Momentum

Definition

Momentum

The product of the mass and the velocity of an object.

Every moving object has momentum, which depends on both its mass and its velocity.
Momentum is calculated with $momentum=mass×velocity\text{momentum}=\text{mass}\times\text{velocity}$ , an equation you must recall.
In symbols the equation is $p = m v$ .
1. $p$ is the momentum in kilogram metres per second, $m/s\text{kg m/s}$ .
2. $m$ is the mass in kilograms, $kg\text{kg}$ .
3. $v$ is the velocity in metres per second, $m/s\text{m/s}$ .
A heavy lorry moving slowly can have the same momentum as a light car moving quickly.
A stationary object has zero momentum, because its velocity is zero.
Momentum is a vector, because velocity is a vector, so momentum always has a direction.
For motion along a line, choose one direction as positive, so an object moving the other way has a negative velocity and a negative momentum.

Example

Momentum of a rugby player

A rugby player with a mass of $kg95\ \text{kg}$ runs at $m/s8.0\ \text{m/s}$ towards the try line.

p = m v

p=95\ \text{kg}\times8.0\ \text{m/s}=760\ \text{kg m/s}

The player has a momentum of $m/s760\ \text{kg m/s}$ towards the try line.

Momentum in collisions

Definition

Conservation of momentum

The principle that the total momentum of colliding objects before a collision equals their total momentum after it, provided no external forces act.

When two objects collide, they push on each other, so the velocity of each object changes and the momentum of each object changes.
One object loses momentum and the other gains exactly the same amount, because the forces on the two objects are equal in size and opposite in direction and act for the same time.
The total momentum of the two objects therefore stays the same, as long as no external forces such as friction act on them.
This is the conservation of momentum: the total momentum before a collision equals the total momentum after the collision.
Momentum is conserved whether the objects bounce apart or stick together.
When you calculate total momentum, add the momenta with their signs, because momenta in opposite directions partly cancel.
Collisions in everyday life show the same pattern.
1. When a moving snooker ball hits a stationary ball head-on, the first ball can stop dead while the second ball moves off with almost the same velocity.
2. When a car runs into the back of a stationary car and the two lock together, they move off together more slowly than the first car was moving.
3. When two dodgem cars collide head-on, each one bounces back, but the total momentum of the pair is unchanged.

Example

Two trolleys that stick together

A $kg2.0\ \text{kg}$ trolley moving at $m/s3.0\ \text{m/s}$ collides with a stationary $kg1.0\ \text{kg}$ trolley, and the two trolleys stick together.

\text{total momentum before}=2.0\times3.0+1.0\times0=6.0\ \text{kg m/s}

The total momentum after the collision is also $m/s6.0\ \text{kg m/s}$ , carried by a combined mass of $kg3.0\ \text{kg}$ .

v=\dfrac{p}{m}=\dfrac{6.0\ \text{kg m/s}}{3.0\ \text{kg}}=2.0\ \text{m/s}

The joined trolleys move off at $m/s2.0\ \text{m/s}$ in the original direction.

Example

A head-on collision

Trolley A, of mass $kg1.5\ \text{kg}$ , moves right at $m/s2.0\ \text{m/s}$ , and trolley B, of mass $kg1.0\ \text{kg}$ , moves left at $m/s1.0\ \text{m/s}$ .
Take right as positive, so the velocity of trolley B is $m/s-1.0\ \text{m/s}$ .

\text{total momentum before}=1.5\times2.0+1.0\times(-1.0)=3.0-1.0=2.0\ \text{kg m/s}

After the collision trolley A moves right at $m/s0.40\ \text{m/s}$ , so the momentum of trolley B is found from the conserved total.

p_{\text{B}}=2.0-1.5\times0.40=1.4\ \text{kg m/s}

Trolley B moves off to the right with a momentum of $m/s1.4\ \text{kg m/s}$ , which is a velocity of $m/s1.4\ \text{m/s}$ .

Practical

Momentum of colliding trolleys

Aim: to measure the momentum of two trolleys before and after a collision and test whether the total momentum is conserved.
Apparatus: two dynamics trolleys, one fitted with a pin and the other with a cork so they stick together, a $m2\ \text{m}$ runway, wooden blocks, two interrupt cards of measured length, two light gates connected to a data logger, a top-pan balance, slotted masses and a box of foam at the end of the runway.
Method:
- Measure the mass of each trolley, including its interrupt card and any added masses, with the top-pan balance.
- Raise one end of the runway until a trolley given a gentle push rolls at a steady speed, so that friction is balanced.
- Set up light gate 1 near the start of the runway and light gate 2 further down, and place trolley B at rest just beyond light gate 1.
- Give trolley A a gentle push so that its card passes through light gate 1 before it hits trolley B.
- Let the joined trolleys pass through light gate 2, and catch them in the foam.
- Use the blocking times and the card lengths to find the velocity of trolley A before the collision and of the joined trolleys after it.
- Calculate the momentum before the collision as $mAvAm_{\text{A}}v_{\text{A}}$ and the momentum after as $(mA+mB)v(m_{\text{A}}+m_{\text{B}})v$ .
- Repeat three times, then add slotted masses to trolley B and repeat for several different masses.
Variables: the mass of trolley B is the independent variable, the momentum after the collision is the dependent variable, and the runway slope, the trolleys and the card lengths are control variables.
Results: the total momentum after each collision is very close to the momentum before it, and the joined trolleys move more slowly when trolley B is heavier.
Conclusion: the total momentum is conserved in the collision, within the uncertainty of the measurements.
Watch out: if the runway is not friction-compensated, friction removes momentum between the gates, so the momentum after the collision comes out noticeably smaller than before.
Safety: catch the trolleys in the foam at the end of the runway so that they do not fall off the bench.

