8 Momentum, Change in Momentum & Impulse

Resources

Equations Introduced and Used for this Topic: (All equations can be written and solved as both scalar and vector and all equations are generally solved as vectors)

  • Impulse [latex](J) = F t = \Delta p[/latex]
  • Change in Momentum [latex](\Delta p) = m \Delta v[/latex]
  • Momentum [latex](p) = m v[/latex]
  • [latex]\Delta p = p_f - p_i[/latex]
  • [latex]\Delta v = v_f - v_i[/latex]

Where…

  • [latex]J[/latex] is the impulse or the product of force × time acting on a system, measured in either newton-seconds (Ns) or kilogram-metres per second (kg, m/s)
  • [latex]p[/latex] (momentum) is the product of mass times velocity and is measured in either newton-seconds (Ns) or kilogram-metres per second (kg, m/s)
  • [latex]\Delta p[/latex] is the change in momentum ([latex]p_f - p_i[/latex] or [latex]m(v_f - v_i)[/latex], measured in either newton-seconds (Ns) or kilogram-metres per second (kg, m/s)
  • [latex]m[/latex] is mass, commonly measured in kilograms (kg), grams (g) or tonnes (t)
  • [latex]F[/latex] is force, measured in newtons (N)
  • [latex]t[/latex] is time, commonly measured in seconds (s) or hours (h)
  • [latex]\Delta v[/latex] is the change in an object’s speed generally taken as the difference between the final and initial speed, measured in metres per second (m/s) or kilometres per hour (km/h)
  • [latex]v[/latex] is average speed, commonly measured in metres/second (m/s) or kilometres/hour (km/h)

Impulse and momentum are two very important concepts that can be easily derived from Newton’s Second Law and the equation defining acceleration.  Specifically using:

[latex]F = m a[/latex] & [latex]a = \dfrac{\Delta v}{t}[/latex]

When you substitute for a in Newton’s Second Law, you are left with:

[latex]F = m \dfrac{\Delta v}{t}[/latex]

Which is more commonly written as:

[latex]F t = m \Delta v[/latex]

Written in this form we now have both the equations for Impulse ([latex]F t[/latex]) and change in momentum ([latex]m \Delta v[/latex]). Both impulse and change in momentum are important physics concepts and have important implications in multiple areas such as analyzing collisions, predicting crash injuries and even helping to analyze rockets and space travel.

If you check out the units used in either F t or in m ∆v, you get two new unit combinations:

Ns and kg m/s (or kg ms-1)

These units are equivalent and can be used for either measure, with the most common unit being Ns because it is quicker to write down.  They are interchangeable and you should expect them to be switched around and should require no explanation.

In this chapter we are also adding these two measures of Impulse ([latex]J = F t[/latex]) and Change in Momentum ([latex]\Delta p = m \Delta v[/latex]) to Chapters 1 – 6 (our study of mechanics) and as such will be using all previous equations and variables that have been used up to this point.

8.1 Linear Momentum

Extra Help: Momentum

Equations Introduced or Used for this Section:

Momentum [latex](p) = m v[/latex]

Momentum is used in common discussions and generally refers to the concept that once something is in motion or is moving then it becomes difficult to change. You can hear this referred to the momentum of a game, where one team starts to dominate the other team, or in politics, where one politician or party is gaining strength and can challenge the leader or is simply increasing its lead over the others.

In physics, momentum is a very specific term that is used in both vector and scalar situations.

It is the product of the mass of the object times the speed or the velocity of the object. This product gives us a very specific measure of this object and allows us to calculate multiple analyses.

We all have a logical understanding of momentum. All you need to ask yourself is:

Would you rather be struck by a fly traveling at 15 km/h or bicyclist traveling at 15 km/h?

Your answer is telling you that momentum is the most likely the reason for your choice.

These first few examples are restricted to working with a constant or an average momentum.

Example 8.1.1

What is the momentum of a 35.0 kg skateboarder when she is traveling at 4.50 m/s South?

