Momentum and impulse
Why can a high jumper fall backwards over the bar onto a thick mat without getting hurt, even though a similar fall onto hard ground would be dangerous?
Prior knowledge
- The net force determines the acceleration, $\Sigma\overline{F} = m\overline{a}$chapter Newton's second law
- Acceleration is the change in velocity divided by the time, $a = \tfrac{\Delta v}{\Delta t}$chapter Acceleration and uniformly accelerated motion
Core path: follow the steps in order. Tasks marked Optional give extra practice. Optional tasks: 4.
Think about it
A high jumper takes off and arches backwards over a bar more than two metres high. Behind the bar the jumper lands on a thick mat and gets straight up. Without the mat, the same fall onto grass could break bones. The jumper hits both surfaces at the same velocity and stops on both.
What is it about the mat that protects the jumper, when the jumper's velocity changes by the same amount in both cases?
Optional Review: Newton's second law
Predict
Make your predictions before investigating 2 questions
Prediction 1: hard or soft bumper
Prediction 2: stopping or rebounding
Investigate
A lab cart hits a bumper on a wall. The top shows the situation with arrows, the bottom the force of the bumper over time. A force to the left is negative, and the area between the curve and the time axis is the impulse.
Simulation
How to use
Choose the initial velocity $v_0$, the mass of the cart, the hardness of the bumper and the type of collision. Press Start. When the cart has stopped or rebounded, the readout shows the duration of the collision $\Delta t$, the greatest force $F_\text{max}$, the impulse $I$, the final velocity $v$ and the momenta. The graph keeps the three previous collisions in grey so that you can compare the curves. Reset returns the cart to its starting place. Speed slows down or speeds up the motion without changing the readings; the normal speed is already slowed down so that you have time to see the collision.
Assumptions of the model
The values are model values calculated by the simulation. The track is horizontal, and friction and air resistance are neglected. The weight of the cart and the normal force of the track cancel each other, so in the horizontal direction only the force of the bumper acts on the cart, and it is the net force on the cart. The positive direction is to the right, towards the wall. The force of the elastic bumper is proportional to how far the bumper is compressed, and the hard bumper is $16$ times as stiff as the soft one. The deformable bumper stays squashed, so the cart stops against it, and its kinetic energy becomes the bumper's permanent deformation and internal energy. The force of the deformable bumper grows smoothly to its largest value half-way through the collision and falls smoothly to zero just as the cart stops. The pulse has the shape of a sine curve, so its average force is about $0.64\,F_\text{max}$, not half of the largest force. Its largest force and the duration of the collision are the same as for an elastic bumper of the same stiffness while the cart presses into it. The elastic bumper springs back and sends the cart back without any loss of energy.
How the calculation works
The motion of the cart is calculated from Newton's second law in small time steps, $0.1\,\text{ms}$ of simulated time at a time. The force of the elastic bumper is $F = -kc$, where $c$ is the compression and $k$ is the spring constant of the bumper; the minus sign says that the force points to the left. The force of the deformable bumper is $F = -F_\text{max}\sin(\pi t/\Delta t)$ during the collision time $\Delta t$. During it the velocity and position of the cart are calculated directly from this force. The time $t$ of the graph is counted from the start of contact, and the graph shows the force with its sign, so the curve lies below the time axis. The impulse $I$ is calculated as the area between the curve and the time axis, and it is negative. The momenta are calculated from the velocities, $p_0 = mv_0$ and $p = mv$, and $\Delta p = p - p_0$. So the impulse and the change in momentum are calculated independently of each other.
Experiment 1: hardness of the bumper
Test the first prediction. The starting state chooses a $1.0\,\text{kg}$ cart, an initial velocity of $2.0\,\text{m/s}$ and a deformable bumper that the cart stops against. Change only the hardness of the bumper: hard, medium and soft. Press Start each time and record the measurement when the cart has stopped.
Initial state: mass $m$: $1.0\,\text{kg}$, initial velocity $v_0$ = $2\,\text{m/s}$, collision: cart stops, bumper: hard. Kept constant: mass $m$, initial velocity $v_0$, collision. Varied: bumper.
| bumper | $\Delta t$ $(\text{s})$ | $F_\text{max}$ $(\text{N})$ | $I$ (N·s) |
|---|
Experiment 2: stopping and rebounding
Test the second prediction. The starting state chooses a $1.0\,\text{kg}$ cart, an initial velocity of $2.0\,\text{m/s}$ and a medium-hard bumper. Change only the type of collision: first the cart stops, then it rebounds. Record a measurement of each when the cart has stopped or returned to its starting place. Compare the columns $I$ and $\Delta p$.
Initial state: mass $m$: $1.0\,\text{kg}$, initial velocity $v_0$ = $2\,\text{m/s}$, bumper: medium, collision: cart stops. Kept constant: mass $m$, initial velocity $v_0$, bumper. Varied: collision.
| collision | $v$ $(\text{m/s})$ | $\Delta p$ (kg·m/s) | $I$ (N·s) |
|---|
Optional Experiment 3: mass of the cart
Next, investigate the effect of mass. The starting state chooses an initial velocity of $2.0\,\text{m/s}$, a medium-hard bumper and a cart that stops. Change only the mass of the cart. How does the impulse depend on the mass when the velocity stays the same?
Initial state: initial velocity $v_0$ = $2\,\text{m/s}$, bumper: medium, collision: cart stops, mass $m$: $0.50\,\text{kg}$. Kept constant: initial velocity $v_0$, bumper, collision. Varied: mass $m$.
| mass $m$ | $p_0$ (kg·m/s) | $I$ (N·s) | $F_\text{max}$ $(\text{N})$ |
|---|
Conclusion
Using your measurements, explain what determines how large an impulse the bumper gives the cart, and what determines the greatest force of the bumper. Use the numbers from your tables.
