Linear Momentum Study Guide

Impulse-Momentum Theorem

Newton's second law of motion can be used to reveal the connection between the impulse and the momentum. When the application of an external net force changes the object's speed, the equation of motion is:

v = v₀ + a · t

Then, the uniform acceleration can be found to be:

a = (v – v₀)/ t

But, according to Newton's second law, the net force on the object is proportional to the acceleration and the mass of the object:

F = m · a

F = m · (v – v_o)/t

F = (m· v – m · v_o)/t

F = (p – p_o)/t

F = Δ p/t

F· t = Δ p

J = Δ p

As one can see, the momentum measures the state of motion at one moment, whereas when defining impulse, we talk about a change in the motion of the object. If the object is at rest or moving with constant velocity, impulse is zero because momentum does not change.

Impulse

Momentum theorem says that when a net force acts on an object the impulse of the force is equal to the change in momentum:

J = Δ p

In this expression, both the momentum and the impulse are vector quantities.

Example

Consider the previous example with the soccer ball and imagine you are again in a case where motion is one dimension. Your ball is directed toward a fence and in the process of interacting with the fence, the ball changes its direction of motion and returns toward you with less speed: The ball moves away with 0.8 m/s and returns to you with 0.6 m/s. Find out the impulse if the ball's mass is 810 g.

Solution

Setting the equation right means to look for the clues, and in this problem, we include the 1-D motion, which means we can give up the vector signs. The second factor is the motion of the ball itself: Going away will be a positive number while returning will be a negative number.

v_o = 0.80 m/s

v = –0.60 m/s

m = 810 g = 0.81 kg

J = ?

According to the impulse momentum theorem:

J = Δ p

J = m · (v – v_o)

J = 0.81 kg · (–0.60 – 0.80) m/s

J = –l.l N · s

Conservation of Momentum

Two types of forces can act on a system:

• Internal forces (within the system)
• External forces (outside the system)

When we discuss internal forces, we talk about the interaction between the atomic components and chemical bonds, whereas for external forces, we talk about tension, friction, and different types of push and pull forces. There are also systems we call isolated systems. A system for which the result of all the external forces is zero is called an isolated system. For these systems, the principle of conservation of linear momentum says that the total linear momentum of an isolated system remains constant, and at different times, the momentum of the system has the same value.

Example

Now consider yourself on rollerblades. You hold a ball in your hands. Your mass and the ball's is 75 kg, while the mass of the rollerblades is 3.5 kg. There is no friction with the asphalt. You step down on the rollerblades with a speed of 0.50 m/s, and the rollerblade starts moving in the opposite direction. Find the speed of the rollerblade immediately after your descent.

Solution

First, we need to see what is given, check units, and then try to solve for the unknown, which is the speed of the blade (see Figure 6.3).

m = 3.5 kg

m_you = 75 kg

v_you = +0.50 m/s

v_blade = ?

Because there is no friction or other external force acting on the system, the linear momentum for this system (you and rollerblades) is conserved. So before and after you step down, there will be the same momentum:

P_total = P + P_blade = constant

This is a vectorial relationship. The total momentum stays the same for the system, and because the system starts at rest, velocity is 0, then momentum is also 0. After you step down, the total momentum should be found to be 0:

Before jump:

p_total = 0

After jump:

p_total = m v_blade + m_you v_you

Then the final equation is:

p_total = 0

0 = m v_blade + m_you v_you

m v_blade = –m_you v_you

v_blade = –(m_you /m) v_you

v_blade = –11 m/s

Conservation of Linear Momentum

As long as there is no net external force acting on the system, the linear momentum before and after the collision will be the same.

Elastic and Perfectly Inelastic Collisions

The following two types of collisions can be considered:

• Elastic collisions: the collisions in which the total energy of motion (which we will call kinetic) of the system before collision equals the total kinetic energy after the collision (conservation of kinetic energy).
• Perfectly inelastic collisions: the collisions in which the kinetic energy before and after collision is different. This happens when the two objects stick together after the collision.

Example 1

Consider two objects in an elastic collision (one of mass 25 kg and the other of mass 5 kg). Also, let's say that object 1 is at rest and object 2 is moving at a speed of 1.5 m/s. The two objects will collide head on, and the second object will recoil at a speed of 1.2 m/s. What is the speed of the heavy object? There is no external force acting on the system. See Figure 6.4.

Solution 1

In this case, both the energy and momentum are conserved.

m₁ = 25 kg

m₂ = 5.0 kg

v_1initial = 0 m/s

v_2initial = 1.5 m/s

v_1final = ?

v_2final = –1.2 m/s (minus at v₂ final because we are told that it recoils)

The momentum equation is:

m₁ v_{1 initial} + m₂ v_{2 initial} = m₁ v_{1 final} + m₂ v_{2 final}

25 · 0 + 5 · 1.5 = 25 · v_{1 final} + 5 · (– 1.2)

v_1final = 0.54 m/s

Example 2

A ball of clay moving on a horizontal plane with a momentum of 0.8 kg · m/s hits a plastic ball, head on. The plastic ball has a mass larger than the mass of the clayball with 0.2 kg. Find out the initial speed of the plastic ball if, after the perfect inelastic collision, the system moves at a speed of 0.5 m/s in the same direction as the incoming clay ball. We are also able to calculate that the final momentum is 1.6 times smaller than the momentum of the clay ball. Determine the mass of the two balls.

