Lesson Explainer: Newton’s Second Law: Variable Mass Mathematics

In this explainer, we will learn how to use differentiation for Newton’s second law of motion of a particle with variable mass.

Let us first recall Newton’s second law of motion by defining Newton’s second law of motion for a body of constant mass.

Definition: Newton’s Second Law of Motion for Constant Mass

When a net force acts on a body, the body accelerates in the direction of the force. The magnitude of the acceleration depends on the magnitude of the force and on the mass of the body according to the formula 𝐹=𝑚𝑎, where 𝑚 is the constant mass of the body and 𝑎 is the acceleration of the body.

Newton’s second law of motion can also be expressed in terms of the rate of change of the momentum of a body. The momentum, 𝑝, of a body is defined as follows.

Definition: The Momentum of a Body

The momentum of a body is given by, 𝑝=𝑚𝑣, where 𝑚 is the mass of the body and 𝑣 is the velocity of the body.

Expressing Newton’s second law of motion in terms of rate of change of momentum gives 𝐹=𝑝𝑡=𝑡(𝑚𝑣).dddd

The mass and velocity terms are both bracketed as either term can be time varying or constant, so the rate of change of either term may need to be considered.

If the mass of an accelerating body is constant, the rate of change of the momentum of the body is given by 𝐹=𝑡(𝑚𝑣)=𝑚𝑣𝑡=𝑚𝑎,dddd which is the constant mass form of Newton’s second law of motion. Equivalently, for a body that moves uniformly when a force acts on it, the rate of change of momentum of the body is given by 𝐹=𝑡(𝑚𝑣)=𝑣𝑚𝑡.dddd

Suppose that a body changes in both velocity and mass with respect to time. If a body increases in both velocity and mass, then the force acting on the body must account for both the increase in velocity and the increase in mass.

When two variables change with respect to a third variable, the product rule can be used to differentiate the product of the two variables.

How To: Using the Product Rule to Differentiate a Product

Consider the variables 𝑎 and 𝑏 that both change with respect to time. The rate at which 𝑎𝑏 changes with respect to time is given by dddddd𝑡(𝑎𝑏)=𝑎𝑏𝑡+𝑏𝑎𝑡.

If we apply the product rule to the rate of change of momentum of a body, we find that 𝐹=𝑡(𝑚𝑣)=𝑚𝑣𝑡+𝑣𝑚𝑡.dddddd

The force acting on a body can be defined as follows.

Definition: Newton’s Second Law of Motion in terms of Momentum Change

When a net force acts on a body, the momentum of the body changes. The rate of change of the mass and velocity of the body depends on the magnitude of the force and on the mass and velocity of the body according to the formula 𝐹=𝑚𝑣𝑡+𝑣𝑚𝑡,dddd where 𝑚 is the mass of the body and 𝑣 is the velocity of the body.

Let us consider an example in which a body changes mass while a force acts on it.

Example 1: Finding the Force Acting on a Body with Variable Mass Moving at a Constant Velocity

Fill in the blank: The force acting on a mass varying according to the function 𝑚(𝑡)=(5+2𝑡)kg and moving with a constant velocity of 4 m/s is .

Answer

A force acting on a body produces a change of momentum of the body: 𝐹=𝑡(𝑚𝑣)=𝑚𝑣𝑡+𝑣𝑚𝑡.dddddd

The body is stated to have a constant velocity, so dd𝑣𝑡=0.

Hence, 𝐹=𝑣𝑚𝑡.dd

The value of 𝑣 is stated to be 4 m/s, which can be substituted to give 𝐹=4𝑚𝑡.dd

For a force to act on a body to move the body with constant velocity and hence with zero acceleration, the mass of the body must change over the time that the force acts on it.

The question expresses the mass of the body as a function of time. The mass in kilograms as a function of time, 𝑚(𝑡), is given by 𝑚(𝑡)=(5+2𝑡), the value of which does indeed increase with increasing 𝑡.

Differentiating 𝑚(𝑡) with respect to time gives dddd𝑚𝑡=𝑡(5+2𝑡)=2.

The mass of the body increases by 2 kilograms per second.

The rate of change of mass can be substituted into the formula as follows: 𝐹=4𝑚𝑡𝐹=4(2)=8.ddN

Therefore, a body that is moving at a constant velocity of 4 m/s that starts to increase in mass by 2 kilograms per second must be acted on by a force of 8 N to maintain that constant velocity.

The mass of the body does not affect the force required to maintain its uniform motion, only the rate of change of its mass.

Let us look at an example of a force acting on a body, where both the mass and the velocity of the body vary with time.

Example 2: Finding the Force Acting on a Body with Variable Mass at Any Time

A body moves in a straight line. At time 𝑡 seconds, its displacement from a fixed point is given by 𝑠=6𝑡+9𝑡m. Its mass varies with time such that 𝑚=(8𝑡+9)kg. Write an expression for the force acting on the body at time 𝑡.

Answer

Both the velocity and the mass of the body vary with time, so the force acting on the body is found using 𝐹=𝑚𝑣𝑡+𝑣𝑚𝑡.dddd

The rate of change of mass of the body can be found by differentiating the function for the mass of the body with respect to time as follows: dddd𝑚𝑡=𝑡(8𝑡+9)=8.

