CCEA GCE Mathematics (2210) · Impulse and momentum
Mini-Lesson
Impulse and momentum
1 mission. Guess, learn, then check. Your place is saved.
Mission 1A 1 kg trolley at 4 m s⁻¹ hits a still 3 kg one. They stick. How fast?
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Mission 1
A 1 kg trolley at 4 m s⁻¹ hits a still 3 kg one. They stick. How fast?
Make a guess: A 1 kg trolley at 4 m s⁻¹ hits a still 3 kg one. They stick. How fast?
Mechanics · impulse and momentum (A2 only)
Momentum and impulse
Momentum=mv=mv. It is a vector: in one dimension, choose a positive direction and give velocities signs. Units: N s (or kg m s⁻¹).
Impulse of a constant force FF acting for time tt is FtFt, in N s.
Impulse = change in momentum:
I=mv−muI=mv−mu
The impulse has the same direction as the change in velocity.
Worked exampleMechanics · impulse and momentum (A2 only)
A ball rebounding from a wall
Question
A ball of mass 0.2 kg hits a vertical wall at 15 m s⁻¹ and rebounds along the same line at 10 m s⁻¹. Find the impulse on the ball, and the average force if contact lasts 0.04 s.
1
Take towards the wall as positive: u=15u=15, v=−10v=−10
2
Use I=mv−muI=mv−mu: I=0.2(−10)−0.2(15)=−5I=0.2(−10)−0.2(15)=−5
3
Interpret the sign: 5 N s away from the wall
4
Average force: F=tI=0.045=125F=I/t=5/0.04=125
Answer
impulse 5 N s away from the wall; average force 125 N
⚠️ Watch out
0.2(15−10)=10.2(15−10)=1 N s ignores the reversal; the velocity changes by 25 m s⁻¹, not 5.
Calculate
?A ball of mass 0.4 kg moving at 6 m s⁻¹ is hit by a bat and moves back along the same line at 14 m s⁻¹. Find the magnitude of the impulse on the ball.
N s
Mechanics · impulse and momentum (A2 only)
Conservation of linear momentum
When two bodies collide, they exert equal and opposite impulses on each other (Newton's third law).
So, with no external impulse, total momentum before = total momentum after:
m1u1+m2u2=m1v1+m2v2m₁u₁+m₂u₂=m₁v₁+m₂v₂
Coalesce: the bodies stick and share one velocity, (m1+m2)v(m₁+m₂)v.
Explosion from rest: total momentum is zero before, so the parts move in opposite directions.
Worked exampleMechanics · impulse and momentum (A2 only)
A head-on collision
Question
Particle AA (3 kg) moving at 5 m s⁻¹ collides directly with particle BB (2 kg) moving at 4 m s⁻¹ towards AA. After the collision AA moves at 1 m s⁻¹ in its original direction. Find the velocity of BB after the collision.
1
Take AA's original direction as positive, so BB's velocity is −4−4
2
Momentum before: 3(5)+2(−4)=73(5)+2(−4)=7
3
Momentum after: 3(1)+2vB=73(1)+2v_B=7, so vB=2v_B=2
4
Check impulses: on AA, 3(1−5)=−123(1−5)=−12; on BB, 2(2−(−4))=122(2−(−4))=12
Answer
BB moves at 2 m s⁻¹ in AA's original direction (it reverses)
⚠️ Watch out
writing BB's initial velocity as +4+4 gives vB=10v_B=10, which is wrong.
Quick check
?A shell of mass 5 kg is at rest. It explodes into two parts of mass 2 kg and 3 kg. The 2 kg part moves off at 12 m s⁻¹. Find the speed of the 3 kg part.
Calculate
?Particle AA (4 kg) moving at 6 m s⁻¹ collides directly with particle BB (mm kg) moving at 3 m s⁻¹ towards AA. They coalesce and move at 1.5 m s⁻¹ in AA's original direction. Find mm.
kg
Calculate
?Hence find the magnitude of the impulse exerted on AA in this collision.
N s
Mission check
?A 1 kg trolley at 4 m s⁻¹ hits a stationary 3 kg trolley and they stick together. How fast do they move?
Total momentum before is 1×4=41 × 4=4 N s, and after the collision the combined 4 kg must carry the same momentum: 4v=44v=4, so v=1v=1 m s⁻¹. Kinetic energy falls from 8 J to 2 J; it is momentum, not energy, that is conserved in the collision.
Recap
Big ideas
Momentum =mv=mv (a vector); impulse =Ft=mv−mu=Ft=mv−mu, in N s.
With no external impulse, total momentum is conserved in collisions, coalescence and explosions.
Choose a positive direction first and give every velocity its sign.
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Mini-lesson complete!
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You've worked through Impulse and momentum for CCEA GCE Mathematics (2210). 🎉
Mission 1A 1 kg trolley at 4 m s⁻¹ hits a still 3 kg one. They stick. How fast?