Weird doubt on momentum

Question

Let us assume there is a car of mass $m_1$ on top of which there is a block of mass $m_2$. The car is moving with velocity $v$. If suddenly,the block falls down,then it is common sense that the speed of the car will increase since the car has become light. But how do we feel this using rigorous mathematics and conservation of momentum? Please allow me to express myself:

Suppose we consider the car without the block as the system. So that means the momentum of the car will be conserved. So $m_1v=m_1v_{\mathrm{final}}$ meaning final and initial velocities will be same, however that is clearly not true. That means in the only car system,some kind of external force is acting on the only car system. But what is this external force? If we draw the free body diagram of the car,what will be the direction of this external force? What does a block having to fall from the car,got to do with external force? In other words,how do we define external force in this system? Is this force the normal vector that we use to represent force on a body?If that's case,could you please show this force as vector?

I am extremely sorry if my question sounds dumb. But i can't feel the conservation of momentum due to the vague definition of external force here. Or a rigorous mathematical calculation proving that the momentum of the car and block system will be conserved here will be very helpful(like we deduce the formula in collision problems).

Answer 1

When the object over the car falls down, it keeps at first the same horizontal velocity. The total linear momentum (car + object) doesn't change. So there is no reason to expect a change of the car velocity.

Of course, as soon as the object hits the road, it will stop due to the friction, but that is another story.

Answer 2

When you look at a change in mass you always have to consider the conservation of the total momentum. Let's look at the car, but let's make it a car in space for simplicity, going at $ vM$, (this includes the brick with mass $m$). The mass you subtract will always have a momentum before and after the separation and you always have to consider both.

The only way to make your scenario sensible would be something like this:

Let's say the brick was just attached with a little cable (that is weightless and magically does not enact any forces) which you now cut so that the brick will continue to float next to your car with $ v m$, for ever if you don't do anything else.

Suppose we consider the car without the block as the system. So that means the momentum of the car will be conserved.

That is really the crux, this kind of just "cutting things out" of your system can only be done with things that don't interact with anything in your system. Really the cutting of the magic cable is not a physical act, as it changes nothing, it's just a change in perspective.

The car now has smaller momentum $v (M-m)$, the brick will have $ v m$, together still $v (M-m) + v m =vM$ right?

BUT that only works because they don't interact the cable was never really there to begin with.

That you are asking now which force changed the momentum of the car, is like going from an inertial system in which the car is moving $ P = vM$ to one inside the car $ P' = 0$ and asking where is the force that stopped the car?

There is none, you just changed perspective.

Answer 3

If suddenly, the block falls down, then it is common sense that the speed of the car will increase since the car has become light.

It's not really common sense.

For a real scenario the car will encounter external dissipative forces, primarily air drag, that needs to be matched by a forward static friction force applied by the road to the wheel (in response to wheel torque) for a net force of zero. Since the block is on top of the car, part of the air drag force is associated with the block. If the block falls off, the air drag force is reduced. Then, if the applied torque to the wheel(s) is unchanged, there will be a net force acting forward resulting in acceleration of the car.

On the other hand, if the block were inside the car it would experience no air drag. If it were dropped out the window, there would be no change in the external forces acting upon the car. The car will continue to move with the same velocity.

Suppose we consider the car without the block as the system. So that means the momentum of the car will be conserved.

Sticking with the "real car scenario" discussed above, momentum is not conserved because the removal of the block results in a reduction of the air drag force acting on the car, resulting in a net forward force and acceleration.

That means in the only car system, some kind of external force is acting on the only car system. But what is this external force?

Again, there are two main external forces acting on the car- air drag and the static friction force the road applies to the drive wheel(s) which is equal and opposite to the force the drive wheel(s) applies to the road per Newton's 3rd law. The loss of the block reduces the air drag force for a net force forward on the car.

If we draw the free body diagram of the car, what will be the direction of this external force?

A free body diagram of the car will show a static friction force applied to the wheel acting forward and an opposing air drag force acting backwards. When the two forces are equal the car is moving at constant velocity. When the block falls off, there will be a reduction in the drag force resulting in a net external force acting forward.

What does a block having to fall from the car, got to do with external force?

