In the exercise 2b when we calculate the force, why is it the mass of the ball in the equation? Shouldn't it be the mass of the golf club?

This is rather confusing... If they are imputing the acceleration and mass of the ball, you find its force. Right. But, if instead of using the mass of the ball, we take the mass of the golf club, the resulting force will be a lot biger (!) Which is not compatible with Newton's third law because the force exerted on the ball by the club will be larger than the force exerted on the club by the ball. Can someone comment on this?

Hello. I believe in 2b acceleration is negative because the final velocity of the club minus the initial velocity is 32-40=-8, which would make the force negative, thus the answer to 2b should be -4kN. Which make sense since the ball applies an impulse force opposite to the swing of the club (Newton's 3rd).

Also, the club is losing force since some of it transferred to the ball. So, in the question the force obtained should be negative.

So I understand the concept of conservation of momentum now, but I have a question: If an object such as a car runs into a very strong wall, the car would completely stop (without bouncing back) but the wall would not move either. Where does the momentum go?

According to the law of conservation of momentum the total momentum of the system remains constant if no external forces act on it. Therefore, the momentum of the car is transferred to the wall, and the wall gains momentum equal in magnitude but opposite in direction to that of the car

In question 1a in bold, it says Pc + Pb = 0. Shouldn't the change in momentum be equal to Pc - Pb = 0? I thought a change is a difference between two magnitudes instead of a sum?

look at the formula up it was pc=-pb so they brought -pb to the other sides and when any integers go to the others side it automatically changes signs so it would become pc+pb=0

Why did we not subtract 20m/s from 25m/s in exercise 3?

In the problem the ball is thrown from person A to person B at 25 m/s then the ball is thrown from person B to person A at 20 m/s. The direction from A to B is the opposite direction than B to A so if you consider the velocity of the ball from A to B to be positive then the velocity from B to A is negative giving you Vi = 25 m/s and Vf = -20 m/s. When you put them into the Vi - Vf part of the equation you have (25 m/s) - (-20 m/s) = 25 m/s + 20 m/s = 45 m/s.

what's smaller than miliseconds?

Microsecond, nanosecond, picosecond and etc.

Is conservation of momentum only applied for two objects that are in motion and are colliding? Is there a conservation of momentum for a ball hitting a wall? How is conservation of momentum differ from Newton's third law of motion?

Conservation of momentum is separate from Newton's third law and it is applicable in any isolated system. So a ball hitting a wall is subject to conservation of momentum. Even when you have a system that seems to violate conservation of momentum you can almost always increase the system to include enough to have momentum conserved.

In the facts about conservation of momentum in third point you said KE is likely not conserved, can you please explain that in detail and state some examples where KE cannot be conserved

Here is a video of an experiment where KE is not conserved but momentum is: https://www.youtube.com/watch?v=N8HrMZB6_dU The KE and momentum of the two bullets should equivalent to each other. The two blocks travel to the same height because of the conservation of liner momentum but the KE of the bullet gets split between heat, sound, block and bullet deformation as well as rotation in one of the blocks.. Since both blocks attain the same height so the KE from vertical motion but the block that was shot off center also has KE from rotation so it has retained more the the bullets KE as KE. In the block that is shot in the center there is more of the KE transferred to heat, sound, block and bullet deformation than in the one shot off center.

Main content

Course: High school physics > Unit 6

Lesson 2: Elastic collisions and conservation of momentum

What is conservation of momentum?

Google Classroom

Learn what conservation of momentum means and how to use it.

What is the principle of conservation of momentum?

In physics, the term conservation refers to something which doesn't change. This means that the variable in an equation which represents a conserved quantity is constant over time. It has the same value both before and after an event.

There are many conserved quantities in physics. They are often remarkably useful for making predictions in what would otherwise be very complicated situations. In mechanics, there are three fundamental quantities which are conserved. These are momentum, energy, and angular momentum. Conservation of momentum is mostly used for describing collisions between objects.

Just as with the other conservation principles, there is a catch: conservation of momentum applies only to an isolated system of objects. In this case an isolated system is one that is not acted on by force external to the system—i.e., there is no external impulse. What this means in the practical example of a collision between two objects is that we need to include both objects and anything else that applies a force to any of the objects for any length of time in the system.

If the subscripts

i

and

f

denote the initial and final momenta of objects in a system, then the principle of conservation of momentum says

p_{1 i} + p_{2 i} + \dots = p_{1 f} + p_{2 f} + \dots

Why is momentum conserved?

Conservation of momentum is actually a direct consequence of Newton's third law.

