Impulse-momentum theorem — derivation: Exercises

The one law we lean on the whole way down: $J = \int_{t_{1}}^{t_{2}} F d t = Δ p = m v_{f} - m v_{i}, \overset{ˉ}{F} = \frac{Δ p}{Δ t}$

Everything below is one of these three faces of the same idea. If you need the derivation itself, return to the parent note.

Level 1 — Recognition

Goal: spot which formula applies and plug in cleanly.

L1·Q1 — Units check

Impulse is measured in $N\cdots$ . Show this equals $kg\cdotm/s$ , the unit of momentum, and say in one sentence why they must match.

Recall Solution

A newton is $kg\cdotm/s^{2}$ (mass × acceleration). Multiply by seconds: $N\cdots = \frac{kg\cdotm}{s ^{2}} \cdot s = \frac{kg\cdotm}{s} = kg\cdotm/s .$ They must match because the theorem says $J = Δ p$ — a quantity can only equal another if their units agree.

L1·Q2 — Constant force, direct plug-in

A constant force of $8 N$ pushes a cart for $3 s$ . What impulse does it deliver?

Recall Solution

Force is constant, so $J = F Δ t$ is legal: $J = 8 \times 3 = 24 N\cdots .$

L1·Q3 — Momentum change from velocities

A $4 kg$ trolley speeds up from $2 m/s$ to $5 m/s$ in a straight line. Find its change in momentum.

Recall Solution

$Δ p = m v_{f} - m v_{i} = 4 (5) - 4 (2) = 20 - 8 = 12 kg\cdotm/s .$ By the theorem this is also the impulse that acted on it: $J = 12 N\cdots$ .

Level 2 — Application

Goal: rearrange the theorem to solve for whatever is missing.

L2·Q1 — Catching a ball (average force)

A $0.15 kg$ ball arrives at $20 m/s$ and is brought to rest in $0.05 s$ . Find the average force your glove exerts.

Recall Solution

Positive = ball's incoming direction. $v_{i} = + 20$ , $v_{f} = 0$ . $Δ p = 0.15 (0) - 0.15 (20) = - 3 kg\cdotm/s .$ $\overset{ˉ}{F} = \frac{Δ p}{Δ t} = \frac{- 3}{0.05} = - 60 N .$ The minus sign says the force points backward, opposing the ball — exactly what a glove does.

L2·Q2 — Softening the blow

Same ball, same speed, but now you pull your hands back so it stops in $0.20 s$ instead. Find the new average force and compare.

Recall Solution

$Δ p$ is unchanged ( $- 3 kg\cdotm/s$ ) because the ball must still go from $20$ to $0$ . $\overset{ˉ}{F} = \frac{- 3}{0.20} = - 15 N .$ Four times the time → one quarter the force. This is exactly why airbags, crumple zones and bent knees work.

L2·Q3 — Finding the stopping time

A $1200 kg$ car moving at $30 m/s$ is stopped by an average braking force of $9000 N$ . How long does braking take?

Recall Solution

$Δ p = 1200 (0) - 1200 (30) = - 36000 kg\cdotm/s$ . The braking force opposes motion, so $\overset{ˉ}{F} = - 9000 N$ . $Δ t = \frac{Δ p}{F ˉ} = \frac{- 36000}{- 9000} = 4 s .$

Level 3 — Analysis

Goal: handle bounces, graphs, and time-varying forces where naive plugging fails.

L3·Q1 — Bounce vs stop

A $0.2 kg$ ball hits a wall at $5 m/s$ and rebounds at $5 m/s$ . (a) Find the impulse from the wall. (b) Compare it to the impulse if the ball had simply stuck to the wall.

Recall Solution

Positive = into the wall. $v_{i} = + 5$ , $v_{f} = - 5$ (reversed). (a) Bounce: $J = m (v_{f} - v_{i}) = 0.2 (- 5 - 5) = 0.2 (- 10) = - 2 N\cdots .$ (b) Stick: $v_{f} = 0$ , so $J = 0.2 (0 - 5) = - 1 N\cdots .$ Bouncing needs twice the impulse of stopping, because the wall must first cancel the incoming motion and then throw the ball back out.

