Force of Momentum due to Mass Change instead of Velocity

Foreword: NOTE: USE MOZILLA FIREFOX BROWSER, SO YOU CAN SEE THE EQUATIONS PROPERLY. OTHER BROWSERS MAY NOT DISPLAY THE EQUATIONS PROPERLY.

Explanation of force, due to change of mass instead of velocity. In this case, velocity is kept constant while mass changes linearly.

By: Chrysanthus Date Published: 25 Apr 2020

Introduction

Newton’s Second Law states that force is proportional (equal) to the rate of change of momentum. In many cases the mass is constant while only the velocity changes. There are cases where both the velocity and mass, change. In this document, what is described, is only the cases where the mass is changing but the velocity is constant during contact (impact). Gun firing of a bullet will also be described.

Before reading this article, you should have passed in middle school (British Secondary School) physics and mathematics, and you should also have studied some differentiation.

Momentum and Force

Momentum is the product of mass and velocity.

Let mass be represented by m.
Let velocity be represented by v.
Let force be represented by F.
Then

$F = \frac{d m v}{d t}$

If the force is constant and the mass is constant while the force acts, then this equation becomes:

F = \frac{m v - m u}{t}

where v is the final velocity and u is the initial velocity in the duration of the force. m is the constant mass which does not change in the duration of the force. t is the time taken while the force acts, as the velocity changes from u to v.

If the force is constant and this time it is the velocity that is constant, while the mass is variable, in the duration of the force, then the first equation instead becomes:

F = \frac{m v - n v}{t}

where m is the final mass quantity and n, the initial mass quantity while the force acts. v is the constant velocity which does not change in the duration of the force. t is the time taken while the force acts, as the mass quantity changes from n to m.

In this article, only the situations for constant force and the last formula (equation) above, is discussed.

Note: This article is about the force during contact; not force acting before or after contact.

v in the last equation above can be factorized out to have:

F = \frac{v (m - n)}{t}

which is the same as,

F = v \frac{(m - n)}{t} --- eqn 1

This can be interpreted as constant force is equal to, constant velocity times mass change per second, when the velocity is constant and the mass is variable.

Let change in mass be ∆m, that is, m - n = ∆m. So equation 1 is still:

F = v \frac{∆ m}{t} --- eqn 2

Both eqn 1 and eqn 2 are the same.

Force due to Water Flow

When water from a horizontal hose-pipe strikes a wall at right angles, a force is exerted on the wall. There is a bit of bouncing of water as the water hits the wall, before it falls downwards. Assume that there is no bouncing and as soon as the water hits the wall, it falls vertically downwards. In this case there is no vertical component of the force while the water hits the wall. In this illustration, the force of interest occurs when the water hits the wall horizontally, and not while the water is moving towards the wall or while it is falling down vertically after hitting the wall.

Assume that the water is moving toward the wall at constant velocity, v. Let the water hit the wall at the rate of ∆m/s kilograms per second. Then the force at the wall as the water contacts the wall, is:

F = v (\frac{∆ m}{t})

Suppose water flows out of the pipe at a rate of 3 kg/s. This means the water hits the wall at 3 kg/s. So at the wall, the mass changes at 3kg/s. Also supposed the water moves to the wall at a constant velocity of 5 m/s. Applying the above equation, the force is:

F = 5 (3)

\Rightarrow F = \underline{15 N}

Rain falling Vertically on Flat Roof

The explanation of rain falling vertically on a flat roof is similar to that of the water hitting the wall, but this time the action is vertical and no horizontal component of the force is assumed. When a drop of rain falls on a flat roof, it spreads out when it reaches the roof; a force is exerted on the roof. Assume that the drop flattens out on the roof; so there is no horizontal component of the force while the rain hits the roof. In this illustration, the force of interest occurs when the rain hits the roof vertically, and not while the drops are moving towards the roof or while they are spreading horizontally after hitting the roof.