Model answer
In Experiment 1 the cart stopped every time, so its momentum changed from $2.0\,\text{kg}\cdot\text{m/s}$ to zero. In the model's values the impulse was $-2.00\,\text{N}\cdot\text{s}$ with every bumper, equal to the change in momentum. The hardness of the bumper decided only how the impulse was distributed: the hard bumper stopped the cart in $0.0393\,\text{s}$ with a greatest force of $80.0\,\text{N}$, the soft one in $0.157\,\text{s}$ with a greatest force of $20.0\,\text{N}$. A time four times as long gave a quarter of the force. In Experiment 2 the rebounding cart received an impulse of $-4.00\,\text{N}\cdot\text{s}$, because its momentum changed from $+2.0$ to $-2.0\,\text{kg}\cdot\text{m/s}$. So the impulse is decided by the change in momentum, and the greatest force is decided in addition by how long the collision lasts.
Assess your own answer
Tick the points your own answer already contains.
Review the theory summary and practise with the exercises.
Theory summary
Key idea
Momentum: mass and velocity together
In the chapter Newton's second law you learned that a force changes the velocity of an object, and that changing the velocity of a heavy object takes a greater force than for a light one. So both the mass and the velocity affect how hard it is to stop or speed up a moving object. Their product is called momentum. A light, fast football can have as much momentum as a heavy, slow bowling ball. Momentum is a vector with the same direction as the velocity. In straight-line motion a positive direction is chosen, and the momentum of an object moving in the opposite direction is negative.
Force and time change momentum
When a net force acts on an object, its velocity changes, and so does its momentum. The size of the change depends on two things: how large the force is and how long it acts. The product of a force and the time it acts is called impulse. A small force for a long time can change the momentum as much as a large force for a short time. A javelin thrower pulls the javelin over a long distance so that the force has time to act for as long as possible and the momentum of the javelin grows large.
The same applies to stopping. When a moving object is stopped, the change in its momentum is fixed: all of the original momentum has to be removed. The stopping force can be large and short or small and long, but the product of force and time is the same in both cases. When the force is drawn as a function of time, the impulse is the area between the curve and the time axis. A hard bumper stops a cart quickly with a large force and a soft bumper slowly with a small force, but the areas are equal.
Rebounding changes momentum more than stopping
If an object does not just stop but bounces back, its momentum changes more. When a ball hits a wall and rebounds straight back at the same speed, its momentum changes from $mv$ to $-mv$. The change is $-2mv$, and its magnitude is twice as large as when stopping. So the wall must give a rebounding ball twice the impulse. When calculating a change in momentum the signs are essential, since even a change in the direction of motion alone changes the momentum.
The answer to the opening question
The jumper hits the surface at the same velocity whether it is hard or soft, and stops in both cases. So the change in the jumper's momentum is the same, and so is the impulse of the net force on the jumper. The force of the surface is larger than this net force by the weight, but it too becomes smaller when the stopping time grows. A thick mat is compressed over tens of centimetres, so stopping takes many times longer than on hard ground, where the body stops within a few centimetres. Since impulse is force times time, a longer stopping time means a smaller average net force, and so a smaller average force from the surface too. A smaller force puts less strain on the body, so the risk of injury is lower. This is also why we bend our knees when landing, boxing gloves are padded and a cycling helmet has compressible foam inside.
the impulse–momentum principle
Momentum and impulse
The momentum of an object is
$$\overline{p} = m\overline{v}$$
where $m$ is the mass of the object ($\text{kg}$) and $\overline{v}$ its velocity ($\text{m/s}$). The unit of momentum is $\text{kg}\cdot\text{m/s}$. In straight-line motion the vectors are replaced by numbers with signs, $p = mv$, and $v$ is negative when the object moves in the negative direction.
When a constant force $\overline{F}$ acts on an object for a time interval $\Delta t$, the impulse of the force is
$$\overline{I} = \overline{F}\,\Delta t$$
where $\overline{F}$ is the force ($\text{N}$) and $\Delta t$ the time it acts ($\text{s}$). The impulse has the direction of the force, and its unit is $\text{N}\cdot\text{s}$. Since $1\,\text{N} = 1\,\text{kg}\cdot\text{m/s}^2$, we have $1\,\text{N}\cdot\text{s} = 1\,\text{kg}\cdot\text{m/s}$: impulse and momentum have the same unit.
The impulse–momentum principle from Newton's second law
When the net force $\Sigma\overline{F}$ is constant, the object accelerates uniformly. In the chapter Acceleration and uniformly accelerated motion you learned that then the acceleration is the change in velocity divided by the time, $\overline{a} = \tfrac{\overline{v} - \overline{v}_0}{\Delta t}$. Substituting this into Newton's second law $\Sigma\overline{F} = m\overline{a}$ and multiplying both sides by the time $\Delta t$ gives
$$\Sigma\overline{F}\,\Delta t = m\overline{v} - m\overline{v}_0$$
On the left is the impulse of the net force and on the right the change in momentum. This is the impulse–momentum principle:
$$\Delta\overline{p} = \overline{I}$$
where $\Delta\overline{p} = \overline{p} - \overline{p}_0$ is the change in momentum ($\text{kg}\cdot\text{m/s}$) and $\overline{I}$ the impulse of the net force acting on the object ($\text{N}\cdot\text{s}$). In words: the momentum of an object changes by as much as the net force gives it impulse. For example, when a $2.0\,\text{kg}$ cart is pushed in the direction of motion with a net force of $4.0\,\text{N}$ for $0.50\,\text{s}$, the impulse is $2.0\,\text{N}\cdot\text{s}$, and the velocity of the cart increases by $1.0\,\text{m/s}$.
Signs in one dimension
Before calculating, a positive direction is chosen. Velocities, momenta and forces that point in the opposite direction get a minus sign. When a ball rebounds straight back from a wall at the same speed and the positive direction is towards the wall, $p_0 = mv$ and $p = -mv$, so $\Delta p = -mv - mv = -2mv$. The change points away from the wall, in the same direction as the force of the wall on the ball.