L3·Q2 — Area under a triangular pulse

During a kick, the force on a $0.45 kg$ football rises linearly from $0$ to $900 N$ over $0.008 s$ , then falls linearly back to $0$ over the next $0.008 s$ . The ball starts at rest. Find its launch speed.

Look at the figure: the impulse is the area of the triangle, not force × total time.

Figure — Impulse-momentum theorem — derivation

Recall Solution

The pulse is a triangle: base $= 0.008 + 0.008 = 0.016 s$ , peak height $= 900 N$ . $J = area = \frac{1}{2} (base) (height) = \frac{1}{2} (0.016) (900) = 7.2 N\cdots .$ Then $J = Δ p = m v_{f}$ : $v_{f} = \frac{7.2}{0.45} = 16 m/s .$ Note: multiplying the peak force by the total time would give $900 \times 0.016 = 14.4 N\cdots$ — exactly double the truth, because a triangle is half its bounding rectangle.

L3·Q3 — Time-varying force, integrate

A force $F (t) = 12 t N$ acts on a $2 kg$ block (initially at rest) from $t = 0$ to $t = 3 s$ . Find the final speed.

The figure shows why we integrate: the "curve" here is a straight ramp, and impulse is the shaded area beneath it.

Recall Solution

Force changes with time, so $F Δ t$ is illegal — integrate: $J = \int_{0}^{3} 12 t d t = [6 t^{2}]_{0}^{3} = 6 (9) - 0 = 54 N\cdots .$ Then $J = Δ p = m v_{f}$ : $v_{f} = \frac{54}{2} = 27 m/s .$ Geometrically the ramp reaches $F (3) = 36 N$ ; the shaded triangle has area $\frac{1}{2} (3) (36) = 54$ — the integral and the area agree.

Level 4 — Synthesis

Goal: combine impulse with momentum conservation, kinematics, or two-stage motion.

L4·Q1 — Two-stage impulse

A $3 kg$ puck moving right at $4 m/s$ receives a rightward impulse of $9 N\cdots$ , and later a leftward impulse of $15 N\cdots$ . Find its final velocity (state direction).

Recall Solution

Positive = right. Initial momentum $p_{i} = 3 (4) = 12 kg\cdotm/s$ . Total impulse $= + 9 - 15 = - 6 N\cdots$ . $p_{f} = p_{i} + J_{total} = 12 + (- 6) = 6 kg\cdotm/s .$ $v_{f} = \frac{6}{3} = 2 m/s (to the right) .$ Impulses add as vectors; the net is what shifts momentum.

L4·Q2 — Recoil (impulse pair)

A $60 kg$ skater standing still throws a $2 kg$ ball forward at $9 m/s$ over $0.30 s$ . (a) Find the average force the skater exerts on the ball. (b) Find the skater's recoil speed, and connect it to Newton's third law.

Recall Solution

(a) Ball: $Δ p = 2 (9) - 0 = 18 kg\cdotm/s$ . $\overset{ˉ}{F} = \frac{18}{0.30} = 60 N (forward on the ball) .$ (b) By Newton's third law the ball pushes back on the skater with $- 60 N$ for the same $0.30 s$ , so the skater's impulse is $- 18 N\cdots$ : $v_{skater} = \frac{- 18}{60} = - 0.30 m/s,$ i.e. $0.30 m/s$ backward. The equal-and-opposite impulses cancel, so total momentum stays zero — this is Conservation of Momentum seen through impulse.

L4·Q3 — Impulse from a graph, then a collision

A $0.5 kg$ cart at rest is pushed by a force whose graph is a rectangle of $10 N$ for $0.4 s$ , immediately followed by a triangle rising to $10 N$ and back to $0$ over $0.6 s$ . Find the cart's final speed.

Recall Solution

Split the area into the two shapes. Rectangle: $10 \times 0.4 = 4 N\cdots$ . Triangle: $\frac{1}{2} (0.6) (10) = 3 N\cdots$ . Total $J = 4 + 3 = 7 N\cdots$ . $v_{f} = \frac{J}{m} = \frac{7}{0.5} = 14 m/s .$

Level 5 — Mastery

Goal: full multi-step problems where you must choose the method yourself.