The mass of the body increases by 8 kilograms per second.

The function for the rate of change of the velocity of the body with respect to time is not given by the question, but the displacement as a function of time is given. The rate of change of displacement is equal to the instantaneous velocity, 𝑣, of the body, and so 𝑣=𝑠𝑡=𝑡6𝑡+9𝑡=(12𝑡+9).dddd

As 𝑣 is known, we can express the rate of change of 𝑣 as follows: dddd𝑣𝑡=𝑡(12𝑡+9)=12.

We can use the following formula: 𝐹=𝑡(𝑚𝑣)=𝑚𝑣𝑡+𝑣𝑚𝑡𝐹=((8𝑡+9)(12))+((12𝑡+9)(8))𝐹=(96𝑡+108)+(96𝑡+72)𝐹=(192𝑡+180).ddddddN

A time-varying force must be expressed as a function of 𝑡 and have an instantaneous value at all instants that the force acts. Let us look at an example in which an instantaneous value of a time-varying force is determined.

Example 3: Finding the Force Acting on a Body with Variable Mass at a Given Time

A body moves in a straight line. At time 𝑡 seconds, its displacement from a fixed point is given by 𝑠=2𝑡+5𝑡+4m. Its mass varies with time such that 𝑚=(6𝑡+5)kg. Determine the force acting upon the body when 𝑡=3s.

Answer

To determine the force acting at an instant, the function representing the change in the force with time must be determined.

Both the velocity and the mass of the body vary with time, so the force acting on the body is found using 𝐹=𝑚𝑣𝑡+𝑣𝑚𝑡.dddd

The rate of change of mass of the body can be found by differentiating the function for the mass of the body with respect to time as follows: dddd𝑚𝑡=𝑡(6𝑡+5)=6.

The function representing the velocity of the body is given by 𝑣=𝑠𝑡=𝑡2𝑡+5𝑡+4=(4𝑡+5).dddd

Differentiating the function representing 𝑣 with respect to time gives us dddd𝑣𝑡=𝑡(4𝑡+5)=4.

As the product of 𝑚 and 𝑣, 𝑚𝑣, is to be differentiated with respect to time, we can use the product rule to differentiate 𝑚𝑣 with respect to 𝑡 as follows: 𝐹=𝑚𝑣𝑡+𝑣𝑚𝑡𝐹=((6𝑡+5)(4))+((4𝑡+5)(6))𝐹=(24𝑡+20)+(24𝑡+30)𝐹=(48𝑡+50).ddddN

Substituting 𝑡=3 gives us 𝐹=48(3)+50=194.N

Let us consider Newton’s second law of motion with variable mass when it is applied to a system where the velocity is given in terms of vectors.

Definition: Newton’s Second Law of Motion for Variable Mass in terms of Vectors

Given a body of mass 𝑚, with velocity 𝑣, the force that is being applied to the body is given by 𝐹=𝑚𝑣𝑡+𝑣𝑚𝑡,dddd where both 𝐹 and 𝑣 are vector quantities.

If we are given the vector for the displacement instead of the velocity, we can differentiate the displacement with respect to time to find the velocity since the relationship between displacement and velocity is the same in vector and scalar forms. We have that 𝑣=𝑠𝑡.dd

We can now look at an example of how this formula can be used.

Example 4: Finding the Force Acting on a Body Using Newton’s Second Law for Variable Mass with Vectors

Complete the following: An object of mass 10 g is moving in a two-dimensional plane with varying mass that increases with a rate of 5 g/s. The velocity vector for the object is given by 𝑣=2𝑡𝑖+12𝑡𝑗/cms. The horizontal component of the force acting on the object to keep it moving at this velocity is dynes.

Answer

We know that the mass of the object starts at 10 g, and it increases at a rate of 5 g/s. Therefore, we can say that 𝑚=10+5𝑡.

Also, since the rate of increase of mass is 5 g/s, we have that dd𝑚𝑡=5.

In order to find the force that is acting on the body, we will need to use Newton’s second law for variable mass in terms of vectors. This tells us that 𝐹=𝑚𝑣𝑡+𝑣𝑚𝑡.dddd

In order to find dd𝑣𝑡, we simply differentiate 𝑣 with respect to 𝑡. This will give us dd𝑣𝑡=4𝑡𝑖+24𝑡𝑗.

We now have all the components we need to find 𝐹. Substituting into our formula, we find 𝐹=(10+5𝑡)4𝑡𝑖+24𝑡𝑗+2𝑡𝑖+12𝑡𝑗×5=40𝑡+20𝑡𝑖+240𝑡+120𝑡𝑗+10𝑡𝑖+60𝑡𝑗=40𝑡+30𝑡𝑖+240𝑡+180𝑡𝑗.

Since the units we have used so far are grams and centimetres per second, the unit for our force will be dynes. We can say that 𝐹=40𝑡+30𝑡𝑖+240𝑡+180𝑡𝑗.dynes

Now, the question has asked us to find the horizontal component of this force, and we are only interested in the 𝑖 component. Therefore, our solution is 40𝑡+30𝑡.