A changing force
In a collision the force is not constant: it grows from zero to its greatest value and falls back. The impulse–momentum principle holds then too, when the impulse is calculated as the area between the $(t, F)$ graph of the force and the time axis. The effect of a changing force is often described with an average force, which would give the same impulse in the same time:
$$F_\text{avg} = \frac{|\Delta p|}{\Delta t}$$
where $F_\text{avg}$ is the average force ($\text{N}$), $|\Delta p|$ the magnitude of the change in momentum ($\text{kg}\cdot\text{m/s}$) and $\Delta t$ the duration of the collision ($\text{s}$). In the graph the average force is the height of the rectangle whose area is the same as the area under the curve. The greatest force is always larger than the average force.
Common misconception
Many people think that a soft mat or an airbag lowers the impulse of a collision or "absorbs the momentum". But the change in momentum is the same whether the object stops against a hard or a soft surface, so the impulse of the net force is the same too. A soft surface lengthens the stopping time, and that is why a smaller average net force is enough for the same impulse.
Another mistake is to forget the signs when an object rebounds. If a ball hits a wall at $10\,\text{m/s}$ and rebounds at $10\,\text{m/s}$, the magnitude of its velocity does not change, but its momentum changes from $mv$ to $-mv$. The change is $-2mv$, not zero: its magnitude is $2mv$. A third mistake is to confuse momentum and kinetic energy: momentum is a vector and proportional to the velocity, kinetic energy is a scalar and proportional to the square of the velocity.
Summary
- Momentum is the product of mass and velocity, $\overline{p} = m\overline{v}$. It is a vector with the direction of the velocity; the unit is $\text{kg}\cdot\text{m/s}$.
- The impulse of a constant force is $\overline{I} = \overline{F}\,\Delta t$. The impulse of a changing force is the area under the $(t, F)$ graph; the unit is $\text{N}\cdot\text{s} = \text{kg}\cdot\text{m/s}$.
- The impulse–momentum principle: the impulse of the net force equals the change in momentum, $\Delta\overline{p} = \overline{I}$.
- The same change in momentum is obtained with a large force in a short time or a small force in a long time. A soft surface, bent knees and the foam of a helmet lengthen the stopping time and lower the force.
- In one dimension a positive direction is chosen and signs are used. For an object rebounding straight back at the same speed the magnitude of the change in momentum is $2mv$, twice as large as for one that stops.
Optional Return to your predictions
Compare your predictions with the readings of the simulation. What surprised you most? Write in your own words what determines the impulse of a collision and what determines the greatest force of a collision.
Would you change your prediction?
Apply
Worked example
Problem
A tennis ball of mass $58\,\text{g}$ flies horizontally at $25\,\text{m/s}$ towards a racket. The racket hits the ball straight back, and the ball leaves the racket horizontally at $35\,\text{m/s}$. The ball is in contact with the racket for $4.0\,\text{ms}$. How large is the average force with which the racket pushes the ball?
Step 1/7
Given and find. $m = 58\,\text{g}$, incoming velocity $25\,\text{m/s}$, outgoing velocity $35\,\text{m/s}$ in the opposite direction and contact time $\Delta t = 4.0\,\text{ms}$. We are asked for the average force of the racket $F_\text{avg}$.
Step 2/7
Principle. The momentum of the ball changes because the racket pushes it. The motion is horizontal, and during contact the force of the racket is so large that the effect of the ball's weight is neglected. Then the force of the racket is the net force, and we use the impulse–momentum principle. The positive direction is chosen as the outgoing direction of the ball, so the incoming velocity is $v_0 = -25\,\text{m/s}$ and the outgoing velocity $v = +35\,\text{m/s}$.
Step 3/7
Formula solved. From the impulse–momentum principle $F_\text{avg}\,\Delta t = mv - mv_0$, so
$$F_\text{avg} = \frac{m(v - v_0)}{\Delta t}$$
Step 4/7
Unit conversions. The mass to kilograms and the time to seconds: $58\,\text{g} = 0.058\,\text{kg}$ and $4.0\,\text{ms} = 0.0040\,\text{s}$.
Step 5/7
Substitution. First we calculate the change in momentum and then divide it by the time:
$$\begin{aligned} &m(v - v_0) \\ &= 0.058\,\text{kg} \cdot (35 + 25)\,\tfrac{\text{m}}{\text{s}} \\ &= 3.48\,\text{kg}\cdot\tfrac{\text{m}}{\text{s}} \\[6pt] &F_\text{avg} = \frac{3.48\,\text{kg}\cdot\text{m/s}}{0.0040\,\text{s}} \\ &= 870\,\text{N} \end{aligned}$$
Step 6/7
Result. The racket pushes the ball with an average force of $8.7 \cdot 10^2\,\text{N}$, to two significant figures like the mass $58\,\text{g}$ and the time $4.0\,\text{ms}$. The force is positive, that is, it points in the outgoing direction of the ball.
Step 7/7
Reasonableness. The weight of the ball is $0.058\,\text{kg} \cdot 9.81\,\text{m/s}^2 \approx 0.57\,\text{N}$, about $1500$ times smaller than the force of the racket, so neglecting the weight was justified. If the sign were forgotten and the velocities were simply subtracted, $35\,\text{m/s} - 25\,\text{m/s}$, the force would come out as only $145\,\text{N}$, even though the racket must first stop the ball and then accelerate it in the opposite direction.
Optional Find the error
Apply to a new situation
A pesäpallo (Finnish baseball) catcher catches a ball of mass $0.16\,\text{kg}$ flying horizontally at $25\,\text{m/s}$. The catcher pulls the hand backwards, and the glove stops the ball in $0.050\,\text{s}$.
How large is the average horizontal force of the glove on the ball? Give the answer in $\text{N}$.
