Free AP Physics 1 Linear Momentum

AP Physics 1 · Linear Momentum · free lesson

Linear Momentum

The idea, then worked examples solved step by step, then practice questions whose numbers change every time you try them.

The idea

Momentum: mass in motion

A loaded truck and a tennis ball can both be moving at 20 m/s, but nobody would try to stop them the same way. Speed alone does not say how hard a moving object is to stop. The mass matters too, and linear momentum puts the two together.

Definition.

The linear momentum of an object is the product of its mass and its velocity:

p⃗=mv⃗\begin{aligned} \vec{p} &= m\vec{v} \end{aligned}

The unit is the kilogram meter per second (kg·m/s). Momentum is a vector, and it points in the same direction as the velocity. When the word "momentum" is used alone, it means linear momentum.

A car of mass 1200 kg1200\,\mathrm{kg} moves at 15.0 m/s15.0\,\mathrm{m/s}, and a cyclist with bicycle, together 80.0 kg80.0\,\mathrm{kg}, moves at 10.0 m/s10.0\,\mathrm{m/s}. Both have a velocity to the east, so both momenta point east.

pc=mcvc=1200 kg×15 m/s=18000 kg⋅m/spb=mbvb=80 kg×10 m/s=800 kg⋅m/s\begin{aligned} p_c &= m_cv_c \\ &= 1200\,\mathrm{kg} \times 15\,\mathrm{m/s} \\ &= 18000\,\mathrm{kg{\cdot}m/s} \\ p_b &= m_bv_b \\ &= 80\,\mathrm{kg} \times 10\,\mathrm{m/s} \\ &= 800\,\mathrm{kg{\cdot}m/s} \end{aligned}

The car has a momentum of 18000 kg⋅m/s18000\,\mathrm{kg{\cdot}m/s} and the cyclist 800 kg⋅m/s800\,\mathrm{kg{\cdot}m/s}. The car's mass is 15.015.0 times as large and its speed 1.501.50 times as large, so its momentum is 22.522.5 times as large. Momentum depends on both factors: double the mass at the same speed, or double the speed with the same mass, and the momentum doubles.

The two arrows are drawn to the same scale, one square for 2000 kg·m/s. The car's arrow is 9.009.00 squares long; the cyclist's is only 0.4000.400 of a square.

Momentum has a direction

Because momentum is a vector, its direction is that of the velocity. Along a line you choose a positive direction and write the momentum as a signed component, px=mvxp_x = mv_x. Mass is never negative, so the sign of the momentum is always the sign of the velocity: an object moving in the negative direction has a negative momentum. Two objects with the same speed and the same mass, moving in opposite directions, have momenta of equal magnitude and opposite signs.

Move the slider to set the velocity of a 2.0 kg cart. The arrow below it is the cart's momentum, drawn at one square for 2 kg·m/s.

Warning.

Momentum is not speed and not energy. It has a direction and a sign, and a moving object always has a momentum that is not zero. A "negative" momentum is not smaller than a positive one in size. A momentum of −6 kg·m/s has the same magnitude as +6 kg·m/s.

The momentum of a system

A system can contain more than one object. The momentum of the system is the vector sum of the momenta of its objects. Along a line that is a sum of signed components. The bar chart shows two carts: A has a mass of 2.0 kg and moves right at 3.0 m/s, and B has a mass of 1.0 kg and moves left at 2.0 m/s. Right is positive.

pA=mAvA=2 kg×3 m/s=6.00 kg⋅m/spB=mBvB=1 kg×(−2 m/s)=−2.00 kg⋅m/sps=pA+pB=6 kg⋅m/s+(−2 kg⋅m/s)=4.00 kg⋅m/s\begin{aligned} p_A &= m_Av_A \\ &= 2\,\mathrm{kg} \times 3\,\mathrm{m/s} \\ &= 6.00\,\mathrm{kg{\cdot}m/s} \\ p_B &= m_Bv_B \\ &= 1\,\mathrm{kg} \times \left(-2\,\mathrm{m/s}\right) \\ &= -2.00\,\mathrm{kg{\cdot}m/s} \\ p_s &= p_A + p_B \\ &= 6\,\mathrm{kg{\cdot}m/s} + \left(-2\,\mathrm{kg{\cdot}m/s}\right) \\ &= 4.00\,\mathrm{kg{\cdot}m/s} \end{aligned}

The momentum of cart A is 6.00 kg⋅m/s6.00\,\mathrm{kg{\cdot}m/s} and that of cart B is −2.00 kg⋅m/s-2.00\,\mathrm{kg{\cdot}m/s}. The system's momentum is 4.00 kg⋅m/s4.00\,\mathrm{kg{\cdot}m/s}, which is less than A's alone, because B's momentum points the other way and takes away from the sum. Add the signed components; never add the speeds.