L5·Q1 — Water jet on a wall

A hose sprays water horizontally at a wall at $20 m/s$ , delivering $3 kg$ of water per second. The water does not bounce (it runs down the wall, ending with zero horizontal velocity). Find the average horizontal force on the wall.

Recall Solution

In one second, mass $Δ m = 3 kg$ arrives at $+ 20 m/s$ and ends at $0$ . Momentum destroyed per second: $\frac{Δ p}{Δ t} = \frac{Δ m ( v _{f} - v _{i} )}{Δ t} = \frac{3 ( 0 - 20 )}{1} = - 60 N .$ That is the force the wall puts on the water; by Newton's third law the water pushes the wall with $+ 60 N$ . So the average force on the wall is $60 N$ .

L5·Q2 — Jet that splashes straight back

Same hose, same $20 m/s$ and $3 kg/s$ , but now the water rebounds straight back at $20 m/s$ . Find the force on the wall and explain the factor of two.

Recall Solution

Now $v_{i} = + 20$ , $v_{f} = - 20$ . $\frac{Δ p}{Δ t} = \frac{3 ( - 20 - 20 )}{1} = \frac{3 ( - 40 )}{1} = - 120 N,$ so the water pushes the wall with $120 N$ — double the sticking case. Reversing the flow needs the wall to first stop the water and then fling it back, exactly the bounce-vs-stop doubling from L3·Q1, now happening continuously.

L5·Q3 — Impulse, then projectile

A $0.25 kg$ ball resting on the ground is struck so that a force graph forms a triangle peaking at $500 N$ over a total contact time of $0.010 s$ , launching it vertically. Ignoring gravity during the tiny contact, find (a) the launch speed and (b) the maximum height reached ( $g = 10 m/s^{2}$ ).

Recall Solution

(a) Impulse = triangle area: $J = \frac{1}{2} (0.010) (500) = 2.5 N\cdots .$ $v = \frac{J}{m} = \frac{2.5}{0.25} = 10 m/s upward .$ (b) Now use kinematics (energy or $v^{2} = u^{2} - 2 g h$ with $v = 0$ at top): $h = \frac{u ^{2}}{2 g} = \frac{1 0 ^{2}}{2 ( 10 )} = 5 m .$ Two tools, each for its own job: impulse converts the messy contact into a clean launch speed; kinematics then handles the smooth flight.

L5·Q4 — Design an airbag

A $70 kg$ crash-test dummy travelling at $15 m/s$ must be stopped. Safety limits the average force to $7000 N$ . (a) What minimum stopping time is required? (b) Without the airbag the dummy stops in $0.05 s$ against the steering wheel — what force is that, and by what factor does the airbag help?

Recall Solution

$Δ p = 70 (0) - 70 (15) = - 1050 kg\cdotm/s$ ; magnitude $1050 N\cdots$ . (a) $Δ t_{m i n} = \frac{∣Δ p ∣}{F _{m a x}} = \frac{1050}{7000} = 0.15 s .$ (b) Hard stop: $\overset{ˉ}{F} = \frac{1050}{0.05} = 21000 N .$ Ratio $= \frac{21000}{7000} = 3$ . The airbag stretches the stopping time threefold, cutting the peak force to one third — same $Δ p$ , gentler landing.

Connections

Newton's Second Law — the source law every solution above quietly uses.
Force–Time Graphs — L3 and L4 read impulse straight off the area.
Conservation of Momentum — the recoil (L4·Q2) and jet (L5) problems are conservation seen through impulse.
Collisions — Elastic and Inelastic — bounce-vs-stick (L3·Q1) is the seed of collision analysis.
Work–Energy Theorem — the L5·Q3 flight phase switches to the distance analogue.
Center of Mass Motion — external impulse shifts the whole system's momentum.

Recall Self-test checklist

Did every L1–L2 answer keep the correct sign? ::: Yes — positive direction chosen first, then $v_{f} - v_{i}$ with signs. When did you integrate instead of using $F Δ t$ ? ::: Whenever the force varied with time (L3·Q2, L3·Q3) — impulse is the area under the curve. What doubles impulse in a bounce? ::: The velocity reverses, so $Δ v = v_{f} - v_{i}$ becomes $- 2 v$ instead of $- v$ . How does a stream become a force? ::: $F = (Δ m /Δ t) Δ v$ — momentum change per second.