Hint 1: a guiding question
By how much does the momentum of the ball change when it stops? Which principle connects the change in momentum and the force?
Hint 2: what is the connection
The impulse–momentum principle:
when the final velocity is zero.
Hint 3: an intermediate step
The change in momentum is $0.16\,\text{kg} \cdot 25\,\text{m/s} = 4.0\,\text{kg}\cdot\text{m/s}$. Divide it by the stopping time.
Show the solution
Given
mass of the ball $m = 0.16\,\text{kg}$, initial velocity $v_0 = 25\,\text{m/s}$, final velocity $v = 0$, stopping time $\Delta t = 0.050\,\text{s}$; the horizontal motion is considered
Find
the average horizontal force of the glove $F_\text{avg}$ in $\text{N}$
Principle
The glove stops the ball, so the ball's momentum changes from $mv_0$ to zero. In the horizontal direction only the force of the glove acts on the ball, since the weight is vertical and does not change the horizontal momentum. So by the impulse–momentum principle the impulse of the force of the glove equals the change in the ball's momentum.
Formula
so $F_\text{avg} = \dfrac{mv_0}{\Delta t}$
Conversions
All quantities are in SI units, so no conversions are needed; $1\,\tfrac{\text{kg}\cdot\text{m/s}}{\text{s}} = 1\,\text{N}$.
Substitution
Answer
$80\,\text{N}$
Two significant figures like the mass $0.16\,\text{kg}$, the velocity $25\,\text{m/s}$ and the time $0.050\,\text{s}$.
Reasonableness
The force corresponds to the weight of an object of about $8\,\text{kg}$, which can be felt in the hand but does not hurt. If the catcher kept the hand rigid and the ball stopped in $0.010\,\text{s}$, the force would be $400\,\text{N}$, five times as large. That is why the hand is pulled backwards: a longer stopping time lowers the force.
Check your understanding
Exit ticket
Answer on your own first and only then open the answer.
- Show that the unit of impulse, $\text{N}\cdot\text{s}$, is the same as the unit of momentum, $\text{kg}\cdot\text{m/s}$.
Answer
By Newton's second law $1\,\text{N} = 1\,\text{kg}\cdot\text{m/s}^2$, so
$1\,\text{N}\cdot\text{s}$$= 1\,\text{kg}\cdot\tfrac{\text{m}}{\text{s}^2}\cdot\text{s}$$= 1\,\text{kg}\cdot\text{m/s}$ - A train brakes steadily and stops. What is the direction of the impulse of the net braking force on the train, and what does its magnitude depend on?
Answer
The impulse points against the motion, in the same direction as the change in momentum. Its magnitude is the train's original momentum $mv$, so it depends on the mass and the initial velocity of the train and not on how long the braking takes.
1. Concept
2. From a graph
A football of mass $0.45\,\text{kg}$ is at rest. The graph shows the force with which a foot kicks the ball horizontally. During the kick the other horizontal forces are neglected.
How large is the velocity of the ball when it leaves the foot? Give the answer in $\text{m/s}$.
Hint 1: a guiding question
The impulse is the area between the graph and the time axis. What shape is the region?
Hint 2: what is the connection
The area of a triangle is $\tfrac{1}{2} \cdot \text{base} \cdot \text{height}$. Convert the milliseconds to seconds.
Hint 3: an intermediate step
The impulse is
and it equals the momentum of the ball $mv$.
Show the solution
Given
mass of the ball $m = 0.45\,\text{kg}$, initial velocity $v_0 = 0$; from the graph the peak value of the force and the duration of the kick
Find
the velocity of the ball $v$ in $\text{m/s}$
Principle
During the kick the force of the foot is the horizontal net force on the ball, when the other horizontal forces are neglected. By the impulse–momentum principle its impulse equals the change in the ball's momentum. The force changes, so the impulse is read from the area of the graph.
Reading the graph
The force rises from zero to its peak value $F_\text{max} = 900\,\text{N}$ at $6\,\text{ms}$ and falls to zero at $12\,\text{ms}$. Between the graph and the time axis is a triangle with base $\Delta t = 12\,\text{ms}$ and height $900\,\text{N}$. The impulse is the area of the triangle, $\tfrac{1}{2} F_\text{max}\,\Delta t$.
Formula
$mv - 0 = \tfrac{1}{2} F_\text{max}\,\Delta t$, so
Conversions
Substitution
Answer
$12\,\text{m/s}$
Two significant figures like the mass $0.45\,\text{kg}$ and the time $12\,\text{ms}$.
Reasonableness
The velocity is about $43\,\text{km/h}$, a typical speed for a brisk kick. The average force of the kick is $5.4\,\text{N}\cdot\text{s} : 0.012\,\text{s} = 450\,\text{N}$, half of the peak value, as it should be for a force that rises and falls steadily.
Suggestion: one answer does not show everything you can do, but review the theory summary and do the core exercises.
3. Explain
Using momentum and impulse, explain why a high jumper is not injured when landing on a thick mat, even though the same fall onto hard ground could be dangerous.
Model answer
The jumper falls from the same height, so we assume that the jumper hits the mat and the hard ground at the same velocity. In both the jumper stops, so the momentum changes by the same amount, from $mv$ to zero. By the impulse–momentum principle the impulse of the net force on the jumper is then the same. The force of the surface is larger than the net force by the weight, but it too becomes smaller when the stopping time grows. Impulse is force times time. Hard ground stops the body within a few centimetres in a very short time, so the force must be large. A thick mat is compressed over tens of centimetres, and stopping takes many times longer. The same impulse is then produced by a net force that is many times smaller on average, and the force from the surface becomes smaller too. A smaller force puts less strain on the body, so the risk of injury is lower. So the mat does not lower the impulse or the change in momentum; it spreads the same impulse over a longer time.
Assess your own answer
Tick the points your own answer already contains.