Collisions and explosions

Momentum is most useful for two kinds of interaction in which objects exert forces on each other for a short time.

Definition.

A collision is a model for an interaction in which the forces the objects exert on each other are much larger than the net external force on those objects during the interaction. An explosion is a model for an interaction in which forces inside the system push objects of the system apart.

The words "much larger" matter. Two carts that hit on a track push on each other with forces of many newtons for a few hundredths of a second, while friction on the wheels is a fraction of a newton. During the hit, the forces between the carts decide what happens, and the outside forces hardly matter. That is what lets us call it a collision. A cart that slides across a rough floor is not a collision, because friction acts the whole time and is not small.

In the collision model you look only at the initial state, just before the objects touch, and the final state, just after they separate. What happens in between, the squashing and rebounding, is not analyzed, so each object can be treated as a single object with one mass and one velocity (the object model). The animation shows a collision. Cart A (2.0 kg) moves right at 3.0 m/s and cart B (1.0 kg) moves left at 2.0 m/s. Afterward A moves right at 1.0 m/s and B moves right at 2.0 m/s. The arrows are momenta, one unit of length for 2 kg·m/s.

Before the collision the momenta are 6.00 kg⋅m/s6.00\,\mathrm{kg{\cdot}m/s} for A and −2.00 kg⋅m/s-2.00\,\mathrm{kg{\cdot}m/s} for B. After it they are 2.00 kg⋅m/s2.00\,\mathrm{kg{\cdot}m/s} for A and 2.00 kg⋅m/s2.00\,\mathrm{kg{\cdot}m/s} for B. Each cart's momentum changed, because each cart pushed the other. Only the numbers just before and just after are used.

An explosion is the opposite kind of event. Two carts at rest touch, with a compressed spring between them. The spring is released and pushes them apart. The forces are inside the system (the spring pushes each cart, and each cart pushes the spring), and again they are much larger than any outside force during the short push.

Cart A (2.0 kg) ends up moving left at 1.5 m/s and cart B (1.0 kg) moves right at 3.0 m/s. At the start both momenta are zero. At the end A has −3.00 kg⋅m/s-3.00\,\mathrm{kg{\cdot}m/s} and B has 3.00 kg⋅m/s3.00\,\mathrm{kg{\cdot}m/s}. Objects in an explosion move apart, and the lighter cart ends up faster.

Rule.

Summary. p⃗=mv⃗\vec{p} = m\vec{v}, in kg·m/s, in the direction of the velocity. The momentum of a system is the vector sum of its objects' momenta. A collision and an explosion are interactions in which the forces between the objects are much larger than the net external force, so only the states just before and just after are analyzed.

Worked examples

Worked examples

Five examples of the kinds the AP exam asks on this topic: a momentum from mass and velocity, signed momenta of a system, a "what changes if" question, a graph read, and a claim about a collision justified.

Example.

Example 1. A freight cart

A cart of mass 850 kg850\,\mathrm{kg} rolls north along a track at 4.00 m/s4.00\,\mathrm{m/s}. What is its momentum?

Solution

The momentum is the mass times the velocity, and it points the way the velocity does.

p=mv=850 kg×4 m/s=3400 kg⋅m/s\begin{aligned} p &= mv \\ &= 850\,\mathrm{kg} \times 4\,\mathrm{m/s} \\ &= 3400\,\mathrm{kg{\cdot}m/s} \end{aligned}
Result.

The momentum is 3400 kg⋅m/s3400\,\mathrm{kg{\cdot}m/s}, to the north. The unit, kilogram meter per second, is just the units of mass and velocity multiplied.

Example.

Example 2. Two objects, one system

On a straight road, a car of mass 1500 kg1500\,\mathrm{kg} moves east at 12.0 m/s12.0\,\mathrm{m/s}, and a car of mass 900 kg900\,\mathrm{kg} moves west at 15.0 m/s15.0\,\mathrm{m/s}. East is positive. What is the momentum of the system of the two cars?

Solution

Find each momentum as a signed component, then add the components.

p1=m1v1=1500 kg×12 m/s=18000 kg⋅m/sp2=m2v2=900 kg×(−15 m/s)=−13500 kg⋅m/sps=p1+p2=18000 kg⋅m/s+(−13500 kg⋅m/s)=4500 kg⋅m/s\begin{aligned} p_1 &= m_1v_1 \\ &= 1500\,\mathrm{kg} \times 12\,\mathrm{m/s} \\ &= 18000\,\mathrm{kg{\cdot}m/s} \\ p_2 &= m_2v_2 \\ &= 900\,\mathrm{kg} \times \left(-15\,\mathrm{m/s}\right) \\ &= -13500\,\mathrm{kg{\cdot}m/s} \\ p_s &= p_1 + p_2 \\ &= 18000\,\mathrm{kg{\cdot}m/s} + \left(-13500\,\mathrm{kg{\cdot}m/s}\right) \\ &= 4500\,\mathrm{kg{\cdot}m/s} \end{aligned}
Result.