Assume that the rain is moving toward the roof at constant velocity, v. Let the rain hit the roof at the rate of ∆m/s kilograms per second. Then the force on the roof as the rain contacts the roof, is:

F = v (\frac{∆ m}{t})

Suppose rain falls vertically at a steady rate of 0.25 kg/s on a flat roof. This means the rain hits the roof at 0.25 kg/s. So on the roof, the mass changes (adds) at 0.25 kg/s. Also supposed the rain is falling at a constant velocity of 10 m/s. Applying the above equation, the force is:

F = 10 (0.25)

\Rightarrow F = \underline{2.5 N}

Sand falling Vertically on a Horizontal Conveyor Belt

Consider a horizontal conveyor belt moving rightward. Assume on the left, there is a funnel dropping sand at a steady rate of ∆m/s kilograms per second, vertically on the belt. Also consider the situation where the funnel has been dropping sand for some time, and continues to drop sand. So, there is some mass of sand on the moving belt and it is increasing. The belt is moving horizontally, rightward with increasing mass of sand. If the belt stops moving abruptly, then the sand on it will flow to the right. So there is a horizontal force of sand, due to its increasing mass, rightward.

If the conveyor belt has been design to move at a constant speed, then the rightward velocity of the sand is constant, at v m/s. The mass is increasing at a steady rate, while the velocity is constant. So the rightward force is constant.

Assume at the beginning of the time interval t, the mass on the belt is n kg. Assume at the end of the same time interval t, the mass on the belt is now m kg; while the funnel continues to drop sand at a steady rate. Then the horizontal force of the sand, pointing rightward, is:

F = v \frac{(m - n)}{t}

Suppose sand is allowed to fall vertically at a steady rate onto a horizontal conveyor belt, moving at a steady velocity of 0.05 m/s. Also suppose, at the beginning of an interval of 3 seconds the sand on the belt was 4kg, and at the end of the interval the sand had increased to 4.6 kg. Applying the above equation, the moving force of the sand is:

$F = 0.05 \frac{(4.6 - 4)}{3}$ $\Rightarrow F = 0.05 (0.2)$ $\Rightarrow F = \underline{0.01 N}$

Rocket Movement

A good analogy for rocket movement is with the balloon. Blow air into a balloon, and release the balloon. The balloon will move away from the direction of the balloon nozzle, while air in the balloon will move out, away from the nozzle. As the balloon contracts, its forces air out.

Newton's third law states, that to every action (force) there is an equal but opposite reaction (opposite force). The action and reaction may be toward each other or may be away from each other. That is, the opposite forces may be toward each other or away from each other. In the case of the released balloon, they are away from each other. So, as the balloon releases air in one direction, it repels in the opposite direction. The releasing of air, is the reducing of mass of the balloon in one direction. The balloon with the reducing air goes in the opposite direction. That is Action and Reaction away from each other.

Rocket movement operates in a similar way. A rocket has gas. The engine of the rocket forces out gas in one direction, and in reaction, by Newton's third law, the rocket with reduced mass goes into the opposite direction. In other words, the rocket forces out, a small part of its mass in one direction and the rocket with its reduced mass, goes into the opposite direction.

Suppose a stationary rocket in space sent out gas, once in one second, and did not send out any more gas.

Let mass of rocket remaining be m,
Let total mass before release be n.
Let the time for the short process of separation of mass, be t (one second).
Then,
$F = v \frac{(m - n)}{t}$

Same as,
$F = v \frac{∆ m}{t}$
where ∆m = m - n (negative value); is the small mass released.

This is the force of the released mass and it is the same force of the remaining mass (rocket), but in the opposite direction.

Now, how do we know the different quantities, m, n, t and v in order to calculate the force?

If we designed the rocket and its engine, then we know the mass, n of the initial size of the rocket. We also know the small mass, that we designed to be release. So we know the mass remaining, m. If we truly designed the engine of the rocket, then we know at what velocity the small mass would be released. So we know v. Time of action (which is the same as time of reaction) has been supposed to be one second. So we know t as 1s.

