Physics 9702 · Paper 2

Impulse & Force-Time Graphs

Rearrange the second law: $F\Delta t = \Delta p$ . A force acting for a time changes momentum by exactly that amount.
The product $F\Delta t$ has its own name: .

Impulse

\text{impulse} = F\Delta t = \Delta p

Symbols

$F$ = average resultant force (N)
$\Delta t$ = time the force acts (s)
$\Delta p$ = change in momentum (kg m s⁻¹ (= N s))

The unit N s equals kg m s⁻¹, because $1\ \text{N} = 1\ \text{kg m s}^{-2}$ .
On a force–time graph, the impulse is the area under the graph, even when the force is not constant.

Worked example

Smallest case: steady push

A steady resultant force of 5.0 N pushes an object for 2.0 s. Find the change in momentum.

$\Delta p = F\Delta t = 5.0 \times 2.0 = 10 \text{ kg m s}^{-1}$ .

Answer

\Delta p = 10 \text{ kg m s}^{-1}

The bounce: the speeds add

When an object bounces, its velocity reverses direction. With up as positive, it arrives at $-u$ and leaves at $+v$ .
So $\Delta p = m(v - (-u)) = m(v + u)$ : the two speeds add inside the bracket.

Worked example

Bouncing ball: change in momentum, then force

A ball of mass 0.25 kg hits the floor moving down at 5.2 m s⁻¹ and leaves moving up at 3.6 m s⁻¹. Contact lasts 0.18 s. Find the change in momentum and the average resultant force. (9702/22/M/J/25 Q4(b))

Up as positive: $\Delta p = 0.25\,(3.6 - (-5.2)) = 0.25 \times 8.8 = 2.2 \text{ kg m s}^{-1}$ upwards.
$F = \dfrac{\Delta p}{\Delta t} = \dfrac{2.2}{0.18} = 12 \text{ N}$ upwards.

Answer

\Delta p = 2.2 \text{ kg m s}^{-1}, \; F = 12 \text{ N}

Common mistake

Writing

\Delta p = m(3.6 - 5.2)

for a bounce. The two velocities point opposite ways, so they need opposite signs. Subtracting the speeds is the most common wrong line in this question family.

Your turn— tap to reveal the worked answer (9702/12/O/N/24 Q9)

A ball of mass 2.0 kg hits a wall at 4.0 m s⁻¹ and rebounds at 2.8 m s⁻¹. Contact lasts 0.150 s. Find the average force on the ball. (9702/12/O/N/24 Q9)

$\Delta p = 2.0\,(4.0 + 2.8) = 13.6 \text{ kg m s}^{-1}$ .
$F = \dfrac{13.6}{0.150} = 91 \text{ N}$ .

Answer: 91 N.

Same momentum change, smaller force

A collision fixes $\Delta p$ : the object arrives with some momentum and must lose it.
But $F = \Delta p / \Delta t$ : making the collision last longer makes the force smaller for the same momentum change.
That is why cars have crumple zones, why you bend your knees on landing, and why helmets have soft padding.

Impulse = area under the force-time graph. Both areas are 3.0 N s, so both give the same momentum change, but the longer collision needs a much smaller peak force.

A car crumpling against a barrier on impact. — The crumple zone makes the collision take longer, so the same momentum change needs a smaller force on the passengers.

How to write it for marks: the change in momentum is the same; the time to stop is longer; since force = rate of change of momentum, the force is smaller. Vague answers with no momentum reasoning score zero (9702/22/F/M/25 Q3(b)(iii)).
The reverse also appears: a golf club or bat acts for a few milliseconds, so even a small $\Delta p$ needs a huge force. Ball X from lesson 3.8 stopped in 2.0 ms: $F = 3.84/0.0020 = 1900 \text{ N}$ (9702/23/O/N/24 Q2(c)).
Chain variant: an impulse gives $\Delta p$ , which then feeds a conservation calculation, for example a spring pushing one part of a spacecraft away from the rest (9702/22/O/N/25 Q2(c)).