Solution

Data:

  • [latex]\vec p = \text{ Find}[/latex]
  • [latex]m = 35.0 \text{ kg}[/latex]
  • [latex]\vec v = 4.50 \text{ m/s South}[/latex]

Solution:

  • [latex]\vec{p} = m \vec{v}[/latex]
  • [latex]\vec{p} = (35.0 \text{ kg})(4.50 \text{ m/s South})[/latex]
  • [latex]\vec{p} = 157.5 \text{ Ns South } (\approx 158 \text{ Ns South})[/latex]

Example 8.1.2

What is the momentum of a 150 metric tonne jet coming in to land at 175 km/h? (1 metric tonne = 1000 kg)

Solution

Data:

  • [latex]p = \text{ Find}[/latex]
  • [latex]m = 150[/latex] metric tonnes (150000 kg)
  • [latex]v = 175[/latex] km/h (48.6 m/s)

Solution:

  • [latex]p = m v[/latex]
  • [latex]p = (150\;000 \text{ kg})(48.6 \text{ m/s})[/latex]
  • [latex]p =[/latex] 7290000 Ns or [latex]7.29 \times 10^6 \text{ Ns}[/latex]

Note: For the example above, it is more convenient to write the solution in scientific notation.

Example 8.1.3

If a 500 g baseball has a momentum of 15 Ns, what is this speed of this ball?

Solution

Data:

  • [latex]p = 15 \text{ Ns}[/latex]
  • [latex]m = 500 \text{ g}[/latex] or [latex]0.50 \text{ kg}[/latex]
  • [latex]v = \text{ Find}[/latex]

Solution:

  • [latex]p = m v[/latex]
  • [latex]15 \text{ Ns } = (0.50 \text{ kg}) v[/latex]
  • [latex]v = 15 \text{ Ns } \div 0.50 \text{ kg}[/latex]
  • [latex]v = 30 \text{ m/s}[/latex]

Since we are only considering constant or an average momentum in these next examples, we can combine momentum with a constant speed or velocity problem. You will see this in the next examples.

Example 8.1.4

A 6400 kg transport truck maintains an average momentum of 160 000 Ns for 0.45 h of driving. What distance does it travel in this 0.45 h?

Solution

When you set up the data for this problem you can see that we need to have a speed so that we can solve for the distance the truck travels.  Therefore, we must first solve for the speed of the truck using the momentum and mass of the truck.

To find the momentum and mass:

  • Data:
    • [latex]p = 160\;000 \text{ Ns}[/latex]
    • [latex]m = 6400 \text{ kg}[/latex]
    • [latex]v = \text{ Find first kg}[/latex]
  • Solution:
    • [latex]p = m v[/latex]
    • [latex]160\;000 \text{ Ns } = (6400 \text{ kg}) v[/latex]
    • [latex]v = 160\;000 \text{ Ns } \div 6400[/latex]
    • [latex]v = 25 \text{ m/s}[/latex] or [latex]90 \text{ km/h}[/latex]

To find the distance:

  • Data:
    • [latex]v = \text{ Find (90 km/h)}[/latex]
    • [latex]d = \text{ Find}[/latex]
    • [latex]t = 0.45 \text{ h}[/latex]
  • Solution:
    • [latex]v = \dfrac{d}{t}[/latex]
    • [latex]90 \text{ km/h } = \dfrac{d}{0.45 \text{ h}}[/latex]
    • [latex]d = (90 \text{ km/h})(0.45 \text{ h})[/latex]
    • [latex]d = 40.5 \text{ km } (\approx 41 \text{ km})[/latex]

Example 8.1.5

What is the average momentum of a 450 kg motorcyclist that travels 40 km East in 25 minutes?

Solution

In solving this problem you need to have the speed of the motorcyclist to find the momentum.  This then requires you to find the average speed of the motorcyclist first to be able to find the momentum.

To find the average speed:

  • Data:
    • [latex]\vec{v} = \text{ Find}[/latex]
    • [latex]\vec{d} = 40 \text{ km East}[/latex]
    • [latex]t = 25 \text{ min } (0.417 \text{ h})[/latex]
  • Solution:
    • [latex]\vec{v} = \dfrac{\vec{d}}{t}[/latex]
    • [latex]\vec{v} = \dfrac{40 \text{ km East}}{0.417 \text{ h}}[/latex]
    • [latex]\vec{v} = 96 \text{ km/h East}[/latex]
    • [latex]\vec{v} = 26.7 \text{ m/s East } (\approx 27 \text{ m/s East})[/latex]

To find the momentum:

  • Data:
    • [latex]\vec{p} = \text{ Find}[/latex]
    • [latex]m = 450 \text{ kg}[/latex]
    • [latex]v = 26.7 \text{ m/s East}[/latex]
  • Solution:
    • [latex]\vec{p} = m \vec{v}[/latex]
    • [latex]\vec{p} = (450 \text{ kg})(26.7 \text{ m/s East})[/latex]
    • [latex]\vec{p} = 12\;000 \text{ Ns East}[/latex]