Review the theory summary and practise with the exercises.
Review the chapter's concepts
Cards you mark Review later collect in your review deck.
- Momentum
- The product of an object's mass and velocity, a vector quantity.$$\overline{p} = m\overline{v}$$
- Impulse
- Describes the total effect of a force over a time; for a constant force:$$\overline{I} = \overline{F}\Delta t$$
- Impulse–momentum principle
- The impulse an object receives equals the change in its momentum.$$\overline{I} = \Delta\overline{p}$$
- Force–time graph
- The area between the graph and the time axis gives the impulse the object receives.
Exercises
Level 1 · basic practice 5 exercises
One formula or one step of reasoning. Core exercises first, the rest is extra practice.
- Exercise 13-1Core
A hockey puck of mass $0.17\,\text{kg}$ slides on the ice at $30\,\text{m/s}$. How large is the momentum of the puck?
Answer
$5.1\,\text{kg}\cdot\text{m/s}$
Show the solution
Given
mass of the puck $m = 0.17\,\text{kg}$, velocity $v = 30\,\text{m/s}$
Find
the momentum of the puck $p$ in $\text{kg}\cdot\text{m/s}$
Principle
Momentum is the product of mass and velocity. It is a vector with the direction of the velocity; here its magnitude is asked.
Formula
$p = mv$
Conversions
The mass and the velocity are in SI units, so no conversions are needed.
Substitution
$p$$= 0.17\,\text{kg} \cdot 30\,\text{m/s}$$= 5.1\,\text{kg}\cdot\text{m/s}$Answer
$5.1\,\text{kg}\cdot\text{m/s}$
Two significant figures like the mass $0.17\,\text{kg}$ and the velocity $30\,\text{m/s}$.
Reasonableness
Estimate with round numbers: $0.2\,\text{kg} \cdot 30\,\text{m/s} = 6\,\text{kg}\cdot\text{m/s}$, so the result is of the right order of magnitude. A light puck has a small momentum even though its velocity is large: the momentum of a walking person is about $100\,\text{kg}\cdot\text{m/s}$.
- Exercise 13-2Core
A shopping cart is pushed horizontally with a force of $12\,\text{N}$ for $0.50\,\text{s}$. How large is the impulse of the pushing force?
Answer
$6.0\,\text{N}\cdot\text{s}$
Show the solution
Given
pushing force $F = 12\,\text{N}$, time of action $\Delta t = 0.50\,\text{s}$
Find
the impulse of the pushing force $I$ in $\text{N}\cdot\text{s}$
Principle
The impulse of a constant force is the product of the force and the time it acts. The impulse has the direction of the force, here the direction of the push.
Formula
$I = F\,\Delta t$
Conversions
The force is in newtons and the time in seconds, so no conversions are needed.
Substitution
$I$$= 12\,\text{N} \cdot 0.50\,\text{s}$$= 6.0\,\text{N}\cdot\text{s}$Answer
$6.0\,\text{N}\cdot\text{s}$
Two significant figures like the force $12\,\text{N}$ and the time $0.50\,\text{s}$.
Reasonableness
Half of $12\,\text{N}$ is $6\,\text{N}$, so the product is correct. If the mass of the cart were $15\,\text{kg}$ and there were no friction, its velocity would increase by $6.0\,\text{N}\cdot\text{s} : 15\,\text{kg} = 0.40\,\text{m/s}$, which fits a short push.
- Exercise 13-3
A volleyball of mass $0.27\,\text{kg}$ is momentarily at rest at the highest point of its path when a player spikes it. The ball leaves the hand at $18\,\text{m/s}$. The effect of the ball's weight during the spike is neglected.
How large an impulse does the hand give the ball?
Answer
$4.9\,\text{N}\cdot\text{s}$
Show the solution
Given
mass of the ball $m = 0.27\,\text{kg}$, initial velocity $v_0 = 0$, final velocity $v = 18\,\text{m/s}$; the effect of the weight is neglected
Find
the impulse the hand gives the ball $I$ in $\text{N}\cdot\text{s}$
Principle
During the spike the force of the hand is the net force on the ball, when the effect of the weight is neglected. By the impulse–momentum principle its impulse equals the change in the ball's momentum.
Formula
$I = \Delta p = mv - mv_0$, and $v_0 = 0$, so $I = mv$
Conversions
The quantities are in SI units, so no conversions are needed.
Substitution
$I$$= 0.27\,\text{kg} \cdot 18\,\text{m/s}$$= 4.86\,\text{N}\cdot\text{s}$Answer
$4.9\,\text{N}\cdot\text{s}$
Two significant figures like the mass $0.27\,\text{kg}$ and the velocity $18\,\text{m/s}$.
Reasonableness
Estimate: $0.3\,\text{kg} \cdot 20\,\text{m/s} = 6\,\text{kg}\cdot\text{m/s}$, and both numbers were rounded up, so the result $4.9$ is suitable. The impulse is in the direction the ball leaves, since the ball started from rest and moved off in that direction.
- Exercise 13-4
A skier of mass $70\,\text{kg}$ has a momentum of $420\,\text{kg}\cdot\text{m/s}$. How large is the skier's velocity?
Answer
$6.0\,\text{m/s}$
Show the solution
Given
mass of the skier $m = 70\,\text{kg}$, momentum $p = 420\,\text{kg}\cdot\text{m/s}$
Find
the velocity of the skier $v$ in $\text{m/s}$
Principle
Momentum is the product of mass and velocity, so the velocity is obtained by dividing the momentum by the mass.
Formula
$p = mv$, so $v = \dfrac{p}{m}$
Conversions
The quantities are in SI units, so no conversions are needed; $\tfrac{\text{kg}\cdot\text{m/s}}{\text{kg}} = \text{m/s}$.