Solution 2

All quantities in the example are expressed in SI. To calculate the speed and the two masses, we have to identify the type of collision and the given information. This is a case of a perfectly inelastic collision, so after the collision, the two objects move as one. We will use a subscript 1 for the clay ball and 2 for the plastic ball.

p_{initial clayball} = m₁ v_{1 initial} = + 0.8 kg · m/s

(we'll consider this object moving to the right)

m₂ = 0.2 kg + m₁

v_final = 0.5 m/s

p_final = p_{initial clay ball} ÷ 1.6 = +0.8 kg · m/s ÷ 1.6 = +0.5 kg · m/s

v_{2 initial} = ?

We apply the principle of conservation of momentum principle and write:

m₁ v_{1 initial} + m₂ v_{2 initial} = (m₁ + m₂)v _final

0.8 kg · m/s + (0.2 kg + m₁) · v_{2 initial} = (m₁ + 0.2 kg + m₁ ) · 0.5 m/s

But the left-hand side of the previous equation is the final momentum, which is given to us as:

(m₁ + 0.2 kg + m₁ ) · 0.5 m/s = +0.5 kg · m/s

And so we can calculate first for the mass m₁ by dividing both sides by 0.5:

(m₁ + 0.2 kg + m₁ ) = 1

2 · m₁ + 0.2 kg = 1

2 · m₁ = 0.8 kg

m₁ = 0.4 kg

m₂ = 0.6 kg

Returning to the principle of conservation of momentum, we have now:

0.8 kg · m/s + 0.6 kg · v_{2 initial}= 0.5 kg · m/s

0.6 kg · v_{2 initial} = –0.3 kg · m/s

v_{2 initial} = –(0.3/0.6) m/s

v_{2 initial} = – 0.5 m/s

The minus sign tells us that the motion of the second ball is in the opposite direction to the motion of the clay ball.

2-D Collisions

Collisions do not always happen in one dimension. In most real-life applications, the encounter is at least a 2-D collision. The need for the vectorial interpretation of velocity, momentum, and impulse is made evident by such cases. This applies to collisions at the atomic level (e.g., between nuclear particles) and at the macroscopic level (e.g., planetary motion).

The most general expression of the principle of conservation of momentum is one that considers the vectors velocity for the objects involved in the collision. Take two objects of masses m₁ and m₂ colliding perfectly elastically:

While for a perfectly inelastic collision:

m₁ · v_1initial + m₂ · v_2initial = (m₁ + m₂) · v_final

The simplification comes from the fact that we can treat independently the equations for each of the three rectangular axes: x, y, and z.

And the same expressions can be written for the y and the z direction.

Perfectly Elastic

m₁ · v_{1 initial} + m₂ · v_{2 initial} = m₁ · v_{1 final} + m₂ · v_{2 final} 

m₁ · v_{1x initial} + m₂ · V_{2x initial} = m₁· v_{1x final} + m₂ · v_{2x final}

Example

Consider a car collision in which one car is driving north at a speed of 16 m/s and the other is moving east at 14 m/s. The cars are of similar mass, 1,500 kg. Find the velocity immediately after the collision if the cars get entangled and move together.

Solution

The units for the data given in the problem are in SI, so there is no need to convert. This is a two-dimensional perfectly inelastic collision, so consider the vector components of the velocities of the two cars.

m₁ = m₂ = 1,500 kg

v_1initial = (0,16) m/s or 16 m/s moving north (we consider positive y direction north and positive x direction east)

v_2initial = (14,0) m/s or 14 m/s moving east

v_final = ?

p_final = ?

The expressions for the conservation of momentum for the two directions are:

m₁ · v_{1x initial} + m₂ · v_{2x initial} = (m₁ + m₂) · v_{x final}

m₁ · v_{1y initial} + m₂ · v_{2y initial} = (m₁ + m₂) · v_{y final}

Because the masses are all the same, they cancel out in both equations, leaving a factor 2 on the righthand side:

v_{1x initial} + v_{2x initial} = 2 · v_{x final}

v_{1y initial} + v_{2y initial} = 2 · v_{y final}

And replacing the known values of the x and y components of the velocity, we get:

0 + 14 m/s = 2 · v_{x final}

16 m/s + 0 = 2 · v_{y final}

And the results are:

v_{x final} = 7 m/s

v_{y final} = 8 m/s

or:

v_final = (7,8) m/s

Or, using Pythagoras's theorem for the sides of the right triangle:

v_final =

√v2_{x final} + v2_{y final} =

√72+82 = 9 m/s

And the direction is given by the angle with x- or y-axes.

α = 48°

So the final velocity is:

v_final= 9 m/s at α = 48° with the +x-axis.

Perfectly Inelastic

m₁ · v_{1x initial} + m₂ · v_{2x initial} = (m₁ + m₂) · v_{x final}

Practice problems of this concept can be found at: Linear Momentum Practice Questions