The system's momentum is 4500 kg⋅m/s4500\,\mathrm{kg{\cdot}m/s}, which is positive, so it points east. The western car has the smaller mass and the greater speed, but the eastern car's momentum is larger, so the sum points east. Adding the speeds, 27 m/s, would have been wrong: the cars move in opposite directions.

Example.

Example 3. What changes if

A 0.600 kg0.600\,\mathrm{kg} ball moves at 8.00 m/s8.00\,\mathrm{m/s}. Its mass stays the same, but its speed becomes three times as large. By what factor does its momentum change?

Solution

The momentum is proportional to the speed when the mass is fixed.

p1=mv=0.6 kg×8 m/s=4.80 kg⋅m/sv3=3v=3×8 m/s=24.0 m/sp3=mv3=0.6 kg×24 m/s=14.4 kg⋅m/s\begin{aligned} p_1 &= mv \\ &= 0.6\,\mathrm{kg} \times 8\,\mathrm{m/s} \\ &= 4.80\,\mathrm{kg{\cdot}m/s} \\ v_3 &= 3v \\ &= 3 \times 8\,\mathrm{m/s} \\ &= 24.0\,\mathrm{m/s} \\ p_3 &= mv_3 \\ &= 0.6\,\mathrm{kg} \times 24\,\mathrm{m/s} \\ &= 14.4\,\mathrm{kg{\cdot}m/s} \end{aligned}
Result.

The momentum goes from 4.80 kg⋅m/s4.80\,\mathrm{kg{\cdot}m/s} to 14.4 kg⋅m/s14.4\,\mathrm{kg{\cdot}m/s}, three times as large, because p=mvp = mv is proportional to vv. Tripling the mass at the original speed would also triple the momentum: momentum depends on the product of mass and speed, not on either one alone.

Example.

Example 4. Reading a graph of momentum against velocity

The graph shows the momentum of a cart as its velocity changes. What is the cart's mass, and what is its momentum at −4.00 m/s-4.00\,\mathrm{m/s}?

Solution

Because p=mvp = mv, the graph is a straight line through the origin, and its slope is the mass. The point (6.00 m/s6.00\,\mathrm{m/s}, 15.0 kg⋅m/s15.0\,\mathrm{kg{\cdot}m/s}) is on the line.

m=pava=15 kg⋅m/s6 m/s=2.50 kgpb=mvb=2.5 kg×(−4 m/s)=−10.0 kg⋅m/s\begin{aligned} m &= \frac{p_a}{v_a} \\ &= \frac{15\,\mathrm{kg{\cdot}m/s}}{6\,\mathrm{m/s}} \\ &= 2.50\,\mathrm{kg} \\ p_b &= mv_b \\ &= 2.5\,\mathrm{kg} \times \left(-4\,\mathrm{m/s}\right) \\ &= -10.0\,\mathrm{kg{\cdot}m/s} \end{aligned}
Result.

The mass is 2.50 kg2.50\,\mathrm{kg}, and the momentum at −4.00 m/s-4.00\,\mathrm{m/s} is −10.0 kg⋅m/s-10.0\,\mathrm{kg{\cdot}m/s}. The negative value means the momentum points in the negative direction, and the line continues below the axis to the left of the origin.

Example.

Example 5. Is it a collision?

Two carts strike each other on a track. During the 0.0200 s0.0200\,\mathrm{s} they are in contact, each pushes on the other with a force of about 40.0 N40.0\,\mathrm{N}. The rolling friction on each cart is about 0.500 N0.500\,\mathrm{N}. A student says that the interaction can be treated as a collision, so the carts need only be looked at just before and just after. Is the claim right?

Solution

The model applies when the forces the objects exert on each other are much larger than the net external force during the interaction.

Compare the two forces: 40.0 N40.0\,\mathrm{N} divided by 0.500 N0.500\,\mathrm{N}.

Result.

The claim is right. The force between the carts is 80.080.0 times the friction, so the friction hardly matters during the 0.0200 s0.0200\,\mathrm{s} of contact, and only the states before and after need to be known. The same carts rolling for a long time over a rough floor would not be a collision, because friction would act the whole time.

Practice

Answer AP-style questions on linear momentum. Each answer is checked, with the full solution, and every try brings new numbers. Practice needs a free account: it keeps your progress.

Practise linear momentum with new numbers