Exercise 8.1

  1. What is the momentum of a 64.0 kg cyclist when he is traveling at 9.50 m/s North?
  2. What is the momentum of a 126 metric tonne barge drifting at 2.0 m/s downstream? (1 metric tonne = 1000 kg)
  3. If a 44.0 g shell has a momentum of 17.64 Ns, what is the speed of this shell?
  4. A 100.0 kg linebacker has a momentum of 80.0 Ns to the right.
    1. What is his velocity?
    2. What would be the velocity of a 0.500 kg football having the same momentum?
  5. A 1400 kg truck maintains an average momentum of 12000 Ns for 0.25 h of driving. What distance does it travel in this 0.25 h?
  6. If two football players have exactly the same average momentum of 950 Ns, what distance would separate them after the faster one has run 200 m? One player has a mass of 90 kg and the other a mass of 95 kg.
  7. What is the average momentum of a 1250 kg truck that travels 100 km in 50 minutes?

8.2 Change in Momentum

Equations Introduced or Used for this Section:

  • Impulse [latex](J) = F t = \Delta p[/latex]
  • Change in Momentum [latex](\Delta p) = m \Delta v[/latex]
  • Momentum [latex](p) = m v[/latex]
  • [latex]\Delta p = p_f - p_i[/latex]
  • [latex]\Delta v = v_f - v_i[/latex]

 

The NASA photo of the space shuttle Endeavour leaving Earth taken from a low orbit observation jet aircraft that was most likely traveling at 10.7 km or lower. Space travel is currently only possible from the ejection of high temperature gases, which can be looked at as impulse causing the spacecraft to move opposite to the direction of the ejected gas.

Change in Momentum,  like change in displacement or velocity, can be found by the difference between the final momentum and the initial momentum. The next few examples cover questions that explore this.

Example 8.2.1

What is the change in momentum of a 1500 kg truck that comes to a complete stop from an initial speed of 75 km/h?

Solution

Data:

  • [latex]\vec{p} = \text{ Find}[/latex]
  • [latex]m = 1500 \text{ kg}[/latex]
  • [latex]\vec{v} = 75[/latex] km/h to a full stop = [latex]0 \text{ m/s } - 20.8 \text{ m/s}[/latex]

Solution:

  • [latex]\Delta p = m \Delta v[/latex]
  • [latex]\Delta p = (1500 \text{ kg})(-20.8 \text{ m/s})[/latex]
  • [latex]\Delta p = -31\;200 \text{ Ns } (\approx -31\;000 \text{ Ns})[/latex]

Example 8.2.2

A loaded SUV (mass = 1750 kg) is traveling at a velocity of 90 km/h North.  What is its change in momentum if it comes to a full stop?

Solution

Data:

  • [latex]\Delta \vec{p} = \text{ Find}[/latex]
  • [latex]m = 1750 \text{ kg}[/latex]
  • [latex]\Delta \vec{v} = 90[/latex] km/h to a full stop = [latex]0 \text{ m/s } - 25 \text{ m/s North}[/latex]

Solution:

  • [latex]\Delta \vec{p} = m \Delta \vec{v}[/latex]
  • [latex]\Delta \vec{p} = (1750 \text{ kg})(- 25 \text{ m/s North})[/latex]
  • [latex]\Delta \vec{p} = -43\;750 \text{ Ns North} (\approx - 44\;000 \text{ Ns North})[/latex] or [latex]43\;750 \text{ Ns South } (\approx 44\;000 \text{ Ns South})[/latex]

Example 8.2.3

A 280 g softball is thrown towards the batter at 96 km/h and is hit directly back at the pitcher traveling at 125 km/h. What is this softball’s change in momentum? (From the batter)

Solution

Data:

  • [latex]\Delta \vec{p} = \text{ Find}[/latex]
  • [latex]m = 0.28 \text{ kg}[/latex]
  • [latex]\Delta \vec{v} = 125[/latex] km/h away, [latex]-96[/latex] km/h towards
  • [latex]\Delta \vec{v} = 221[/latex] km/h away

Solution:

  • [latex]\Delta \vec{p} = m \Delta \vec{v}[/latex]
  • [latex]\Delta \vec{p} = (0.28 \text{ kg})(61.4 \text{ m/s away})[/latex]
  • [latex]\Delta \vec{p} = 17.2 \text{ Ns away } (\approx 17 \text{ Ns away})[/latex]

Example 8.2.4

What is the change in momentum of a 90 kg hockey player that comes to a complete stop if he decelerates at 5.0 m/s2 over 15 m?