It is reasonable to wonder what physical process could produce the increase in the mass of a body. Suppose that a body accumulates some of the matter that it makes contact with that is part of the medium that the body travels in. The mass of the accumulated matter is added to the mass of the body.

Representing such a situation realistically would involve determining the rate at which the area of contact of the body and the medium changed as well as the rate at which the velocity of the body changed. Considering only these variables would require assuming that neither the density of the medium nor the process of accumulation of matter from the medium varied due to the velocity of the body or the area of contact of the body and the medium.

Let us consider an example in which the various processes affecting the accumulation of matter by a body are represented in a simplified way.

Example 5: Finding the Force Acting on a Body with Variable Mass at a Given Time Using Newton’s Second Law

A ball of mass 5 g was moving in a straight line through a medium loaded with dust. The dust was accumulating on its surface at a rate of 1 g/s. Find the magnitude of the force acting on the ball at time 𝑡=5seconds, given that the displacement of the ball is expressed by the relation 𝑠(𝑡)=23𝑡+𝑡+7𝑡+1𝑐, where 𝑐 is a unit vector in the direction of the motion and the displacement is measured in centimetres.

Answer

The force acting on the ball is given by 𝐹=𝑚𝑣𝑡+𝑣𝑚𝑡.dddd

The ball has a mass of 5 g at 𝑡=0, which increases at a rate of 1 g/s. The function representing the change of mass of the ball with time, 𝑚(𝑡), is given by 𝑚(𝑡)=(1𝑡+5)=(𝑡+5).

Differentiating 𝑚(𝑡) gives us dd𝑡(𝑡+5)=1.

The function representing the velocity of the body is given by 𝑣=𝑠𝑡=𝑡23𝑡+𝑡+7𝑡+1=2𝑡+2𝑡+7.dddd

Differentiating the function representing 𝑣 with respect to time gives us dddd𝑣𝑡=𝑡2𝑡+2𝑡+7=(4𝑡+2).

As the product of 𝑚 and 𝑣, 𝑚𝑣, is to be differentiated with respect to time, we can use the product rule to differentiate 𝑚𝑣 with respect to 𝑡: 𝐹=𝑡(𝑚𝑣)=𝑚𝑣𝑡+𝑣𝑚𝑡.dddddd

Recalling that dd𝑚𝑡=1, we have 𝐹=((𝑡+5)(4𝑡+2))+2𝑡+2𝑡+7.

Substituting 𝑡=5 gives us 𝐹=(10(22))+(50+10+7)𝐹=220+67=287/.gcms

The mass is in grams and the displacement is in centimetres, and so the force calculated in newtons is multiplied by the product of the number of centimetres in a metre and the number of grams in a kilogram to give a force in newtons: 𝐹=2871010=287×10.N

One dyne is equal to 10 N, so the force is 287 dynes.

Let us now look at another such example.

Example 6: Finding the Rate of Change of the Mass of a Ball as It Moves Through a Dusty Medium

A metallic ball moves in a straight line with a constant velocity of magnitude 1 m/s. It enters a dusty medium. If the force acting on the ball at any instant is 10 dynes, find the rate of change of the mass of the ball due to the dust adherence to its surface.

Answer

The ball has a constant velocity. So, in the formula 𝐹=𝑚𝑣𝑡+𝑣𝑚𝑡,𝑣𝑡=0.dddddd

The fact that the velocity is constant allows us to express the formula as 𝐹=𝑣𝑚𝑡.dd

The rate of increase of the mass of the ball is directly proportional to the force acting on the ball.

The force acting on the ball is 10 dynes, where 1=1×1=10.dynegcmsN

The force is equal to the rate of change of momentum of the ball, given in newtons by 1010=10=𝑝𝑡=𝑡(𝑚𝑣).dddd

The velocity of the ball is 1 m/s, so the rate of change of momentum can be written as dddddd𝑝𝑡=1𝑚𝑡=𝑚𝑡.

Combining the value of the force with the expression for the rate of change of momentum gives us 10=𝑚𝑡,dd where 𝑚 is in kilograms (kg), the force is in newtons (N), and the velocity is in metres per second (m/s). The rate of change of mass, in grams per second (g/s), is given by ddgs𝑚𝑡=1010=0.1/.

Key Points

  • When a net force acts on a body of constant mass, the body accelerates in the direction of the force. The magnitude of the acceleration depends on the magnitude of the force and on the mass of the body according to the formula 𝐹=𝑚𝑎, where 𝑚 is the mass of the body and 𝑎 is the acceleration of the body.
  • The momentum of a body is given by 𝑝=𝑚𝑣, where 𝑚 is the mass of the body and 𝑣 is the velocity of the body.
  • Expressing Newton’s second law of motion in terms of rate of change of momentum gives 𝐹=𝑝𝑡=𝑡(𝑚𝑣).dddd
    Using the product rule, this can be expressed as 𝐹=𝑚𝑣𝑡+𝑣𝑚𝑡.dddd

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