# Connexions

You are here: Home » Content » Force and invariant mass

### Recently Viewed

This feature requires Javascript to be enabled.

# Force and invariant mass

Module by: Sunil Kumar Singh. E-mail the author

Summary: Different forms of Newton’s laws of motion are consistent with each other in classical mechanics.

Newton’s second law is stated in terms of either time rate of change of momentum or as a product of mass and acceleration.

F = ( m v ) t F = ( m v ) t

and

F = m a F = m a

There is a bit of debate (even difference of opinion) about the forms (momentum .vs. acceleration) in which Newton’s second law is presented – particularly with respect to consideration of mass as invariant or otherwise. In this module, we shall discuss to resolve this issue.

As far as classical mechanics is concerned, we shall see that mass is always considered invariant. The reference to variable mass is actually about redistribution of mass - not the change in the mass of the matter. Such is the case with a leaking balloon or with a rocket. Matter (gas exhaust) is simply let out in the surrounding. So long, mass is not redistributed during the application of external force, we can safely take out mass from the differential of first equation. In that case, expression of force in two forms are same and equal.

However, the two forms of force defining relations are not equivalent in relativistic mechanics. Here, the definition of force in terms of linear momentum stands, but not the form involving mass and acceleration. We shall refer the actual relation in the end of this module.

## Mass invariance

The mass invariance in classical mechanics enables us to take the mass out of the differential equation given by Newton’s second law :

F = p t = ( m v ) t F = p t = ( m v ) t

This is important from the perspective that external force can be expressed in terms of acceleration :

F = m v t = m a F = m v t = m a

A situation, where measurement of force is required, assumes that mass is invariant. We proceed to measure force either in terms of acceleration or change in momentum in equivalent manner.

### Force as product of mass and acceleration

#### Example 1

Problem : A bullet of 10 gram enters into a wooden target with a speed of 200 m/s and comes to stop with constant deceleration. If linear penetration into the wood is 10 cm, then find the force on the bullet.

Solution : The motion of the bullet is resisted by wood, which applies a force in a direction opposite to that of the bullet. We assume wood to be uniform. The resistance offered by it, therefore, is constant. Now, as force is constant, resulting deceleration is also constant. Applying equation of motion,

u = 500 m/s ; v = 0; x = 10/100 = 0.1 m.

v 2 = u 2 + 2 a x 0 = 200 2 + 2 a x 0.1 a = - 40000 0.2 = - 200000 m / s 2 v 2 = u 2 + 2 a x 0 = 200 2 + 2 a x 0.1 a = - 40000 0.2 = - 200000 m / s 2

Force on the bullet,

F = ma = - 0.01 x 200000 = -2000 N

Note that bullet, in turn, applies 2000 N of force on the wooden target. This explains why bullet hits are so fatal for animals. Further note that we measured force as product of mass and acceleration.

### Force as a time rate of change in linear momentum

#### Example 2

Problem : A ball of mass “m” with a speed “v” hits a hard surface as shown in the figure. The ball rebounds with the same speed and at the same angle, θ, with vertical as before. Find average force acting on the ball.

Solution : We can determine force either (i) by measuring acceleration or (ii) by measuring change in linear momentum during the motion. Here, we take the second approach.

The linear momentum (remember that it is a vector) before hitting surface is :

p i = m v sin θ i - m v cos θ j p i = m v sin θ i - m v cos θ j

The linear momentum (remember that it is a vector) after hitting surface is :

p f = m v sin θ i + m v cos θ j p f = m v sin θ i + m v cos θ j

The change in linear during contact with surface, Δp,

Δ p = p f - p i = m v sin θ i + m v cos θ j - m v sin θ i + m v cos θ j Δ p = 2 m v cos θ j Δ p = p f - p i = m v sin θ i + m v cos θ j - m v sin θ i + m v cos θ j Δ p = 2 m v cos θ j

The average force is :

F = Δ p Δ t = 2 m v cos θ j Δ t F = Δ p Δ t = 2 m v cos θ j Δ t

Note that force is acting in vertical upward direction.

## Mass redistribution and Newton's second law

There are certain system in which mass is redistributed with the surrounding. The change in mass of the system results as there is exchange of mass between the system and its surrounding. The motion of a leaking balloon, as shown, is an example of changing mass system.

Similar is the situation in the case of a rocket. Now, does this exchange of mass with surrounding have any bearing on the form of Newton’s second law or about the meaning of force as product of mass and acceleration? This is the question that we need to answer. To investigate this question, let us consider the motion of the rocket. It acquires very high speed quickly as a result of high speed gas escaping in the opposite direction.

Applying Newton’s second law, considering that the mass of the system is changing :

F = ( m v ) t F = m v t + v m t F = m a + v m t F = ( m v ) t F = m v t + v m t F = m a + v m t

This apparent mass variant form of Newton’s second law appears to destroy the well founded meaning of force as "product of invariant mass and acceleration". As a matter of fact, it is not so.

Let us recall that the relation, F = ma, essentially underlines relation between force (cause) and acceleration (effect). Now look closely at the additional term resulting from change in mass in the context of a real time case like of a rocket. Rearranging, we have :

F - v m t = m a F - v m t = m a

Here, we need to be cautious in interpreting this expression. The first question that we need to answer is : what does "v" represent in the expression? Is it the velocity of rocket or escaping gas or is it the relative velocity of rocket with respect to escaping gas? The second question, then, is what does "a" represent?