Solution

First:

  • Data:
    • [latex]a = -5.0 \text{ m/s}^2[/latex]
    • [latex]v_i = \text{ Find first}[/latex]
    • [latex]v_f = 0 \text{ m/s}[/latex]
    • [latex]d = 15 \text{ m}[/latex]
    • [latex]t = \text{ Not mentioned}[/latex]
  • Solution:
    • [latex]2 a d = v_r^2 - v_i^2[/latex]
    • [latex]2(-5.0 \text{ m/s}^2)(15 \text{ m}) = (0 \text{ m/s})^2 - v_i^2[/latex]
    • [latex]- v_i^2 = - 150 \text{ m}^2\text{/s}^2[/latex]
    • [latex]v_i = 12.2 \text{ m/s}[/latex]

Second:

  • Data:
    • [latex]\Delta p = \text{ Find second}[/latex]
    • [latex]m = 90 \text{ kg}[/latex]
    • [latex]\Delta v = \text{ Need to calculate}[/latex]
  • Solution:
    • [latex]\Delta p = m \Delta v[/latex]
    • [latex]\Delta p = (90 \text{ kg})(v_f - v_i)[/latex]
    • [latex]\Delta p = (90 \text{ kg})(0 \text{ m/s} - 12.2 \text{ m/s})[/latex]
    • [latex]\Delta p = - 1100 \text{ Ns}[/latex]

Example 8.2.5

A 1500 kg car comes to a complete stop in 8.0 s.  If it changed its momentum by – 42000 Ns, what distance did it travel?

Solution

Data:

Data to find Distance Data to find Change in an Object’s Speed
[latex]a = \text{ Not mentioned}[/latex] [latex]\Delta p = - 42\;000 \text{ Ns}[/latex]
[latex]v_i = \text{ Need to calculate}[/latex] [latex]m = 1500 \text{ kg}[/latex]
[latex]v_f = 0 \text{ m/s}[/latex] [latex]\Delta v = \text{ Find first}[/latex]
[latex]d = \text{ Find second}[/latex]
[latex]t = 8.0 \text{ s}[/latex]

Solution to find [latex]\Delta v[/latex]:

  • [latex]\Delta p = m \Delta v[/latex]
  • [latex]- 42\;000 \text{ Ns } = (1500 \text{ kg}) \Delta v[/latex]
  • [latex]\Delta v = - 42\;000 \text{ Ns } \div 1500 \text{ kg}[/latex]
  • [latex]\Delta v = - 28 \text{ m/s}[/latex]

Now, to find [latex]v_i[/latex]:

  • [latex]\Delta v = v_f - v_i[/latex]
  • [latex]- 28 \text{ m/s } = 0 \text{ m/s } - v_i[/latex]
  • [latex]v_i = 28 \text{ m/s}[/latex]

Finally, to find [latex]d[/latex]:

  • [latex]d = \dfrac{(v_i + v_f)}{2}t[/latex]
  • [latex]d = \dfrac{(28 \text{ m/s } + 0 \text{ m/s})}{2}(8.0 \text{ s})[/latex]
  • [latex]d = 112 \text{ m } (\approx 110 \text{ m})[/latex]

Exercise 8.2

  1.  What is the change in momentum of a 2500 kg truck that comes to a complete stop from an initial speed of 90 km/h?
  2. A train engine (mass = 18500 kg) is pulling a loaded car (mass = 4250 kg) at a velocity of 22.2 m/s North. What is its change in momentum if it comes to a full stop?
  3. A 200 g billiard ball is rolling towards the wall at 1.25 m/s, strikes the wall and rebounds backwards at 1.25 m/s. What is this billiard ball’s change in momentum? (careful)
  4. What is the change in momentum of a 120 kg rugby player that comes to a complete stop if he decelerates at 17 m/s2 during a 0.5 s tackle?
  5. A 20 000 kg small jet comes to a complete stop in 12.0 s. If the change in its momentum was [latex]- 2.4 \times 10^6[/latex] Ns, what distance did it travel?