Substitution
$v$$= \dfrac{420\,\text{kg}\cdot\text{m/s}}{70\,\text{kg}}$$= 6.0\,\text{m/s}$Answer
$6.0\,\text{m/s}$
Two significant figures like the mass $70\,\text{kg}$.
Reasonableness
$70\,\text{kg} \cdot 6.0\,\text{m/s} = 420\,\text{kg}\cdot\text{m/s}$, so the calculation works backwards. The velocity is about $22\,\text{km/h}$, which fits a skier.
- Exercise 13-5
A badminton racket gives the shuttlecock an impulse of $0.15\,\text{N}\cdot\text{s}$. The racket and the shuttlecock are in contact for $0.0010\,\text{s}$.
How large is the average force of the racket on the shuttlecock?
Answer
$150\,\text{N}$
Show the solution
Given
impulse $I = 0.15\,\text{N}\cdot\text{s}$, contact time $\Delta t = 0.0010\,\text{s}$
Find
the average force of the racket $F_\text{avg}$ in $\text{N}$
Principle
Impulse is force times time. When the force changes during contact, the impulse divided by the contact time gives the average force.
Formula
$I = F_\text{avg}\,\Delta t$, so $F_\text{avg} = \dfrac{I}{\Delta t}$
Conversions
The time is in seconds, so no conversions are needed; $\tfrac{\text{N}\cdot\text{s}}{\text{s}} = \text{N}$.
Substitution
$F_\text{avg}$$= \dfrac{0.15\,\text{N}\cdot\text{s}}{0.0010\,\text{s}}$$= 150\,\text{N}$Answer
$150\,\text{N}$
That is, $1.5 \cdot 10^2\,\text{N}$, two significant figures like the impulse $0.15\,\text{N}\cdot\text{s}$ and the time $0.0010\,\text{s}$.
Reasonableness
Dividing by a thousandth of a second multiplies the number by a thousand: $0.15 \cdot 1000 = 150$. The force is large compared with a light shuttlecock, but it acts for only a thousandth of a second, as in hits generally.
Level 2 · applied practice 4 exercises
Two steps, working with measured data, a unit conversion or rearranging a formula. Core exercises first, the rest is extra practice.
- Exercise 13-6Core
A car of mass $1200\,\text{kg}$ drives on a horizontal road at $54\,\text{km/h}$. The driver brakes, and the car stops in $3.0\,\text{s}$. Air resistance is neglected.
How large is the average braking force on the car? Give the answer in kilonewtons.
Answer
$6.0\,\text{kN}$
Show the solution
Given
mass of the car $m = 1200\,\text{kg}$, initial velocity $v_0 = 54\,\text{km/h}$, final velocity $v = 0$, stopping time $\Delta t = 3.0\,\text{s}$; air resistance is neglected
Find
the average braking force $F_\text{avg}$ in $\text{kN}$
Principle
The road is horizontal, so the weight of the car and the normal force of the road cancel each other, and the horizontal net force is the braking friction force of the road. By the impulse–momentum principle its impulse equals the change in the car's momentum. The positive direction is the direction of motion of the car.
Formula
$F\,\Delta t = mv - mv_0 = -mv_0$, so the magnitude of the force is $F_\text{avg} = \dfrac{mv_0}{\Delta t}$
Conversions
$v_0$$= 54\,\text{km/h}$$= \dfrac{54\,\text{m}}{3.6\,\text{s}}$$= 15\,\text{m/s}$Substitution
$F_\text{avg}$$= \dfrac{1200\,\text{kg} \cdot 15\,\text{m/s}}{3.0\,\text{s}}$$= \dfrac{18\,000\,\text{kg}\cdot\text{m/s}}{3.0\,\text{s}}$$= 6000\,\text{N}$Answer
$6.0\,\text{kN}$
Two significant figures like the velocity $54\,\text{km/h}$ and the time $3.0\,\text{s}$.
Reasonableness
The sign of the force is negative, that is, it points against the motion, as a braking force should. The force is about half of the weight of the car, $1200\,\text{kg} \cdot 9.81\,\text{m/s}^2 \approx 12\,\text{kN}$, which corresponds to ordinary braking on a dry road.
- Exercise 13-7Core
A golf ball of mass $46\,\text{g}$ is at rest. A club hits it, and the ball leaves at $70\,\text{m/s}$. The club and the ball are in contact for $0.50\,\text{ms}$. The effect of the ball's weight during the hit is neglected.
How large is the average force of the club on the ball? Give the answer in kilonewtons.
Answer
$6.4\,\text{kN}$
Show the solution
Given
mass of the ball $m = 46\,\text{g}$, initial velocity $v_0 = 0$, final velocity $v = 70\,\text{m/s}$, contact time $\Delta t = 0.50\,\text{ms}$; the effect of the weight is neglected
Find
the average force of the club $F_\text{avg}$ in $\text{kN}$
Principle
During the hit the force of the club is the net force on the ball, when the effect of the weight is neglected. By the impulse–momentum principle the impulse of the force of the club equals the change in the ball's momentum.
Formula
$F_\text{avg}\,\Delta t$$= mv - mv_0$$= mv$so $F_\text{avg} = \dfrac{mv}{\Delta t}$
Conversions
$m$$= 46\,\text{g}$$= 0.046\,\text{kg}$$\Delta t$$= 0.50\,\text{ms}$$= 0.00050\,\text{s}$Substitution
$F_\text{avg}$$= \dfrac{0.046\,\text{kg} \cdot 70\,\text{m/s}}{0.00050\,\text{s}}$$= \dfrac{3.22\,\text{kg}\cdot\text{m/s}}{0.00050\,\text{s}}$$= 6440\,\text{N}$Answer
$6.4\,\text{kN}$
Two significant figures like the mass $46\,\text{g}$, the velocity $70\,\text{m/s}$ and the time $0.50\,\text{ms}$.