The air drag force acting backwards on the car is reduced when the block falls off.

In other words, how do we define external force in this system? Is this force the normal vector that we use to represent force on a body? If that's case, could you please show this force as vector?

In this example the only forces of interested are the horizontal forces. So the vectors for the static friction force and air drag force are simply vectors in the horizontal direction.

Hope this helps.

Answer 4

if an object is moving with a constant velocity in the absence of any forces, and apart of it self falls off. The momentum of the object decreases, only because of a loss of mass. the velocity of the object will not change as no forces are acting on it.

With a car its slightly different. assuming no air resistance to less complicate things. the engine constantly does work on the car, this causes some constant acceleration F/m = a. If the car loses mass, The same force, is now applied to a smaller mass. Causing a greater acceleration.

with air resistance the effect is an increase to some higher constant velocity.

Answer 5

Here is your wrong assumption: "If suddenly,the block falls down,then it is common sense that the speed of the car will increase since the car has become light"

There are two problems. The first is that Newton's First Law means that the block cannot simply "fall down." There has to be some force on it that causes it to accelerate. The second problem is that the car does not necessarily change its velocity. Whether or not the car changes its velocity depends on what causes the block to move.

The key is whether the force that causes the block to fall is an internal force or an external force. The law of conservation of momentum says $$\Delta P = F_{external} \Delta t$$ The total momentum of a system can only change when there is an external force. That means a force that is between an object in the system and an object outside the system. Forces between the car and the block don't change the total momentum, but forces between the block and some other object can change the total momentum.

Here are two different examples:

Example 1 -- Internal forces only

The car and the block are traveling at $v_{c1} = v_{b1} = v$. The driver of the car pushes the block backwards with a force $F$ for a short time $\Delta t$, which causes the block to fall off the car.

This is an internal force, so the momentum of the car+block is conserved. The block now has velocity $v - F\Delta t/m_b$ and the car now has velocity $v + F\Delta t /m_c$. In terms of Newton's Third Law, the force from the car caused the block to accelerate backward, and the equal and opposite force from the block on the car caused the car to accelerate forward.

Example 2 -- External force

The car and the block are traveling at $v_{c1} = v_{b1}=v$. A bird flies by and hits the block, knocking it off the top of the car. The bird applies a force to the block $F$ for a time $\Delta t$.

The force from the bird is an external force, so it does change the total momentum of the car plus the block. The velocity of the block is now $v - F\Delta t/m_b$ but the velocity of the car is still $v$. The total momentum of the car plus block is now $$P_{final} = P_{initial} - F\Delta t$$

Conclusion

What happens to the car when the block falls depends on what made the block fall. The law of conservation of momentum doesn't guarantee that the car will automatically change velocity just because the block does. We still have to think about which forces are causing the changes.

Answer 6

As explained in some of the answers, the car does not accelerate when the block falls. However there is a "if". The car's velocity does not change if the car is moving without its engine turned on. If the engine is on, providing a forward force F, then the car accelerates when its mass decreases. $ a= \frac{F}{m} $. Reduce m translates into an increase of the acceleration.

Answer 7

The car's environment is not fully indicated in the question. The word falls implies a gravity. Assuming the car travels on a road then the falling of the block is perpendicular to the cars motion. In this case the car and the block can be analyzed as separate systems.

For the car the normal force is reduced thus reducing friction:

1. If the engine applies a constant force then the car will accelerate. At the moment the block is detached, momentum is conserved. After the car will establish a new momentum. This has nothing to do with conservation of momentum.

2. If the engine applies a constant velocity then the car will not accelerate. Momentum is conserved when the block is released. As time goes by the car's momentum is not considered conserved, but rather maintained by inertia while the car's engine is used to deal with friction.

For the block:

In the absence of friction, the block will continue to move at the velocity of release, parallel to the car. So horizontal momentum of the block is conserved. The force of gravity will accelerate the block downward increasing the vertical momentum of the block.

Notice that the combined system horizontal momentum is conserved in the absence of friction.

The only external force acting on the car is friction that is overcome by the engine. In the absence of friction there is no net external force on the car. In the absence of friction the only force acting on the released block is gravity. There is no interaction between the car and the block.