Common Mistake

Do not confuse momentum, measured in $m/s\text{kg m/s}$ , with a moment, which is a turning effect measured in $m\text{N m}$ .
Do not forget the sign of each velocity, because adding two momenta in opposite directions as if they were in the same direction gives the wrong total.
Do not say that each object keeps its own momentum in a collision, because it is the total momentum that is conserved.

Exam technique

Momentum questions

Write $p = m v$ from memory and give the unit $m/s\text{kg m/s}$ with every answer.
For a collision, write a line for the total momentum before and a line for the total momentum after, then set them equal.
State the direction you have chosen as positive, and keep the signs in your working.
When you describe conservation of momentum, say that the total momentum before equals the total momentum after, provided no external forces act.

Self review

Write the equation for momentum and give its unit.
Why is momentum a vector quantity?
What is the momentum of a stationary object?
State the law of conservation of momentum.
Calculate the momentum of a $kg0.16\ \text{kg}$ cricket ball moving at $m/s30\ \text{m/s}$ .

Momentum

Definition

Momentum

The product of the mass and the velocity of an object.

Every moving object has momentum, which depends on both its mass and its velocity.
Momentum is calculated with $momentum=mass×velocity\text{momentum}=\text{mass}\times\text{velocity}$ , an equation you must recall.
In symbols the equation is $p = m v$ .
1. $p$ is the momentum in kilogram metres per second, $m/s\text{kg m/s}$ .
2. $m$ is the mass in kilograms, $kg\text{kg}$ .
3. $v$ is the velocity in metres per second, $m/s\text{m/s}$ .
A heavy lorry moving slowly can have the same momentum as a light car moving quickly.
A stationary object has zero momentum, because its velocity is zero.
Momentum is a vector, because velocity is a vector, so momentum always has a direction.
For motion along a line, choose one direction as positive, so an object moving the other way has a negative velocity and a negative momentum.

Example

Momentum of a rugby player

A rugby player with a mass of $kg95\ \text{kg}$ runs at $m/s8.0\ \text{m/s}$ towards the try line.

p = m v

p=95\ \text{kg}\times8.0\ \text{m/s}=760\ \text{kg m/s}

The player has a momentum of $m/s760\ \text{kg m/s}$ towards the try line.

Momentum in collisions

Definition

Conservation of momentum

The principle that the total momentum of colliding objects before a collision equals their total momentum after it, provided no external forces act.

When two objects collide, they push on each other, so the velocity of each object changes and the momentum of each object changes.
One object loses momentum and the other gains exactly the same amount, because the forces on the two objects are equal in size and opposite in direction and act for the same time.
The total momentum of the two objects therefore stays the same, as long as no external forces such as friction act on them.
This is the conservation of momentum: the total momentum before a collision equals the total momentum after the collision.
Momentum is conserved whether the objects bounce apart or stick together.
When you calculate total momentum, add the momenta with their signs, because momenta in opposite directions partly cancel.
Collisions in everyday life show the same pattern.
1. When a moving snooker ball hits a stationary ball head-on, the first ball can stop dead while the second ball moves off with almost the same velocity.
2. When a car runs into the back of a stationary car and the two lock together, they move off together more slowly than the first car was moving.
3. When two dodgem cars collide head-on, each one bounces back, but the total momentum of the pair is unchanged.

Example

Two trolleys that stick together

A $kg2.0\ \text{kg}$ trolley moving at $m/s3.0\ \text{m/s}$ collides with a stationary $kg1.0\ \text{kg}$ trolley, and the two trolleys stick together.

\text{total momentum before}=2.0\times3.0+1.0\times0=6.0\ \text{kg m/s}

The total momentum after the collision is also $m/s6.0\ \text{kg m/s}$ , carried by a combined mass of $kg3.0\ \text{kg}$ .

v=\dfrac{p}{m}=\dfrac{6.0\ \text{kg m/s}}{3.0\ \text{kg}}=2.0\ \text{m/s}

The joined trolleys move off at $m/s2.0\ \text{m/s}$ in the original direction.