8.3 Impulse and Change in Momentum

In the introduction to chapter 7, we derived both the Impulse and Change in Momentum equations, combining Newton’s second law and Acceleration, resulting in:

[latex]F = m a[/latex] & [latex]a = \dfrac{\Delta v}{t}[/latex] which yielded [latex]F t = m \Delta v[/latex]

It is the combined equation of  [latex]F t = m \Delta v[/latex]that allows one to measure the change in momentum that results from an impulse or the impulse needed to cause a change in momentum.  This is a different approach to finding the change in momentum by using the difference of the final and initial momentum, [latex]\Delta p = p_f - p_i[/latex] or by using the chance in speed/velocity [latex]\Delta p = m(v_f - v_i)[/latex].  This combined equation also uses the versatility of Newton’s second law  to expand tools that are useful in explaining other natural phenomenon.

Impulse and the resulting change in momentum manifest themselves in many different forms in our physical world. In contact sports, most athletes now wear protective headgear, such as helmets, and in many parts of the world the law requires that people wear such helmets when riding scooters, bicycles and motorcycles.  Headgear like helmets will extend the time of a force acting on a person’s head if they encounter a blow. Since the change in the momentum remains the same, if the time of the blow increases, the force of the blow is decreased. This lowers the chance of severe brain injuries.

Cushioning and reducing the impact of blows are not restricted to contact sports but include cushioned surfaces and mats for individuals involved in sports such as gymnastics, cheerleading and wrestling or other martial arts. Injuries increase in severity if surfaces where the individual impacts the surface are solid and have no cushioning surface that can absorb and reduce the forces acting.  Automobiles prior to 1959 did not have seat belts or air bags that are currently used as safety features to reduce driver and passenger injuries. As a result, passenger mortality began to dramatically decrease post 1959 due to increased focus on ways one could reduce the impact injuries on vehicle occupants.

Example 8.3.1

An unbalanced force of 1200 N acts on an object for 4.2 s. What impulse acted on this object?

Solution

Data:

  • [latex]J = \text{ Find}[/latex]
  • [latex]F = 1200 \text{ N}[/latex]
  • [latex]t = 4.2 \text{ s}[/latex]

Solution:

  • [latex]J = F t[/latex]
  • [latex]J = (1200 \text{ N})(4.2 \text{ s})[/latex]
  • [latex]J = 5040 \text{ Ns } (\approx 5000 \text{ Ns})[/latex]

While the above example was elementary in solution, the next few questions use the relationship between impulse and change in momentum to arrive at a solution. Using the relationship,

Impulse = Change on Momentum

or

[latex]F t = m \Delta v[/latex]

we now have two equations that can be used to find either the impulse of change in momentum. They can be combined to find the force, change in velocity (or speed), time or mass, all of which depends on what is being asked and if enough data is there to use.

Example 8.3.2

What unbalanced force would cause a 25 kg mass to change its speed by 8.2 m/s in 2.0 s?

Solution

Data:

  • [latex]F = \text{ Find}[/latex]
  • [latex]m = 25 \text{ kg}[/latex]
  • [latex]\Delta v = 8.2 \text{ m/s}[/latex]
  • [latex]t = 2.0 \text{ s}[/latex]

Solution:

  • [latex]F t = m \Delta v[/latex]
  • [latex]F (2.0 \text{ s}) = (25 \text{ kg})(8.2 \text{ m/s})[/latex]
  • [latex]F = 102.5 \text{ N } (\approx 100 \text{ N})[/latex]

Example 8.3.3

A 300 N unbalanced force from a storm is directed North and acts on an 5.0 kg ball for 0.750 s. What is the change in velocity of this ball?

Solution

Data:

  • [latex]\vec{F} = 300 \text{ N North}[/latex]
  • [latex]m = 5.0 \text{ kg}[/latex]
  • [latex]\Delta \vec{v} = \text{ Find}[/latex]
  • [latex]t = 0.75 \text{ s}[/latex]

Solution:

  • Use [latex]\vec{F} t = m \Delta \vec{v}[/latex]
  • [latex](300 \text{ N})(0.750 \text{ s}) = (5.0 \text{ kg}) \Delta \vec{v}[/latex]
  • [latex]\Delta \vec{v} = 45 \text{ m/s North}[/latex]

Example 8.3.4

A 5000 N braking force acts on a 1250 kg car for 4.0 s. If the car was initially traveling at 80 km/h south, what is the final velocity of this car, or was it stopped?