Reasonableness
The weight of the ball is only $0.046\,\text{kg} \cdot 9.81\,\text{m/s}^2 \approx 0.45\,\text{N}$, so the force of the club is over $10\,000$ times larger and neglecting the weight was justified. A large force is possible because it acts for only half a millisecond.
- Exercise 13-8
A ball of mass $60\,\text{g}$ is thrown horizontally at a wall. The ball hits the wall at $14\,\text{m/s}$ and rebounds straight back at $11\,\text{m/s}$.
How large an impulse does the wall give the ball? Give the magnitude of the impulse.
Answer
$1.5\,\text{N}\cdot\text{s}$
Show the solution
Given
mass of the ball $m = 60\,\text{g}$, incoming velocity $14\,\text{m/s}$ towards the wall, outgoing velocity $11\,\text{m/s}$ away from the wall
Find
the magnitude of the impulse the wall gives the ball $|I|$ in $\text{N}\cdot\text{s}$
Principle
The force of the wall changes the ball's momentum, and by the impulse–momentum principle the impulse equals the change in momentum. The ball rebounds in the opposite direction, so the velocities need signs. The positive direction is chosen towards the wall: $v_0 = +14\,\text{m/s}$ and $v = -11\,\text{m/s}$.
Formula
$I = \Delta p = mv - mv_0 = m(v - v_0)$
Conversions
$m = 60\,\text{g} = 0.060\,\text{kg}$
Substitution
$I$$= 0.060\,\text{kg} \cdot (-11 - 14)\,\text{m/s}$$= 0.060\,\text{kg} \cdot (-25\,\text{m/s})$$= -1.5\,\text{N}\cdot\text{s}$Answer
$1.5\,\text{N}\cdot\text{s}$
Two significant figures like the mass $60\,\text{g}$ and the velocities; the impulse points away from the wall.
Reasonableness
The impulse is larger than the $0.060\,\text{kg} \cdot 14\,\text{m/s} = 0.84\,\text{N}\cdot\text{s}$ needed just to stop the ball, because the wall also accelerates the ball back. If the velocities were subtracted without signs, the result would be only $0.18\,\text{N}\cdot\text{s}$, which is too small even to stop the ball.
- Exercise 13-9
A hockey puck of mass $0.17\,\text{kg}$ slides at $8.0\,\text{m/s}$. A player hits the puck with a stick in the direction of motion with an average force of $120\,\text{N}$, and the contact lasts $0.025\,\text{s}$. Friction from the ice is neglected.
How large is the velocity of the puck after the hit?
Answer
$26\,\text{m/s}$
Show the solution
Given
mass of the puck $m = 0.17\,\text{kg}$, initial velocity $v_0 = 8.0\,\text{m/s}$, average force of the stick $F = 120\,\text{N}$ in the direction of motion, contact time $\Delta t = 0.025\,\text{s}$; friction from the ice is neglected
Find
the velocity of the puck $v$ after the hit in $\text{m/s}$
Principle
When friction is neglected, the force of the stick is the horizontal net force on the puck. By the impulse–momentum principle its impulse equals the change in the puck's momentum. The force is in the direction of motion, so the momentum increases.
Formula
$F\,\Delta t = mv - mv_0$, so $v = v_0 + \dfrac{F\,\Delta t}{m}$
Conversions
The quantities are in SI units, so no conversions are needed; $\tfrac{\text{N}\cdot\text{s}}{\text{kg}} = \text{m/s}$.
Substitution
$v$$= 8.0\,\text{m/s} + \dfrac{120\,\text{N} \cdot 0.025\,\text{s}}{0.17\,\text{kg}}$$= 8.0\,\text{m/s} + 17.65\,\text{m/s}$$= 25.65\,\text{m/s}$Answer
$26\,\text{m/s}$
Two significant figures like the velocity $8.0\,\text{m/s}$, the mass $0.17\,\text{kg}$ and the time $0.025\,\text{s}$.
Reasonableness
An impulse of $3.0\,\text{N}\cdot\text{s}$ increases the velocity of a light puck a lot, by about $18\,\text{m/s}$. The final velocity $26\,\text{m/s}$, about $92\,\text{km/h}$, fits a hard hit.
Level 3 · challenging 3 exercises
Optional: application, two ideas together or a new situation where you reason for yourself.
- Exercise 13-10
A child sits on a sled on horizontal snow, and the combined mass of the child and the sled is $24\,\text{kg}$. The sled is first gliding at $1.0\,\text{m/s}$ when an adult pushes it in the direction of motion. The graph shows the pushing force as a function of time. Friction from the snow is neglected.
How large is the velocity of the sled after the push?
Answer
$3.0\,\text{m/s}$
Show the solution
Given
combined mass of the child and the sled $m = 24\,\text{kg}$, initial velocity $v_0 = 1.0\,\text{m/s}$; from the graph the pushing force as a function of time; friction is neglected
Find
the velocity of the sled $v$ after the push in $\text{m/s}$
Principle
When friction is neglected, the pushing force is the horizontal net force on the sled; the weight and the normal force of the snow cancel each other. By the impulse–momentum principle the impulse of the pushing force equals the change in momentum. The force changes, so the impulse is read from the area of the graph.
Reading the graph
The force rises from zero to $60\,\text{N}$ by $0.20\,\text{s}$, stays at $60\,\text{N}$ until $0.80\,\text{s}$ and falls to zero at $1.00\,\text{s}$. Under the graph is a trapezoid with height $60\,\text{N}$ and parallel sides $1.00\,\text{s}$ (base) and $0.60\,\text{s}$ (top). The area is the height times the average of the sides.
Formula
$I$$= F_\text{max} \cdot \dfrac{t_\text{base} + t_\text{top}}{2}$$I = mv - mv_0$, so $v = v_0 + \dfrac{I}{m}$
Conversions
The quantities are in SI units, so no conversions are needed.