Impulse = area under the force–time graph = change in momentum. Longer contact time, smaller force. That single line answers every crumple-zone, seatbelt, helmet and soft-landing question.

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Impulse & Force-Time Graphs

Rearrange the second law: $F\Delta t = \Delta p$ . A force acting for a time changes momentum by exactly that amount.
The product $F\Delta t$ has its own name: .

Impulse

\text{impulse} = F\Delta t = \Delta p

Symbols

$F$ = average resultant force (N)
$\Delta t$ = time the force acts (s)
$\Delta p$ = change in momentum (kg m s⁻¹ (= N s))

The unit N s equals kg m s⁻¹, because $1\ \text{N} = 1\ \text{kg m s}^{-2}$ .
On a force–time graph, the impulse is the area under the graph, even when the force is not constant.

Worked example

Smallest case: steady push

A steady resultant force of 5.0 N pushes an object for 2.0 s. Find the change in momentum.

$\Delta p = F\Delta t = 5.0 \times 2.0 = 10 \text{ kg m s}^{-1}$ .

Answer

\Delta p = 10 \text{ kg m s}^{-1}

The bounce: the speeds add

When an object bounces, its velocity reverses direction. With up as positive, it arrives at $-u$ and leaves at $+v$ .
So $\Delta p = m(v - (-u)) = m(v + u)$ : the two speeds add inside the bracket.

Worked example

Bouncing ball: change in momentum, then force

Up as positive: $\Delta p = 0.25\,(3.6 - (-5.2)) = 0.25 \times 8.8 = 2.2 \text{ kg m s}^{-1}$ upwards.
$F = \dfrac{\Delta p}{\Delta t} = \dfrac{2.2}{0.18} = 12 \text{ N}$ upwards.

Answer

\Delta p = 2.2 \text{ kg m s}^{-1}, \; F = 12 \text{ N}

Common mistake

Writing

\Delta p = m(3.6 - 5.2)

for a bounce. The two velocities point opposite ways, so they need opposite signs. Subtracting the speeds is the most common wrong line in this question family.

Your turn— tap to reveal the worked answer (9702/12/O/N/24 Q9)

A ball of mass 2.0 kg hits a wall at 4.0 m s⁻¹ and rebounds at 2.8 m s⁻¹. Contact lasts 0.150 s. Find the average force on the ball. (9702/12/O/N/24 Q9)

$\Delta p = 2.0\,(4.0 + 2.8) = 13.6 \text{ kg m s}^{-1}$ .
$F = \dfrac{13.6}{0.150} = 91 \text{ N}$ .

Answer: 91 N.

Same momentum change, smaller force

A collision fixes $\Delta p$ : the object arrives with some momentum and must lose it.
But $F = \Delta p / \Delta t$ : making the collision last longer makes the force smaller for the same momentum change.
That is why cars have crumple zones, why you bend your knees on landing, and why helmets have soft padding.

Impulse = area under the force-time graph. Both areas are 3.0 N s, so both give the same momentum change, but the longer collision needs a much smaller peak force.

How to write it for marks: the change in momentum is the same; the time to stop is longer; since force = rate of change of momentum, the force is smaller. Vague answers with no momentum reasoning score zero (9702/22/F/M/25 Q3(b)(iii)).
The reverse also appears: a golf club or bat acts for a few milliseconds, so even a small $\Delta p$ needs a huge force. Ball X from lesson 3.8 stopped in 2.0 ms: $F = 3.84/0.0020 = 1900 \text{ N}$ (9702/23/O/N/24 Q2(c)).
Chain variant: an impulse gives $\Delta p$ , which then feeds a conservation calculation, for example a spring pushing one part of a spacecraft away from the rest (9702/22/O/N/25 Q2(c)).

Impulse = area under the force–time graph = change in momentum. Longer contact time, smaller force. That single line answers every crumple-zone, seatbelt, helmet and soft-landing question.