Solution

Data:

  • [latex]\vec{F} = - 5000 \text{ N}[/latex]
  • [latex]m = 1250 \text{ kg}[/latex]
  • [latex]\vec{v_f} = \text{ Find}[/latex]
  • [latex]\vec{v_i} = 80 \text{ km/h South}[/latex] (22.2 m/s South)
  • [latex]t = 4.0 \text{ s}[/latex]

Solution:

  • Use [latex]\vec{F} t = m \Delta \vec{v}[/latex]
  • [latex](- 5000 \text{ N})(4.0 \text{ s}) = (1250 \text{ kg}) \Delta \vec{v}[/latex]
  • [latex]\Delta \vec{v} = - 16 \text{ m/s}[/latex]

Working with this, [latex]- 16[/latex] m/s change in velocity requires one to subtract 16 m/s from the initial velocity of 22.2 m/s south. Since 22.2 m/s south is greater than the 16 m/s that is removed due to braking, the car is still in motion.

The final Velocity is:

  • [latex]\Delta \vec{v} = \vec{v_f} - \vec{v_i}[/latex]
  • [latex]\vec{v_f} = \vec{v_i} + \Delta \vec{v}[/latex]
  • [latex]\vec{v_f} = 22.2 \text{ m/s South } - 16 \text{ m/s South}[/latex]
  • [latex]\vec{v_f} = 6.2 \text{ m/s South}[/latex]

Had the change in velocity been greater than the initial 22.2 m/s south, then the car would have been brought to a full stop.

This final example uses the relationship between Impulse and Change in Momentum to add other possible methods to solving the four kinematic equations covered in Chapter 3. As before, the variables are defined exactly the same as in the previous chapters and the chart introduced in Chapter 3 remains of use in solving these types of problems.

The short synopsis of this is that you now have more tools at your disposal to solve these kinematic problems.

Consider the following example:

Example 8.3.5

A 1350 kg car traveling at 108 km/h slows to 36 km/h in 8.0 s. Find:

  1. The change in momentum,
  2. The impulse acting on the car, and
  3. The average braking force.

Solution

Data for Question (i):

  • [latex]\Delta p = \text{ Find}[/latex]
  • [latex]m = 1350 \text{ kg}[/latex]
  • [latex]\Delta v = 108 \text{ to } 36 \text{ km/h}[/latex]

Solution for Question (i):

  • [latex]\Delta p = m \Delta v[/latex]
  • [latex]\Delta p = (1350 \text{ kg})(10 \text{ m/s } - 30 \text{ m/s})[/latex]
  • [latex]\Delta p = -27\;000 \text{ Ns}[/latex]

 

Data for Question (ii):

  • [latex]J = \text{ Find}[/latex]
  • [latex]F = \text{ No data}[/latex]
  • [latex]t = 8.0 \text{ s}[/latex]

Solution for Question (ii):

  • Since [latex]J = \Delta p[/latex]
  • [latex]J = - 27\;000 \text{ Ns}[/latex]

 

Data for Question (iii):

  • [latex]\vec{F} = \text{ Find}[/latex]
  • [latex]m = 1350 \text{ kg}[/latex]
  • [latex]\Delta v = -20 \text{ m/s}[/latex]
  • [latex]t = 8.0 \text{ s}[/latex]

Solution for Question (iii):

  • [latex]F t = m \Delta v[/latex]
  • [latex]F (8.0 \text{ s}) = (1350 \text{ kg})(- 20 \text{ m/s})[/latex]
  • [latex]F = -27\;000 \text{ Ns } \div 8.0 \text{ s}[/latex]
  • [latex]F = -3375 \text{ N } (\approx - 3400 \text{ N})[/latex]

