Conservation of Momentum
- The principle of conservation of momentum states that:
The total momentum before an interaction is equal to the total momentum after an interaction, if no external forces are acting on the objects
- In this context, an interaction can be either:
- A collision i.e. where two objects collide with each other
- An explosion i.e. where a stationary object explodes into two (or more) parts
Collisions
- For a collision between two objects:
The total momentum before a collision = the total momentum after a collision
- Before the collision:
- The momentum is only generated by mass m because it is the only moving object
- If the right is taken as the positive direction, the total momentum of the system is m × u
- After the collision:
- Mass M also now has momentum
- The velocity of m is now -v (since it is now travelling to the left) and the velocity of M is V
- The total momentum is now the momentum of M + the momentum of m
- This is (M × V) + (m × -v) or (M × V) – (m × v) written more simply
Momentum Before and After a Collision
The momentum of a system before and after a collision where the momentum before equals the momentum after
Newton's Third Law in Collisions
- According to Newton's Third Law:
- When two objects collide, both objects will react, generally causing one object to speed up (gain momentum) and the other object to slow down (lose momentum)
Equal and Opposite Forces on Two Colliding Objects
Newton's third law can be applied to collisions. Here the force of trolley A on B is equal and opposite to the force of trolley B on A.
- Consider the collision between two trolleys, A and B:
- Trolley A exerts a force on trolley B
- Trolley B exert an equal force on trolley A in the opposite direction
- This can be written as:
Higher Tier
Explosions
- Examples of explosions include:
- A person jumping off a skateboard
- A cannon or bullet being fired
- A firework exploding
- A person being thrown from a vehicle
- The separation of space vehicles
- As with collisions, the principle of conservation of momentum equations can be applied:
- The mass of the separate objects = the mass of the original object
- The initial velocity (and therefore, momentum) of the object is 0
- Remember that velocity is a vector, so it considers the direction of the speed of each component
- Some will travel in a positive direction and others in a negative direction
- When they are all added together the resultant velocity = 0
Momentum of a Dynamite Explosion
The sum of the masses of the exploded pieces equals the mass of the original dynamite and the sum of the velocities of the separate pieces equals zero.
Worked example
Higher Tier
The diagram shows a car and a van, just before and after the car collides with the van, which is initially at rest.
The car initially moves at a speed of 10 m/s, but this reduces to 2 m/s after the collision.
The mass of the car is 990 kg and the mass of the van is 4200 kg.
Calculate the velocity of the van when it is pushed forward by the collision.
Answer:
Step 1: State the principle of the conservation of momentum
Total momentum before a collision = total momentum after a collision
Step 2: Calculate the total momentum of the car and van before the collision
Momentum: p = mv
- Initial momentum of the car:
pcar = 990 × 10 = 9900 kgm/s
- Initial momentum of the van:
pvan = 0 (the van is at rest, so p = v = 0)
- Total momentum before collision:
pbefore = pcar + pvan
pbefore = 9900 + 0 = 9900 kg m/s
Step 3: Calculate the total momentum of the car and van after the collision
- Final momentum of the car:
pcar = 990 × 2 = 1980 kg m/s
- Final momentum of the van:
pvan = 4200 × v
- Total momentum after collision:
pafter = 1980 + 4200v
Step 4: Rearrange the conservation of momentum equation for the velocity of the van
pbefore = pafter
9900 = 1980 + 4200v
9900 − 1980 = 4200v
= 1.9 m/s
Worked example
A gun of mass 3 kg fires a bullet of 30 g at a speed of 175 m/s.
As the bullet is fired, the gun moves back with a recoil speed v.
Calculate the recoil speed of the gun v.
Answer:
Step 1: List the known quantities:
- Velocity of gun/bullet before firing = 0 m/s
- Mass of gun, mgun = 3 kg
- Mass of bullet, mbullet = 30 g = 0.03 kg
- Velocity of bullet after firing , vbullet = 175 m/s
- Velocity of gun after firing = v
Step 2: State the conservation of momentum equation
momentum before firing = momentum after firing
momentum before firing = velocity of gun/bullet before firing = 0
0 = (mgun × v) + (mbullet × vbullet)
Step 3: Substitute the known quantities into the equation
0 = (3 × v) + (0.03 × 175)
Step 4: Rearrange the equation to find the recoil speed of the gun
−3v = 5.25
v = −1.75 m/s
- Therefore, the recoil speed of the gun is 1.75 m/s
Examiner Tip
Remember any question with rearranging to obtain velocity is a higher-tier one.
If it is not given in the question already, drawing a diagram of before and after helps keep track of all the masses and velocities (and directions) in the conservation of momentum questions.
Remember that velocity is a speed with a direction. The question asks for the speed, so you do not need to include the final speed in your answer.