Can Conservation of Momentum be Used for Inelastic Collisions?

[sachin123]

Hi.
I found this question:
A ball is suspended by a string and a bullet is shot at it from underneath.Given the masses and velocity of the bullet,can we conserve the momentum for this inelastic collision?
My doubt is isn't there a vertical downward external force acting on the system(gravity).
But my textbook has conserved the momentum and got the answer.How?

And for a box moving on a surface which meets another box on a surface with lot of friction?Can we conserve momentum here,for this collision?
Thank You

 

[collinsmark]

A ball is suspended by a string and a bullet is shot at it from underneath.Given the masses and velocity of the bullet,can we conserve the momentum for this inelastic collision?

If you're only concerned about the velocities the instant before the collision and the instant after, then yes. (And assuming that the collision itself takes place within an instant.)

My doubt is isn't there a vertical downward external force acting on the system(gravity).
But my textbook has conserved the momentum and got the answer.How?

If you're only concerned about the instants before and after the collision, momentum is conserved.

Most textbook problems of this type specify the horizontal velocity of the bullet, and specify that the bullet strikes the ball (pendulum) horizontally -- perpendicular to the force of gravity -- which makes things easier. But either way, if you're only concerned about the instants before and after the collision, momentum is conserved.

In a more general sense, momentum is conserved at all other times too, if you make the system big enough. For example, if you consider the momentum of the entire Earth itself, as well as the bullet and ball, then momentum is conserved at all times.

And for a box moving on a surface which meets another box on a surface with lot of friction?Can we conserve momentum here,for this collision?

Same as before. If you're only concerned about the velocities of the blocks at the instants before and after the collision, momentum is conserved (and assuming that the dynamics of the collision happen quickly.)

And like before, conservation of momentum applies at other times too if you include not only the momentum of the blocks but also the momentum of the surface itself in the system. As long as you make the system large enough (such that all forces are within the system [i.e no external forces]), conservation of momentum applies whether or not there is friction.

 

[sachin123]

Can you tell me how my equations would change if I took Earth in the system?
I mean Earth doesn't collide with anything,so how can we write momentum eqs for Earth?
I don't get this right.

And if I am interested in only the instantaneous states after and before collision I can blindly use COM right?

 

[collinsmark]

Can you tell me how my equations would change if I took Earth in the system?
I mean Earth doesn't collide with anything,so how can we write momentum eqs for Earth?
I don't get this right.

I wasn't really suggesting that you take the momentum of the Earth into consideration for this particular problem. I was speaking hypothetically. If you absolutely had to you could, and momentum would be conserved at all times. But I'm not suggesting that you actually do it that way. There are simpler ways of solving such problems.

Just use conservation of momentum when dealing the instant of the collision itself. That's the easy way.

But if you're curious, here is what I meant about momentum being conserved at all times, if everything involved is included in the system: It is a result of Newton's third law. For every action there is an equal and opposite reaction. Consider an example where you jump into the air as high as you can. When you leap off the ground, you move away from the ground with an upwards velocity. At the same time, you push the entire Earth away from you with a downward velocity. The entire momentum of both you and the Earth (together) remains zero at all times. After you reach your maximum height, you fall back toward the Earth, and at the same time the Earth moves back up toward you. Your momentum in one direction is equal and opposite the momentum of the Earth in the other direction. The force of gravity is between you and the Earth in this case, which is included within the system (no external force). Momentum is conserved.

(Of course the velocity of the Earth is much smaller than your own because the Earth is so much more massive. But the magnitude of your momentum is identical to the magnitude of the Earth's momentum, in this example.)

And if I am interested in only the instantaneous states after and before collision I can blindly use COM right?

For the problems like the ones you've described in your first post, pretty much yes. But only use conservation of momentum at the moment of the collision itself.

 

[sachin123]

Thank You for your explanation.
So you meant we would have to assume a velocity for the Earth and include its momentum in the correct equation right?

Lets see this:two masses m1 m2 connected by a spring,if we take spring inside the system,we can easily write the eqs.But how if we don't?And yes the spring is light and exerts force on the bodies which become external forces.Then?

 

[tiny-tim]

hi sachin123!

