Momentum
Section: Motion, Forces & Energy | Syllabus: Cambridge IGCSE Physics (0625)
Momentum
- momentum = mass × velocity
- momentum is a vector quantity - it acts in the same direction as the velocity
- a stationary object has zero momentum
- unit: kg m/s (equivalent to N s)
- Formula:
- p = mv
- Where:
- p = momentum (kg m/s)
- m = mass (kg)
- v = velocity (m/s)
Different combinations of mass and velocity can give the same momentum. The direction of the momentum arrow always matches the direction of the velocity.
Worked Example: Calculating Momentum
- Muumbe drives a car of mass 600 kg at a velocity of 15 m/s. calculate the momentum of the car.
- Step 1: write down the equation. p = mv
- Step 2: substitute the values. p = 600 × 15
- Step 3: calculate the answer. p = 9000 kg m/s
Impulse
- impulse = force × the time for which the force acts
- impulse = FΔ t = Δ (mv)
- impulse is equal to the change in momentum of the object
- unit: N s (equivalent to kg m/s)
- the bigger the force and the longer it acts for, the greater the change in momentum produced
- for a given change in momentum, a longer time taken means a smaller force is needed
Worked Example: Calculating Impulse
- Museli applies a force of 30 N to a car for 10 s. the car initially has a momentum of 9000 kg m/s. calculate the impulse and the final momentum of the car.
A force of 30 N applied for 10 s produces an impulse of 300 N s, increasing the car's momentum from 9000 kg m/s to 9300 kg m/s.
- Step 1: write down the equation for impulse. impulse = FΔ t
- Step 2: substitute the values. impulse = 30 × 10
- Step 3: calculate the impulse. impulse = 300 N s
- Step 4: add the impulse to the initial momentum to find the final momentum. p_final = 9000 + 300 = 9300 kg m/s
Safety Applications
- seatbelts, airbags and crumple zones all increase the time taken for a person to stop in a crash
- for the same change in momentum, a longer stopping time results in a smaller force acting on the passenger, which reduces injury
Conservation of Momentum
- whenever two objects interact, the total momentum before the interaction equals the total momentum after the interaction, provided no external forces act
- total momentum before = total momentum after
- if one object gains momentum, the other object loses an equal amount of momentum
- this only applies in one dimension at IGCSE - take care with direction by choosing a positive direction and using signs for velocities in the opposite direction
Total momentum before a collision equals total momentum after a collision.
Worked Example: Objects Sticking Together After a Collision
- Muchindu's trolley of mass 2 kg is moving at 3 m/s when it collides with, and sticks to, a stationary trolley of mass 3 kg. find their common velocity after the collision.
- Step 1: calculate the total momentum before the collision. p = (2 × 3) + (3 × 0) = 6 kg m/s
- Step 2: find the total mass after the collision. m = 2 + 3 = 5 kg
- Step 3: use conservation of momentum to find the final velocity. v = (p/m) = (6/5) = 1.2 m/s, in the original direction
When the 2 kg and 3 kg trolleys stick together, they move off with a common velocity of 1.2 m/s.
Worked Example: Finding an Unknown Velocity After a Collision
- Muumbe hits a stationary ball of mass 0.25 kg with a tennis racket of mass 3.0 kg. before the collision the racket moves at 20 m/s, and after the collision the racket moves at 18 m/s. find the velocity of the ball immediately after the collision.
The racket slows from 20 m/s to 18 m/s as it transfers momentum to the ball, which moves off at 24 m/s.
- Step 1: calculate the momentum of the racket before the collision. p = 3.0 × 20 = 60 kg m/s
- Step 2: calculate the momentum of the racket after the collision. p = 3.0 × 18 = 54 kg m/s
- Step 3: find the momentum gained by the ball, using conservation of momentum. Δ p = 60 - 54 = 6.0 kg m/s
- Step 4: calculate the velocity of the ball. v = (p/m) = (6.0/0.25) = 24 m/s
Resultant Force as Rate of Change of Momentum
- resultant force is defined as the change in momentum per unit time
- F = (Δ p/Δ t)
- F = resultant force (N)
- Δ p = change in momentum (kg m/s)
- Δ t = time taken for the change (s)
- this is an alternative form of F = ma
- a larger force produces a faster rate of change of momentum
- this explains why increasing the time taken to stop reduces the force experienced
Worked Example: Force in a Car Crash
- Museli is analysing a crash test dummy of mass 70 kg travelling at 15 m/s that comes to rest in a crash. in Test 1 (no airbag) it stops in 0.03 s. in Test 2 (with an airbag) it stops in 0.15 s. calculate the force acting on the dummy in each test.
The airbag increases the stopping time from 0.03 s to 0.15 s, reducing the force on the dummy from 35,000 N to 7,000 N.
- Step 1: calculate the change in momentum (the same in both tests). Δ p = 70 × 15 = 1050 kg m/s
- Step 2: calculate the force in Test 1. F = (Δ p/Δ t) = (1050/0.03) = 35,000 N
- Step 3: calculate the force in Test 2. F = (Δ p/Δ t) = (1050/0.15) = 7,000 N
- the airbag increases the stopping time by a factor of 5, which reduces the force by a factor of 5
Interactive revision notes, videos and practice questions load below.