentum of this
tennis ball initially is the mass of the tennis
ball, .058, times the initial velocity, would be positive 40. Positive because it's
directed to the right, and I'm gonna consider
rightwards positive. Plus the initial momentum of the golf ball would be .045 times the initial
velocity of the golf ball and that'd be negative 50. Again, you have to be careful
with the negative signs. Momentum is also a vector,
so if these velocities are ever negative, you've
got to plug them in with their negative sign. And that initial momentum
should equal the final momentum. So the final momentum of the tennis ball is gonna be 0.058 times our final velocity of the tennis ball. And I'm gonna use the same nomenclature. I'm gonna use the same symbol over here that I used over here. This Vt final, final
velocity of the tennis ball is the same as this Vt
final, final velocity of the tennis ball, plus
.045 times same thing, final velocity of the golf ball. I'll use the same symbol
which is gonna be Vg final. I could multiply out this
entire left hand side, and what I get is 0.07
kilogram meters per second. And that equals on the right hand side, this entire expression right here. I'll just copy that. So we've got two unknowns
in this equation as well, so we can't solve this directly or either of the final velocities. But we do have two equations
and two unknowns now. And whenever you have that situation, you can solve one of the
equations for one of the variables and plug that expression
into the other equation. In other words, I know
that the final velocity of the tennis ball is
equal to the final velocity of the golf ball minus 90, so I can take this entire term right here,
since it's equal to Vt final, and just plug that in for Vt final. And what that would do for
me is give me one expression, all in terms of the final
velocity of the golf ball. So let's do that. We've still got 0.07 kilogram
meters per second on the left. And that's gonna equal 0.058,
that's still here, kilograms, times Vt final. I'm plugging in this entire
expression for Vt final. So that gets multiplied by
Vg final, the final velocity of the golf ball, minus
90 meters per second. That was the term I
plugged in for Vt final, and it got multiplied by this mass here, so I can't forget about that mass. And then I still have to
add this final momentum of the golf ball, .045 kilograms
times the final velocity of the golf ball. So at this point you might
be feeling ripped off. You might be like, easy way to do this? This isn't easy, this is hard. I've gotta plug one equation
into another, and then solve? Well, this easy approach
does not avoid having to plug one equation into the other, that's true. But the reason that it's easy is because the equations that we're
plugging into each other are a whole lot simpler than
the kinetic energy formula that you would have to use if you didn't know this expression here. Because we have this one,
we do not have to plug conservation of momentum
into conservation of energy. That would square the term we put in, that would get nasty, the
algebra would be a lot worse. These formulas that we're
dealing with in this process only have velocity. None of these velocities are squared, so the algebra doesn't get nearly as bad. And we're actually almost done over here. Let me show you how close we
are to finishing this thing. I just need to multiply
through this term right here. So what am I going to get, this left hand side stays the same. And then when I multiply
through .058, I'm gonna get 0.058 times the final
velocity of the golf ball, and then negative 90 times
.058 is negative 5.22, and that would be units of
kilogram meters per second. And I still just have
this term right here, so I'll copy that one. And now here's a key step. The whole reason we plugged
one formula into the other is so that we had the same
unknown in that formula. Only one unknown, there's only
one unknown variable in here, which is the final
velocity of the golf ball. It's located in two spots, but at least it's the same variable. And what that allows us to
do is to combine these terms. If on this right hand
side I have this much of the final velocity of the golf ball, and that much of the final
velocity of the golf ball, when I add them up, I can just add these two factors out front. In other words I'll have 0.07 equals, I can rewrite this as 0.058 kilograms plus 0.045 kilograms times the final
velocity of the golf ball. And then I still have this minus 5.22. So if that looked like
mathematical witchcraft, all I did was I combined
the terms that had Vg final. Cause if you had A times Vg
final plug B times Vg final, that's the same as A
plus B, times Vg final. When you multiply this through, you'll just get both of
these terms back again. So we keep going, don't
divide by this first, sometimes people try to divide
by this whole term right now. You don't want to do that, you
gotta go in the right order. I need to add this 5.22 to both sides to get rid of it first. So when I do that, when
I add 5.22 to both sides it'll cancel this term,
and on the left hand side I'll get 5.29 kilogram meters per second. And that's gonna equal, if I
add these two terms together, if I just add .058 and
.045, I get 0.103 kilograms times the final velocity of the golf ball. And let me move these initial
velocities down so we can see. At this point, if I
divide both sides by .103, I'll get 51.36 meters per
second on the left hand side. And that equals the final
velocity of the golf ball. So we did it, we found one
of the final velocities of these objects. We found the final
velocity of the golf ball. But what about the final
velocity of the tennis ball? How do we figure out what that is? 'Cause that's also an
unknown in this problem. Well that one's really easy. Now that we know the final
velocity of the golf ball, I can just take that value,
plug it right back into here, this expression that we plugged into conservation of momentum. And I can figure out what the velocity of the tennis ball is. So in other words, the
velocity of the tennis ball I know, finally, should
equal the final velocity of the golf ball, which was
51.36 meters per second, and then minus 90 meters per second, and I'll get the final
velocity of the tennis ball was negative 38.64 meters per second. And it came out to be
negative, that means that this tennis ball got deflected backwards. It was heading leftward,
38.64 meters per second after the collision. So recapping, we used this nice formula to get one equation that
involved the velocities that we didn't know for
an elastic collision, which you can only use
for an elastic collision. If you want to see where
this equation comes from, we derived it in the last video. We solved for one of the
velocities and plugged it into conservation of momentum. We combined the like unknowns,
and solved algebraically to get one of the final
unknown velocities. Then we plugged that back
into the first equation to get the other unknown velocity. And this is easier than the alternative, because the alternative
involves kinetic energies, which means when you take
one of these expressions, you'd have to square them, and the algebra would be significantly more difficult.