Substitution
$I$$= 60\,\text{N} \cdot \dfrac{(1.00 + 0.60)\,\text{s}}{2}$$= 48\,\text{N}\cdot\text{s}$$v$$= 1.0\,\text{m/s} + \dfrac{48\,\text{N}\cdot\text{s}}{24\,\text{kg}}$$= 1.0\,\text{m/s} + 2.0\,\text{m/s}$$= 3.0\,\text{m/s}$Answer
$3.0\,\text{m/s}$
Two significant figures like the mass $24\,\text{kg}$ and the velocity $1.0\,\text{m/s}$.
Reasonableness
If the force had been $60\,\text{N}$ for the whole second, the impulse would be $60\,\text{N}\cdot\text{s}$; because of the rising and falling edges it is slightly smaller, $48\,\text{N}\cdot\text{s}$. An increase in velocity of $2.0\,\text{m/s}$ from a push of one second is believable. In reality the friction of the snow lowers the final velocity slightly.
- Exercise 13-11
A car hits a fixed obstacle and stops. The seat belt stops a $75\,\text{kg}$ passenger in $0.12\,\text{s}$, and in the horizontal direction only the force of the belt acts on the passenger. Assume that the force of the belt on the passenger may be at most $10\,\text{kN}$ on average.
From what greatest velocity can the passenger be stopped under these assumptions? Give the answer in $\text{km/h}$.
Answer
$58\,\text{km/h}$
Show the solution
Given
mass of the passenger $m = 75\,\text{kg}$, stopping time $\Delta t = 0.12\,\text{s}$, average force of the belt at most $F = 10\,\text{kN}$, final velocity $v = 0$
Find
the greatest initial velocity $v_0$ from which the passenger can be stopped, in $\text{km/h}$
Principle
In the horizontal direction only the force of the belt acts on the passenger, so it is the net force. By the impulse–momentum principle the impulse of the force of the belt equals the change in the passenger's momentum. When the force and the time are at their greatest, the momentum to be stopped, and so the initial velocity, is also at its greatest.
Formula
$F\,\Delta t = mv_0 - 0$, so $v_0 = \dfrac{F\,\Delta t}{m}$
Conversions
$F = 10\,\text{kN} = 10\,000\,\text{N}$; finally the answer is converted to $\text{km/h}$ by multiplying by $3.6$, since $1\,\text{m/s} = 3.6\,\text{km/h}$.
Substitution
$v_0$$= \dfrac{10\,000\,\text{N} \cdot 0.12\,\text{s}}{75\,\text{kg}}$$= \dfrac{1200\,\text{N}\cdot\text{s}}{75\,\text{kg}}$$= 16\,\text{m/s}$$= 16 \cdot 3.6\,\text{km/h}$$= 57.6\,\text{km/h}$Answer
$58\,\text{km/h}$
Two significant figures like the mass $75\,\text{kg}$ and the time $0.12\,\text{s}$.
Reasonableness
The velocity corresponds to an ordinary speed on a country road or a main city street, and seat belt tests use velocities of the same order of magnitude. The force of the belt, $10\,\text{kN}$, is about $14$ times the weight of the passenger, $75\,\text{kg} \cdot 9.81\,\text{m/s}^2 \approx 740\,\text{N}$, which shows why a higher collision velocity would be dangerous.
- Exercise 13-12
Two balls, each of mass $0.20\,\text{kg}$, are thrown horizontally at a wall at $8.0\,\text{m/s}$. The clay ball sticks to the wall. The rubber ball rebounds straight back at $6.0\,\text{m/s}$. Each ball is in contact with the wall for $0.020\,\text{s}$.
By how much is the average force of the rubber ball on the wall greater than that of the clay ball?
Answer
$60\,\text{N}$
Show the solution
Given
mass of each ball $m = 0.20\,\text{kg}$, incoming velocity $8.0\,\text{m/s}$, outgoing velocity of the rubber ball $6.0\,\text{m/s}$ back, the clay ball stays in place, contact time $\Delta t = 0.020\,\text{s}$ for both
Find
the difference $\Delta F$ between the average forces of the rubber ball and the clay ball on the wall, in $\text{N}$
Principle
The force of the wall is the horizontal net force on the ball, and by the impulse–momentum principle its impulse is the change in the ball's momentum. By Newton's third law the ball pushes the wall with a force equal in magnitude to the force of the wall on the ball. The positive direction is chosen towards the wall: $v_0 = 8.0\,\text{m/s}$, for the clay ball $v = 0$ and for the rubber ball $v = -6.0\,\text{m/s}$.
Formula
The magnitude of the force is $F = \dfrac{m\,|v - v_0|}{\Delta t}$, so
$\Delta F$$= \dfrac{m(v_0 + v_\text{r})}{\Delta t} - \dfrac{mv_0}{\Delta t}$$= \dfrac{m v_\text{r}}{\Delta t}$where $v_\text{r} = 6.0\,\text{m/s}$ is the outgoing speed of the rubber ball
Conversions
The quantities are in SI units, so no conversions are needed.
Substitution
$\Delta F$$= \dfrac{0.20\,\text{kg} \cdot 6.0\,\text{m/s}}{0.020\,\text{s}}$$= \dfrac{1.2\,\text{kg}\cdot\text{m/s}}{0.020\,\text{s}}$$= 60\,\text{N}$Answer
$60\,\text{N}$
Two significant figures like the mass $0.20\,\text{kg}$ and the velocity $6.0\,\text{m/s}$.
Reasonableness
Calculated separately, the force of the clay ball is $\tfrac{0.20 \cdot 8.0}{0.020}\,\text{N} = 80\,\text{N}$ and that of the rubber ball $\tfrac{0.20 \cdot 14}{0.020}\,\text{N} = 140\,\text{N}$, and the difference is $60\,\text{N}$. The rebounding ball pushes the wall harder, because in addition to stopping the ball the wall must accelerate it back.
Photographs
- Strong woman punching bag in gym. Annushka Ahuja. source, Pexels License