Example

A head-on collision

Trolley A, of mass $kg1.5\ \text{kg}$ , moves right at $m/s2.0\ \text{m/s}$ , and trolley B, of mass $kg1.0\ \text{kg}$ , moves left at $m/s1.0\ \text{m/s}$ .
Take right as positive, so the velocity of trolley B is $m/s-1.0\ \text{m/s}$ .

\text{total momentum before}=1.5\times2.0+1.0\times(-1.0)=3.0-1.0=2.0\ \text{kg m/s}

After the collision trolley A moves right at $m/s0.40\ \text{m/s}$ , so the momentum of trolley B is found from the conserved total.

p_{\text{B}}=2.0-1.5\times0.40=1.4\ \text{kg m/s}

Trolley B moves off to the right with a momentum of $m/s1.4\ \text{kg m/s}$ , which is a velocity of $m/s1.4\ \text{m/s}$ .

Practical

Momentum of colliding trolleys

Aim: to measure the momentum of two trolleys before and after a collision and test whether the total momentum is conserved.
Apparatus: two dynamics trolleys, one fitted with a pin and the other with a cork so they stick together, a $m2\ \text{m}$ runway, wooden blocks, two interrupt cards of measured length, two light gates connected to a data logger, a top-pan balance, slotted masses and a box of foam at the end of the runway.
Method:
- Measure the mass of each trolley, including its interrupt card and any added masses, with the top-pan balance.
- Raise one end of the runway until a trolley given a gentle push rolls at a steady speed, so that friction is balanced.
- Set up light gate 1 near the start of the runway and light gate 2 further down, and place trolley B at rest just beyond light gate 1.
- Give trolley A a gentle push so that its card passes through light gate 1 before it hits trolley B.
- Let the joined trolleys pass through light gate 2, and catch them in the foam.
- Use the blocking times and the card lengths to find the velocity of trolley A before the collision and of the joined trolleys after it.
- Calculate the momentum before the collision as $mAvAm_{\text{A}}v_{\text{A}}$ and the momentum after as $(mA+mB)v(m_{\text{A}}+m_{\text{B}})v$ .
- Repeat three times, then add slotted masses to trolley B and repeat for several different masses.
Variables: the mass of trolley B is the independent variable, the momentum after the collision is the dependent variable, and the runway slope, the trolleys and the card lengths are control variables.
Results: the total momentum after each collision is very close to the momentum before it, and the joined trolleys move more slowly when trolley B is heavier.
Conclusion: the total momentum is conserved in the collision, within the uncertainty of the measurements.
Watch out: if the runway is not friction-compensated, friction removes momentum between the gates, so the momentum after the collision comes out noticeably smaller than before.
Safety: catch the trolleys in the foam at the end of the runway so that they do not fall off the bench.

Common Mistake

Do not confuse momentum, measured in $m/s\text{kg m/s}$ , with a moment, which is a turning effect measured in $m\text{N m}$ .
Do not forget the sign of each velocity, because adding two momenta in opposite directions as if they were in the same direction gives the wrong total.
Do not say that each object keeps its own momentum in a collision, because it is the total momentum that is conserved.

Exam technique

Momentum questions

Write $p = m v$ from memory and give the unit $m/s\text{kg m/s}$ with every answer.
For a collision, write a line for the total momentum before and a line for the total momentum after, then set them equal.
State the direction you have chosen as positive, and keep the signs in your working.
When you describe conservation of momentum, say that the total momentum before equals the total momentum after, provided no external forces act.

Self review

Write the equation for momentum and give its unit.
Why is momentum a vector quantity?
What is the momentum of a stationary object?
State the law of conservation of momentum.
Calculate the momentum of a $kg0.16\ \text{kg}$ cricket ball moving at $m/s30\ \text{m/s}$ .

2.2.11 Momentum in collisions

Momentum

Momentum of a rugby player

Momentum in collisions

Two trolleys that stick together

A head-on collision

Momentum of colliding trolleys

Momentum questions

Review 2.2.11 Momentum in collisions by teaching Genie

2.2.11 Momentum in collisions

Momentum

Momentum of a rugby player

Momentum in collisions

Two trolleys that stick together

A head-on collision

Momentum of colliding trolleys

Momentum questions

Review 2.2.11 Momentum in collisions by teaching Genie

2.2.11 Momentum in collisions

Momentum

Momentum of a rugby player

Momentum in collisions

Two trolleys that stick together

A head-on collision

Momentum of colliding trolleys

Momentum questions

Review 2.2.11 Momentum in collisions by teaching Genie

2.2.11 Momentum in collisions Revision Guide

2.2.11 Momentum in collisions

Momentum

Momentum of a rugby player

Momentum in collisions

Two trolleys that stick together

A head-on collision

Momentum of colliding trolleys

Momentum questions

Review 2.2.11 Momentum in collisions by teaching Genie

2.2.11 Momentum in collisions Revision Guide