Consider a collision between two objects, object A and object B. When the two objects collide, there is a force on A due to B—

F_{AB}

—but because of Newton's third law, there is an equal force in the opposite direction, on B due to A—

F_{BA}

F_{AB} = - F_{BA}

The forces act between the objects when they are in contact. The length of time for which the objects are in contact—

t_{AB}

and

t_{BA}

—depends on the specifics of the situation. For example, it would be longer for two squishy balls than for two billiard balls. However, the time must be equal for both balls.

t_{AB} = t_{BA}

Consequently, the impulse experienced by objects A and B must be equal in magnitude and opposite in direction.

F_{AB} \cdot t_{AB} = - F_{BA} \cdot t_{BA}

If we recall that impulse is equivalent to change in momentum, it follows that the change in momenta of the objects is equal but in the opposite directions. This can be equivalently expressed as the sum of the change in momenta being zero.

\begin{aligned} m_{A} \cdot Δ v_{A} & = - m_{B} \cdot Δ v_{B} \\ m_{A} \cdot Δ v_{A} + m_{B} \cdot Δ v_{B} & = 0 \end{aligned}

What is interesting about conservation of momentum?

There are at least four things that are interesting—and sometimes counter-intuitive—about momentum conservation:

Momentum is a vector quantity, and therefore we need to use vector addition when summing together the momenta of the multiple bodies which make up a system. Consider a system of two similar objects moving away from each other in opposite directions with equal speed. What is interesting is that the oppositely-directed vectors cancel out, so the momentum of the system as a whole is zero, even though both objects are moving.
Collisions are particularly interesting to analyze using conservation of momentum. This is because collisions typically happen fast, so the time colliding objects spend interacting is short. A short interaction time means that the impulse, $F \cdot Δ t$ ‍, due to external forces such as friction during the collision is very small.
It is often easy to measure and keep track of momentum, even with complicated systems of many objects. Consider a collision between two ice hockey pucks. The collision is so forceful that it breaks one of the pucks into two pieces. Kinetic energy is likely not conserved in the collision, but momentum will be conserved.
Provided we know the masses and velocities of all the pieces just after the collision, we can still use conservation of momentum to understand the situation. This is interesting because by contrast, it would be virtually impossible to use conservation of energy in this situation. It would be very difficult to work out exactly how much work was done in breaking the puck.

Collisions with "immovable" objects are interesting. Of course, no object is really immovable, but some are so heavy that they appear to be. Consider the case of a bouncy ball of mass $m$ ‍ traveling at velocity $v$ ‍ towards a brick wall. It hits the wall and bounces back with velocity $- v$ ‍. The wall is well attached to the earth and doesn't move, yet the momentum of the ball has changed by $2 m v$ ‍ since velocity went from positive to negative.

If momentum is conserved, then the momentum of the earth and wall also must have changed by

2 m v

. We just don't notice this because the earth is so much heavier than the bouncy ball.

What kind of problems can we solve using momentum conservation?

Exercise 1a: The recoil of a cannon is probably familiar to anyone who has watched pirate movies. This is a classic problem in momentum conservation. Consider a wheeled, 500 kg cannon firing a 2 kg cannonball horizontally from a ship. The ball leaves the cannon traveling at 200 m/s. At what speed does the cannon recoil as a result?

Exercise 1b: Suppose the cannon was raised to fire at an angle

α = 30^{\circ}

to the horizontal. What is the recoil speed in this case? Where did the additional momentum go?

Because the cannon can roll only along the deck of the ship, we are only interested in the horizontal component of the momentum. Drawing a triangle and using trigonometry allows us to find the horizontal component of the velocity

v_{h}

\begin{aligned} v_{h} & = v_{b} \cdot \cos (α) \\ = 173 m / s \end{aligned}

We can then proceed as before:

\begin{aligned} \frac{m_{b} v_{h}}{m_{c}} & = - v_{c} \\ \frac{2 kg \cdot 173 m / s}{500 kg} & = - 0.69 m / s \end{aligned}

The vertical component of the momentum acts to momentarily push the boat down into the water.

Exercise 2a: A golf club head of mass

m_{c} = 0.25 kg

is swung and collides with a stationary golf ball of mass

m_{b} = 0.05 kg

. High speed video shows that the club is traveling at

v_{c} = 40 m / s

when it touches the ball. It remains in contact with the ball for

t = 0.5 ms

; after that, the ball is traveling at a speed of

v_{b} = 40 m / s

. How fast is the club traveling after it has hit the ball?

Exercise 2b: What is the average force on the club due to the golf ball in the previous problem?