My doubt is isn't there a vertical downward external force acting on the system(gravity).
But my textbook has conserved the momentum and got the answer.

conservation of momentum can apply in one direction at a time

from good ol' Newton's second law, F = d(mv)/dt,

and although this is a vector equation, it obviously also applies in each direction separately …

if F has no component in the x direction, then the x component of mv must be constant …

in simple language, momentum is conserved in any direction in which there is no external force: so, in your question, the only external force is vertical, and so momentum in any horizontal direction must be conserved

 

[collinsmark]

Thank You for your explanation.
So you meant we would have to assume a velocity for the Earth and include its momentum in the correct equation right?

Yes, its relative velocity. It's often convenient to define the velocity relative to the center of mass of the system.

But anyway, like I mentioned before, defining the system to include the entire Earth is overkill for problems like the one involving the bullet and the ball. It would make the problem really, really complicated, for no good reason. All I'm saying is that you could do it (not that you should do it). When the gun is fired, a bullet shoots out in one direction, but it also imparts a torque on the Earth (due to the recoil) in the other direction. After the bullet hits the ball, the ball raises up and also imparts a horizontal force on the pendulum support system, which creates a torque on the Earth in the opposite direction that the recoil did earlier. If you're careful enough, the vector sum of all momentums (bullet, ball, Earth and anything else) is zero at any given instant in time. In the beginning, since nothing is moving, all the momentums are zero. In the end, after everything comes to rest eventually, again all the momentums are zero. And in the middle, when things are moving, all the momentums still add up to zero as long as you consider everything. Now that's not the best way to solve problems like this. My only point is that momentum is always conserved as long as nothing is left out of the system (such that there are no external forces).

Lets see this:two masses m1 m2 connected by a spring,if we take spring inside the system,we can easily write the eqs.But how if we don't?And yes the spring is light and exerts force on the bodies which become external forces.Then?

You've described a classic (almost textbook example) mechanics problem. And we could solve for the detailed equations of motion if we wanted to.

But for the point of this thread, suffice it to say that

m₁v₁(t) + m₂v₂(t) = 0​

assuming the center of mass is stationary, and the spring mass system is on a frictionless horizontal surface such that we can ignore gravity. (Here v₁(t) and v₂(t) are the respective velocities of each mass.)

Now if instead you have a mass attached to a spring, such that the other end is attached to the ceiling, you generally wouldn't use conservation of momentum. Sure, you could if you really wanted to (by bringing the whole Earth into the system), but it would make things more complicated than necessary.

 

[sachin123]

@tiny_tim,the bullet goes vertically.Thank You collinsmark for you help.:)
But its kind of confusing because the C.O.Momentum is derived from constancy of position of Centre of Mass(which you used to get that equation) isn't it?

 

[collinsmark]

@tiny_tim,the bullet goes vertically.

(By the way, in my previous post I was sort of assuming that the bullet hits the ball at some angle to the vertical, even if it is a small angle. But even if the bullet hits the ball completely in a vertical direction, my above post still applies: it's just that instead of a torque being applied to the Earth, a completely downward force is applied to the Earth via the gun's recoil. The concepts are still the same.)

Thank You collinsmark for you help.:)
But its kind of confusing because the C.O.Momentum is derived from constancy of position of Centre of Mass(which you used to get that equation) isn't it?

Yes, in a closed system with no external forces acting upon it, the center of mass of the system travels at a constant velocity. And if we define all velocities as being relative to the center of mass, then the center of mass never moves at all. This is the result of Newton's first law. In this case you can phrase Newton's first law of motion as, "If the center of mass of a system is at rest, the center of mass tends to stay at rest unless acted upon by an outside force."

But once again, as I mentioned earlier, bringing in the entire mass of the Earth into the system, so that conversation of momentum can be used at all times, makes the problem more complicated than it needs to be. All I'm saying is that if you wanted to, you could do it. Not that you should do it.

But you should use conservation of momentum at least at the moments immediately before and immediately after the collision, even for the bullet and ball problem. (This assumes that the dynamics of the collision itself happen very quickly).

• Find the velocity of the bullet immediately before the collision.
• Find the velocity of the bullet/ball combination at the moment immediately after the collision, using conservation of momentum.
• After that, you don't need to use conservation of momentum anymore. Standard kinematics can get you the rest of the way. And you might even be able to use conservation of energy after the collision, depending on the details of the problem. [Perhaps involving something of the form,
  ½(m_bul + m_ball)v² = (m_bul + m_ball)gΔh.]