Exercise 8.3

  1. An unbalanced force of 450 N acts on an object for 4.0 s. What impulse acted on this object?
  2. What unbalanced force would cause a 13.5 kg object to change its speed by 6.50 m/s in 1.30 s?
  3. A 156 N unbalanced force directed East acts on an 18 kg rock for 0.150 s. What is the change in velocity of this rock?
  4. A 2250 N braking force acts on a 750 kg car for 2.5 s. If the car was initially traveling at 16.5 m/s South, what is the final velocity of this car?
  5. An unbalanced force accelerates a 25 kg mass from 12 m/s to 27 m/s in 5.0 s.
    1. What is the acceleration?
    2. What is the magnitude of this unbalanced force?
  6. An unbalanced force of 68 N acts on a 17 kg mass for 2.0 s. If this mass was originally traveling at 10 m/s:
    1. What final speed did it reach?
    2. How far did it travel during these 2.0 s?
  7. A 0.50 kg kite changes its velocity from 12 m/s down to 6.0 m/s up in a time of 1.5 s. What unbalanced force acts on this kite?
  8. What impulse is produced by a force of 21 N East acting for 0.30 s?
  9. What impulse is produced by a force of 340 N West acting for 5.0 µs?
  10. What is the amount of time that a force of 4.0 N East must last to produce an impulse of:
    1. 0.80 Ns East?
    2. [latex]2.0 \times 10^4[/latex] Ns East?
  11. A 4.44 x 103 kg railway car coasts along an East/West track at 8.0 m/s East.  What impulse is necessary to cause this railway car to travel:
    1. at 9.0 m/s East?
    2. at 2.0 m/s West?
  12. A marble ([latex]3.30 \times 10^{-3}[/latex] kg), moving at a uniform velocity of 12 m/s experiences a retarding force for [latex]2.0 \times 10^{-3}[/latex] s.  This force causes this marble to rebound at 2.0 m/s.  Calculate:
    1. the change in momentum
    2. the force acting on the object.
  13.  What unbalanced force will bring a car of weight 10000 N from rest to a speed of 108 km/h in a time of 15 s?

Exercise 8.4.1: Airbag Failure[1]

During a collision between two cars, both the driver’s and the passenger’s head received identical impulses.  The driver’s head strikes the airbag with an average force of 48 N with the force of this collision lasting for 0.60 s.  For the passenger, the airbag failed to inflate and the impact between this passenger’s head and the unpadded dashboard lasted for [latex]3.0 \times 10^{-3}[/latex] s.  What is the average reaction force the passenger receives from striking the unpadded dashboard?

Exercise 8.4.2: The Antarctic Iceberg B-15A

Icebergs have a density of around 900 kg/m3 and only about 10% can be seen above the surface (90% is below the surface). Calved from the Ross Ice Shelf in March 2000, B-15 broke up into several pieces in 2000, 2002 and 2003, the largest of which, B-15A, was the world’s largest free floating object at [latex]27 \times 122[/latex] km with an area of 3100 km2 (approximately the size of Luxembourg) with a height averaging 37 m above the water’s surface.

Question: If ocean currents generally move these icebergs at an average of 0.5 m/s, what would be its momentum?

Exercise Answers

8.1 Momentum

  1. 608 Ns North
  2. [latex]2.52 \times 10^5[/latex] Ns downstream
  3. 401 m/s towards target
    1. 0.800 m/s right
    2. 160 m/s right
  5. 7.7 km
  6. 10.5 m (≈ 11 m)
  7. 42000 Ns

8.2 Change in Momentum

  1. − 62500 Ns
  2. [latex]5.05 \times 10^5[/latex] Ns South
  3. 0.50 Ns opposite to original direction
  4. − 1020 N
  5. 720 m

8.3 Change in Momentum and Impulse

  1. 1800 Ns
  2. 67.5 N
  3. 1.3 m/s East
  4. 9.0 m/s South
    1. 3.0 m/s2
    2. 75 N
    1. 18 m/s
    2. 28 m
  7. 6 N upwards
  8. 6.3 Ns East
  9. [latex]1.7 \times 10^{-3}[/latex] Ns West
    1. 0.20 s
    2. 5000 s
    1. [latex]4.4 \times 10^3[/latex] East
    2. [latex]4.4 \times 10^4[/latex] Ns West
    1. 0.046 Ns opposite
    2. 23.1 N opposite
  13. 2040 N

8.4.1 Airbag Failure

  1. 9600 N away from the dash

8.4.2 Antarctic Iceberg B-15A

  1. [latex]p = 5.48 \times 10^{14}[/latex] Ns

Media Attributions

  1. Today's cars are regulated stringently for safety and, in theory, to reduce toxic emissions: they're fast, efficient, and comfortable. We can thank legally mandated regulations for this, of course. Go back to the 1970s and cars didn't have crumple zones. Go back to the 1950s and cars didn't come with seat belts as standard. In the 1930s, indicators, turn signals and brakes on all four wheels were optional, and your best hope of surviving a 50km/h crash was to be thrown clear of the car and land somewhere without breaking your neck.” Charles Stross (2017) ↵

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