We know that the club velocity changes by 8 m/s as it strikes the ball, so we can find the acceleration:

\begin{aligned} a & = \frac{Δ v}{Δ t} \\ = \frac{32 \frac{m}{s} - 40 \frac{m}{s}}{0.5 \cdot 10^{- 3} s} \\ = \frac{- 8.0 m / s}{0.5 \cdot 10^{- 3} s} \\ = - 16 \cdot 10^{3} m / s^{2} \end{aligned}

Thus, we know the the force is

\begin{aligned} F & = m_{c} a \\ = (0.25 kg) (- 16 \cdot 10^{3} m / s^{2}) \\ = - 4.0 kN \end{aligned}

It might be surprising that a small golf ball can exert such a large force. The force is also applied through a very small contact area, making the pressure very high. This is why golf balls and clubs need to be made so strong.

Exercise 3: Suppose a 100 kg football player is at rest on an ice rink. A friend throws a 0.4 kg football towards him at a speed of 25 m/s. In a smooth motion he receives the ball and throws it back in the same direction at a speed of 20 m/s. What is the speed of the player after the throw?

In this problem, all the motion of interest is close to parallel to the ground, so we can use the scalar form of the momentum and define positive velocity as the direction from the friend towards the player. Here we will use the subscripts

b

and

p

for the ball and player and

i

and

f

for initial and final. We'll begin by writing the sum of the momenta before and after the event:

\begin{aligned} m_{b} v_{bi} + m_{p} v_{pi} & = m_{b} v_{bf} + m_{p} v_{pf} \\ m_{b} (v_{bi} - v_{bf}) & = m_{p} v_{pf} \end{aligned}

When putting in the numbers, we need to make sure that the final velocity of the ball is in the negative direction.

\begin{aligned} v_{pf} & = \frac{m_{b}}{m_{p}} (v_{bi} - v_{bf}) \\ = \frac{0.4 kg}{100 kg} (25 m / s + 20 m / s) \\ = 0.18 m / s \end{aligned}

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Eka Terina
Posted 8 years ago. Direct link to Eka Terina's post “In the exercise 2b when w...”
In the exercise 2b when we calculate the force, why is it the mass of the ball in the equation? Shouldn't it be the mass of the golf club?
Button navigates to signup pageButton navigates to signup page
(22 votes)
Answer
- Juan Rojas
  Posted 8 years ago. Direct link to Juan Rojas's post “This is rather confusing....”
  This is rather confusing... If they are imputing the acceleration and mass of the ball, you find its force. Right. But, if instead of using the mass of the ball, we take the mass of the golf club, the resulting force will be a lot biger (!) Which is not compatible with Newton's third law because the force exerted on the ball by the club will be larger than the force exerted on the club by the ball. Can someone comment on this?
  Comment on Juan Rojas's post “This is rather confusing....”
  (3 votes)
Robert Ireland
Posted 8 years ago. Direct link to Robert Ireland's post “Hello. I believe in 2b ac...”
Hello. I believe in 2b acceleration is negative because the final velocity of the club minus the initial velocity is 32-40=-8, which would make the force negative, thus the answer to 2b should be -4kN. Which make sense since the ball applies an impulse force opposite to the swing of the club (Newton's 3rd).
Button navigates to signup pageButton navigates to signup page
(15 votes)
Answer
- Surya Bhushan Tripathi
  Posted 7 years ago. Direct link to Surya Bhushan Tripathi's post “Also, the club is losing ...”
  Also, the club is losing force since some of it transferred to the ball. So, in the question the force obtained should be negative.
  Button navigates to signup page
  (6 votes)
aca92bg00
Posted 8 years ago. Direct link to aca92bg00's post “in exercise 2a: can we do...”
in exercise 2a: can we do it like this:

mc*deltaVc= - mb*deltaVb
mc*(Vcf-Vci)= - mb*(Vbf-Vbi)
0.25kg*(Vcf - 40m/s) = - 0.05kg*(40m/s - 0m/s)
0.25kg*Vcf - 0.25kg*40m/s = - 0.05*40 kg*m/s
0.25Vcf = -0.05*40 + 0.25*40
0.25Vcf = -2 + 10
Vcf= 8/0.25
Vcf= 32m/s
Button navigates to signup pageButton navigates to signup page
(15 votes)
Answer
JPhilip
Posted 7 years ago. Direct link to JPhilip's post “So I understand the conce...”
So I understand the concept of conservation of momentum now, but I have a question: If an object such as a car runs into a very strong wall, the car would completely stop (without bouncing back) but the wall would not move either. Where does the momentum go?
Button navigates to signup pageComment on JPhilip's post “So I understand the conce...”
(12 votes)
Answer
- Matthew Oatmeal Brown
  Posted 5 months ago. Direct link to Matthew Oatmeal Brown's post “According to the law of c...”
  According to the law of conservation of momentum the total momentum of the system remains constant if no external forces act on it. Therefore, the momentum of the car is transferred to the wall, and the wall gains momentum equal in magnitude but opposite in direction to that of the car
  Button navigates to signup page
  (2 votes)
Milen Paul
Posted 4 years ago. Direct link to Milen Paul's post “Why did we not subtract 2...”
Why did we not subtract 20m/s from 25m/s in exercise 3?
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Charles LaCour
  Posted 4 years ago. Direct link to Charles LaCour's post “In the problem the ball i...”
  In the problem the ball is thrown from person A to person B at 25 m/s then the ball is thrown from person B to person A at 20 m/s.
  
  The direction from A to B is the opposite direction than B to A so if you consider the velocity of the ball from A to B to be positive then the velocity from B to A is negative giving you Vi = 25 m/s and Vf = -20 m/s.
  
  When you put them into the Vi - Vf part of the equation you have (25 m/s) - (-20 m/s) = 25 m/s + 20 m/s = 45 m/s.
  Comment on Charles LaCour's post “In the problem the ball i...”
  (11 votes)
Tierney Donahue
Posted 8 years ago. Direct link to Tierney Donahue's post “what types of situations ...”
what types of situations does conservation of momentum apply to?
Button navigates to signup pageComment on Tierney Donahue's post “what types of situations ...”
(6 votes)
Answer
Jons9
Posted 10 months ago. Direct link to Jons9's post “In question 1a in bold, i...”
In question 1a in bold, it says Pc + Pb = 0. Shouldn't the change in momentum be equal to Pc - Pb = 0? I thought a change is a difference between two magnitudes instead of a sum?
Button navigates to signup pageComment on Jons9's post “In question 1a in bold, i...”
(4 votes)
Answer
- Nalimun Alana
  Posted 10 months ago. Direct link to Nalimun Alana's post “look at the formula up it...”
  look at the formula up it was pc=-pb so they brought -pb to the other sides and when any integers go to the others side it automatically changes signs so it would become pc+pb=0
  Button navigates to signup page
  (2 votes)
Violet
Posted 8 months ago. Direct link to Violet's post “what's smaller than milis...”
what's smaller than miliseconds?
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- fRy
  Posted 8 months ago. Direct link to fRy's post “Microsecond, nanosecond, ...”
  Microsecond, nanosecond, picosecond and etc.
  Button navigates to signup page
  (5 votes)
Pranav Nayak
Posted 8 years ago. Direct link to Pranav Nayak's post “In the facts about conser...”
In the facts about conservation of momentum in third point you said KE is likely not conserved, can you please explain that in detail and state some examples where KE cannot be conserved
Button navigates to signup pageComment on Pranav Nayak's post “In the facts about conser...”
(0 votes)
Answer
- Charles LaCour
  Posted 8 years ago. Direct link to Charles LaCour's post “Here is a video of an exp...”
  Here is a video of an experiment where KE is not conserved but momentum is: https://www.youtube.com/watch?v=N8HrMZB6_dU
  
  The KE and momentum of the two bullets should equivalent to each other. The two blocks travel to the same height because of the conservation of liner momentum but the KE of the bullet gets split between heat, sound, block and bullet deformation as well as rotation in one of the blocks..
  
  Since both blocks attain the same height so the KE from vertical motion but the block that was shot off center also has KE from rotation so it has retained more the the bullets KE as KE. In the block that is shot in the center there is more of the KE transferred to heat, sound, block and bullet deformation than in the one shot off center.
  Comment on Charles LaCour's post “Here is a video of an exp...”
  (9 votes)
alona maria
Posted 2 years ago. Direct link to alona maria's post “Is conservation of moment...”
Is conservation of momentum only applied for two objects that are in motion and are colliding? Is there a conservation of momentum for a ball hitting a wall? How is conservation of momentum differ from Newton's third law of motion?
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Charles LaCour
  Posted 2 years ago. Direct link to Charles LaCour's post “Conservation of momentum ...”
  Conservation of momentum is separate from Newton's third law and it is applicable in any isolated system. So a ball hitting a wall is subject to conservation of momentum.
  
  Even when you have a system that seems to violate conservation of momentum you can almost always increase the system to include enough to have momentum conserved.
  Button navigates to signup page
  (2 votes)