A particle of mass $4\ \text{kg}$ moves along a straight line (one-dimensional motion). Its position $x$ (in metres) varies with time $t$ (in seconds) as follows: for $t 4\ \text{s}$ it stays constant at $x = 3\ \text{m}$. Find (a) the force on the particle for $t 4\ \text{s}$, and for $0 < t < 4\ \text{s}$; and (b) the impulse at $t = 0$ and at $t = 4\ \text{s}$. ![Figure 4.16](/figures/laws-of-motion/position-time-graph-particle-uniform-velocity-then-rest-fig-4-16.svg)

A particle of mass 4kg moves along a straight line (one-dimensional motion). Its position x (in metres) varies

Exercises · 4.14

4\ \text{kg}

(a) the force on the particle for

t < 0

, for

t > 4\ \text{s}

, and for

0 < t < 4\ \text{s}

; and

(b) the impulse at

t = 0

and at

t = 4\ \text{s}

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Concept understanding — Impulse Momentum Theorem

The Intuition: Why Do We Need a "New" Idea?

Imagine you're catching a cricket ball. If you let your hands stay rigid, the ball stings and might bounce off. But if you give with the ball — pulling your hands back as you catch — the catch feels soft and the ball stops gently.

Same ball, same speed, same change in momentum. But the force you feel is completely different. Why?

The answer is time. When you pull your hands back, you increase the time over which the ball slows down. A longer time means a smaller force — even though the total "oomph" needed to stop the ball is the same. That "oomph" is called impulse.

Note

Impulse is not a mysterious new quantity. It's just force multiplied by the time it acts. If you push gently for a long time, or push hard for a short time, you can produce the same effect.

The Precise Statement

The Impulse-Momentum Theorem says:

The impulse delivered to an object equals the change in its momentum.

In symbols:

\vec{J} = \Delta \vec{p}

Where:

$\vec{J}$ is the impulse (a vector)
$\Delta \vec{p}$ is the change in momentum (also a vector)

And since impulse is force times time:

\vec{F}_{\text{avg}} \, \Delta t = m \vec{v}_f - m \vec{v}_i

\vec{J} = \vec{F}_{\text{avg}} \, \Delta t = \Delta \vec{p}

Breaking It Down Piece by Piece

Momentum ( $\vec{p}$ ) is mass times velocity: $\vec{p} = m\vec{v}$ . It's a measure of how hard it is to stop a moving object. A truck moving slowly has large momentum; a bullet moving fast has large momentum too.

Impulse ( $\vec{J}$ ) is the product of the average force and the time interval over which it acts: $\vec{J} = \vec{F}_{\text{avg}} \, \Delta t$ .

The theorem connects them: the net impulse changes the momentum. If you apply a net force to an object for some time, its momentum changes by exactly that amount.

Watch out

A common mistake is to think impulse is just force. It's force × time. A huge force acting for a tiny time (like a bat hitting a ball) can produce the same impulse as a tiny force acting for a long time (like a gentle push).

Why This Matters: Real-World Examples

Catching a ball (soft vs. hard hands)

Hard hands: $\Delta t$ is small → $\vec{F}_{\text{avg}}$ is large (it hurts)
Soft hands: $\Delta t$ is large → $\vec{F}_{\text{avg}}$ is small (it's comfortable)
In both cases, $\Delta \vec{p}$ is the same (ball goes from moving to stopped)

Airbags in cars

Without airbag: your head hits the dashboard in ~0.01 s → huge force
With airbag: your head decelerates over ~0.1 s → force is 10 times smaller
Same change in momentum, but the airbag extends the time

A cricket bat hitting a ball

The bat is in contact with the ball for a few milliseconds
The force during that contact is enormous (hundreds of Newtons)
The impulse changes the ball's momentum from one direction to another

The Mathematical Derivation (Short)

Start from Newton's second law:

\vec{F}_{\text{net}} = m \vec{a} = m \frac{d\vec{v}}{dt}

Multiply both sides by $dt$ :

\vec{F}_{\text{net}} \, dt = m \, d\vec{v}

Integrate over the time interval:

\int_{t_i}^{t_f} \vec{F}_{\text{net}} \, dt = m \int_{\vec{v}_i}^{\vec{v}_f} d\vec{v} = m\vec{v}_f - m\vec{v}_i

The left side is the impulse (the area under the force-time graph). The right side is the change in momentum. …

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On a position-time graph the velocity is the slope. The particle is at rest for $t<0$ , moves with the constant velocity $0.75\ \text{m s}^{-1}$ for $0<t<4\ \text{s}$ , and is again at rest for $t>4\ \text{s}$ . Constant velocity means zero acceleration, so the net force is zero in every interval. The velocity jumps at $t=0$ and $t=4\ \text{s}$ , and each jump corresponds to an impulse of magnitude $3\ \text{N s}$ .

Concept

Newton's second law gives the net force as $F = ma = m\dfrac{dv}{dt}$ . The velocity itself is the slope of the position-time graph, $v = \dfrac{dx}{dt}$ . If $x$ increases linearly with $t$ , the slope (velocity) is constant, so $a = 0$ and $F = 0$ . Impulse is the change in momentum, $J = \Delta p = m\,\Delta v$ , and it is what tells us about the abrupt velocity changes at the corners of the graph.

(a) Force in each interval

Read the slope of the graph in each region:

For $t < 0$ : $x = 0$ (constant), so $v = 0$ . The particle is at rest, $a = 0$ .

F = ma = 4 \times 0 = 0\ \text{N}

For $0 < t < 4\ \text{s}$ : $x$ rises uniformly from $0$ to $3\ \text{m}$ , so

v = \frac{\Delta x}{\Delta t} = \frac{3-0}{4-0} = 0.75\ \text{m s}^{-1}\ \text{(constant)}

A constant velocity means $a = 0$ , hence

F = ma = 4 \times 0 = 0\ \text{N}

For $t > 4\ \text{s}$ : $x = 3\ \text{m}$ (constant), so $v = 0$ and $a = 0$ .

F = ma = 4 \times 0 = 0\ \text{N}

So the net force is zero in all three intervals. (The force is undefined exactly at the sharp corners $t=0$ and $t=4\ \text{s}$ , where the slope changes suddenly — that is where the impulses act.)

(b) Impulse at $t = 0$ and $t = 4\ \text{s}$

Use $J = m\,\Delta v = m(v_{\text{after}} - v_{\text{before}})$ .

At $t = 0$ : velocity changes from $0$ (just before) to $0.75\ \text{m s}^{-1}$ (just after). …

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Concept

(a) Force in each interval

(b) Impulse at t=0t = 0t=0 and t=4 st = 4\ \text{s}t=4 s

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(b) Impulse at $t = 0$ and $t = 4\ \text{s}$

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