What we do not know is the velocity at which the rocket of remaining mass, would move away. Do not confuse between velocity and force. The two forces are equal in magnitude but opposite in direction. This does not mean that the two velocities have to be equal in magnitude. However, the velocities are opposite in direction.

To know the velocity of the remaining mass (rocket), we need another principle, called the Principle of Conservation of Linear Momentum. This principle is actually derived from Newton's second and third laws. It is:

$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
which means, total momentum before separation is equal to total momentum after separation. Here, m₁ is the remaining mass; m₂ is the small mass; u₁ is the velocity before separation of the remaining mass, which is zero; u₂ is the velocity before separation of the small mass, which is also zero. v₁ is the velocity after separation of the remaining mass, which is still to be determined; v₂ is the velocity after separation of the small mass, which is v in the equation for the force above.

Substituting, we have:
$m_{1} (0) + m_{2} (0) = m_{1} v_{1} + m_{2} v_{2}$ $\Rightarrow 0 = m_{1} v_{1} + m_{2} v_{2}$ $\Rightarrow m_{1} v_{1} + m_{2} v_{2} = 0$ $\Rightarrow m_{1} v_{1} = - m_{2} v_{2}$

This is a linear equation in one unknown, because we know m₁, m₂, and v₂. Since m₁ is bigger than m₂, then v₁ is smaller than v₂.

That was v₁ in one second. If the rocket engine continues to pump the same small amount of gas every second, then the rocket would be accelerating, according to Newton's first law, which is, that an object will remain at constant velocity unless a force acts on it. In this case, the reaction force for each second, adds more velocity.

Explosive Force

There are many cases where momentum changes are produced by explosive forces. Firing a bullet from a gun is an example. When the gun is fired, as the bullet leaves the gun, the gun recoils (suddenly moves backwards). Before the bullet was fired, the gun had a total mass, which included the mass of the bullet. As the bullet left, there was separation of mass. The remaining mass is the gun and the small mass leaving, is the bullet.

Now, if the gun somehow fired itself in space, where there is no resistance to movement, then the bullet will leave, moving ahead with a constant velocity, while the gun will recoil behind with a different constant velocity.

Suppose the gun fired itself in space, and the short time for the explosion is t.

Let remaining mass (gun) be m,
Let total mass before firing be n.
Then,
$F = v \frac{(m - n)}{t}$

Same as,
$F = v \frac{∆ m}{t}$
where ∆m = m - n (negative value); is the small mass released, mass of bullet (ignore the negative sign for now).

This is the force of the bullet (released mass) and it is the same force of the gun (remaining mass), but in the opposite direction.

The Principle of Conservation of Linear Momentum is actually derived from Newton's second and third laws. It is:

$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
which means, total momentum before separation is equal to total momentum after separation. Here, m₁ is the remaining mass; m₂ is the small mass; u₁ is the velocity before separation of the remaining mass, which is zero; u₂ is the velocity before separation of the small mass, which is also zero. v₁ is the velocity after separation of the remaining mass; v₂ is the velocity after separation of the small mass, which is v in the equation for the force above.

Since m₁ is bigger than m₂, then v₁ is smaller than v₂.

Another thing you should know about explosion of this sort is the total energy after explosion. It is given by:

Total Energy = \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2}

Conclusion

Force is the rate of change of momentum. It is not only due to the rate of change of velocity with constant mass. It can also be due to the rate of change of mass with constant velocity. For Water Flow at a wall, the mass hitting the wall keeps increasing, though falling off. For Rain on a Flat Roof the mass dropping on the roof keeps increasing, though flattening out. For Sand falling on Conveyor Belt, the mass keeps increasing, as the belt moves. For the rocket, the mass decreases in opposite directions. For explosion, the mass decreases in opposite directions.

Broad Network

Series Articles