We see that "v" should represent the velocity of the rocket in the ground inertial reference. This is the way we interpret Newton's second law for a constant mass body? The situation here, however, is different in that a part of the body is continuously being transferred to other system, which itself is moving. In order to simplify the analysis, we consider that rocket is ejecting gas at constant rate and with constant relative velocity. We can then say that ejected gas is applying force on the rocket in the reference of ejected gas, which is moving at uniform velocity. As such moving mass of gas constitute an inertial frame in which force is applied. The velocity, "v", therefore, represents the velocity of the rocket with respect to ejected mass of gas. We, then, rewrite the equation as :

F - v r m t = m a F - v r m t = m a

where " v r v r " represents the velocity of rocket with respect to gas being ejected.

Now, we turn to answer second question about acceleration. We know that acceleration is time rate of change of velocity. But, as we discussed, we measure velocity of the rocket with reference to ejected gas. Therefore, it follows that rate of change should also be associated with relative velocity - not the absolute velocity with respect to ground. However, there is an important difference between velocity and acceleration, as measured in two inertial frames viz. ground and ejected gas. Though, measurement of velocities are different, but measurement of "change" in velocity remains same in all inertial frames. Hence, we can interpret "a" as acceleration measured in either of two references without any distinction.

Now, we are in position to correctly interpret the additional term " - v r m t - v r m t " by answering the following question : "How could a high speed ejection of gas make the rocket accelerate in the absence of any other external force? We should remember that gravitational pull and air resistance, as a matter of fact, retards the motion of rocket. The answer lies in the fact that the additional term " - v r m t - v r m t " actually represents a force called "thrust" on the rocket. For rocket to accelerate, this thrust is an external force on the rocket. If we represent thrust by symbol "T", then :

T = - v r m t T = - v r m t

Substituting in the equation of Newton's law,

F + T = m a F + T = m a

The additional term, therefore, is not an “effect” term like “ma”, but a "cause" term representing “external force”, which results from the exchange of mass with the surrounding.

Clearly, there is no other external force other than thrust that gives such a great acceleration to a rocket. Gravitational and air resistance actually retards the motion of a rocket moving vertically upward. Thus, we can safely say that thrust is the only force that imparts acceleration in the direction of the motion of the rocket. In case rocket is fired from a region where other external forces like that of gravity and air resistance can be neglected, then we can put , F = 0. Then,

T = m a T = m a

In the nutshell, we see that a varying mass does not change the fundamental aspect of Newton’s second law of motion. It only modifies the external force on the body. The net external force is still equal to the product of mass and acceleration.

We shall reinforce these concepts with another approach that makes use of conservation of linear momentum in a separate module on rocket.

### Example 3

Problem : Air from a leaking balloon of mass 15 gm gushes out at a constant speed of 10 m/s. The balloon shrinks completely in 10 seconds and reduces to a mass of 5 gm. Find the average force on the balloon.

Solution : There is no external force on balloon initially. However, gushing air constitutes an exchange of mass between balloon and its surrounding. This generates an external force on the balloon given by :

F avg = - v r x Δ m Δ t F avg = - 10 x 0.005 - 0.015 10 F avg = 0.01 N F avg = - v r x Δ m Δ t F avg = - 10 x 0.005 - 0.015 10 F avg = 0.01 N

We summarize the discussion held, in the context of classical mechanics, so far as :

1. Given mass is invariant.
2. In some instances, the the variance of the mass of body system observed is actually a redistribution of mass from one body system to more than one body systems.
3. Irrespective of the nature of mass of a body system (varying or constant), external force is equal to time rate change of momentum.
4. If there is redistribution of mass, then it results into an external force on the original body, modifying the external force on it.
5. Irrespective of the nature of mass of a body system (varying or constant), the net external force is equal to the product of mass and acceleration.
6. Irrespective of the nature of mass of a body system (varying or constant), momentum and acceleration forms of Newton’s second law are equivalent and consistent to each other.

## Mass variance in relativistic mechanics

The detailed discussion of this topic is not part of the course. For information sake, we shall only outline the features of force law in relativistic mechanics. Here, momentum and acceleration forms are not equivalent. As a matter of fact, the momentum form is valid even in relativistic mechanics i.e.

F = p t F = p t

However, the acceleration form is not valid. For relativistic mechanics, this form is written, in terms of rest mass, m 0 m 0 , as :

F = m 0 a ( 1 - v 2 c 2 ) 1 2 + ( 1 - v 2 c 2 ) 3 2 m 0 v . a c 2 v F = m 0 a ( 1 - v 2 c 2 ) 1 2 + ( 1 - v 2 c 2 ) 3 2 m 0 v . a c 2 v

Yet another important aspect of force here is that force and acceleration vectors need not be parallel or in the same direction.

We summarize the discussion held, in the context of relativistic mechanics, so far as :

1. Given mass is not invariant.
2. The momentum and acceleration forms of Newton’s second law are not equivalent.
3. Momentum form of Newton’s second law is valid.
4. Acceleration form of Newton’s second law is not valid.
5. Force and acceleration need not be in the same direction.

## Content actions

PDF | EPUB (?)

### What is an EPUB file?

EPUB is an electronic book format that can be read on a variety of mobile devices.

My Favorites (?)

'My Favorites' is a special kind of lens which you can use to bookmark modules and collections. 'My Favorites' can only be seen by you, and collections saved in 'My Favorites' can remember the last module you were on. You need an account to use 'My Favorites'.

| A lens I own (?)

#### Definition of a lens

##### Lenses

A lens is a custom view of the content in the repository. You can think of it as a fancy kind of list that will let you see content through the eyes of organizations and people you trust.

##### What is in a lens?

Lens makers point to materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.

##### Who can create a lens?

Any individual member, a community, or a respected organization.

##### What are tags?

Tags are descriptors added by lens makers to help label content, attaching a vocabulary that is meaningful in the context of the lens.

| External bookmarks