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This lesson covers momentum and impulse at the Further Mathematics level. These are foundational concepts for the Further Mechanics 1 module, requiring rigorous treatment of vector quantities, variable forces, and sign conventions.
The linear momentum of a particle of mass m moving with velocity v is:
p = mv
| Property | Detail |
|---|---|
| Type | Vector (same direction as v) |
| SI unit | kg m s^(-1) (equivalently, N s) |
| Depends on | Mass and velocity |
Because momentum is a vector, its sign (in 1-D) or its component form (in 2-D) matters. Always define a positive direction before solving a problem.
The impulse of a force F acting over a time interval is:
J = F * delta t (for a constant force)
For a variable force F(t):
J = integral from t1 to t2 of F(t) dt
The SI unit of impulse is N s (newton-seconds), which is the same as kg m s^(-1).
The key relationship connecting impulse and momentum is:
Impulse = Change in momentum
J = mv - mu
where u is the initial velocity and v is the final velocity.
This follows from Newton's Second Law: F = ma = m(dv/dt), so F dt = m dv, and integrating gives the impulse-momentum theorem.
A particle of mass 3 kg is moving at 8 m s^(-1). Find its momentum.
Solution:
p = mv = 3 * 8 = 24 kg m s^(-1)
A force F = 4t N acts on a 2 kg particle initially at rest, from t = 0 to t = 3 s. Find the impulse and the final velocity.
Solution:
Impulse = integral from 0 to 3 of 4t dt = [2t^2] from 0 to 3 = 18 N s
By the impulse-momentum theorem: 18 = 2v - 0, so v = 9 m s^(-1)
A 0.5 kg ball travelling at 6 m s^(-1) to the right is brought to rest. Find the impulse on the ball.
Solution:
Taking right as positive:
J = mv - mu = 0.5(0) - 0.5(6) = -3 N s
The magnitude of the impulse is 3 N s, directed to the left.
A 0.1 kg ball moving at 20 m s^(-1) to the right is hit back at 15 m s^(-1) to the left. Find the impulse.
Solution:
Take right as positive: u = 20, v = -15.
J = 0.1(-15) - 0.1(20) = -1.5 - 2 = -3.5 N s
The impulse is 3.5 N s to the left.
On a force-time graph, the impulse is the area under the curve.
For a constant force F acting for time T: impulse = FT (a rectangle).
For a linearly varying force: impulse = area of a triangle or trapezium.
For a general F(t): impulse = integral of F(t) dt.
A constant force of 50 N acts for 0.02 s on a ball during a collision. Find the impulse.
Solution:
J = F * t = 50 * 0.02 = 1 N s
The average force during a collision is:
F_avg = Impulse / Duration = (mv - mu) / delta t
A 0.15 kg cricket ball is bowled at 30 m s^(-1) and hit back at 40 m s^(-1). The bat is in contact with the ball for 0.01 s. Find the average force.
Solution:
Take the direction of the hit as positive: u = -30 (bowled towards the batsman), v = 40.
J = 0.15(40) - 0.15(-30) = 6 + 4.5 = 10.5 N s
F_avg = 10.5 / 0.01 = 1050 N
In two dimensions:
p = m(v_x i + v_y j)
Each component is treated separately.
A particle of mass 4 kg has velocity (2i + 3j) m s^(-1). Find its momentum.
p = 4(2i + 3j) = (8i + 12j) kg m s^(-1)
The magnitude is sqrt(64 + 144) = sqrt(208) = 4 sqrt(13) approximately 14.4 kg m s^(-1).
A 2 kg particle with velocity (5i - 3j) m s^(-1) receives an impulse of (-4i + 6j) N s. Find the final velocity.
Solution:
J = m(v - u), so mv = mu + J.
mv = 2(5i - 3j) + (-4i + 6j) = (10i - 6j) + (-4i + 6j) = (6i + 0j)
v = (6i)/2 = (3i) m s^(-1)
The particle is now moving at 3 m s^(-1) in the i-direction.
| Concept | Formula |
|---|---|
| Momentum | p = mv |
| Impulse (constant force) | J = F delta t |
| Impulse (variable force) | J = integral of F dt |
| Impulse-momentum theorem | J = mv - mu |
| Average force | F_avg = J / delta t |
AQA Exam Tip: Always define a positive direction clearly at the start. In 2-D problems, work with components separately. Show the impulse-momentum equation explicitly before substituting numbers.