Unit 4: Linear Momentum and Impulse

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Useful variables

• $\vec{p}$ = linear momentum (Units: $\text{kg}\cdot\text{m/s}$)
• $\vec{J}$ = impulse (Units: $\text{N}\cdot\text{s}$)
• $\vec{F}_{\text{net}}$ = net external force
• $M$ = total mass of a system
• $\vec{r}_{\text{cm}}$ = center-of-mass position
• $\vec{v}_{\text{cm}}$ = center-of-mass velocity

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Linear Momentum

The linear momentum of a particle is

$$\vec{p} = m\vec{v}.$$

Momentum is a vector, so the components must be conserved independently. Newton’s second law can be written in its more general form as

$$\vec{F}_{\text{net}} = \frac{d\vec{p}}{dt}.$$

For constant mass, this reduces to $\vec{F}_{\text{net}} = m\vec{a}$. However, the momentum form is the more fundamental statement of Newton’s second law (in fact, it was the original statement of the law!), and it is essential whenever mass is not constant, such as a rocket burning fuel or a rope piling onto the ground.

Intuitively, momentum measures how hard it is to stop something because of both mass and velocity. A slow truck and a fast baseball can both have large momentum, but for different reasons. Force changes momentum over time, which is why stopping the same object gently means spreading the same momentum change over a longer time.

Example. A $0.15\ \text{kg}$ baseball moving at $40\ \text{m/s}$ and a $1200\ \text{kg}$ car moving at $0.0050\ \text{m/s}$ have the same speed? The same kinetic energy? The same momentum? Compare their momenta and kinetic energy.

The baseball’s momentum is

$$p_{\text{ball}}=mv=(0.15)(40)=6.0\ \text{kg}\cdot\text{m/s}.$$

The car’s momentum is

$$p_{\text{car}}=(1200)(0.0050)=6.0\ \text{kg}\cdot\text{m/s}.$$

They have the same momentum even though their speeds are very different. Momentum depends on both mass and velocity, so both matter when analyzing a collision.

Their kinetic energies are very different:

$$K_{\text{ball}}=\frac{1}{2}(0.15)(40)^2=120\ \text{J},$$

while

$$K_{\text{car}}=\frac{1}{2}(1200)(0.0050)^2=0.015\ \text{J}.$$

So equal momentum does not mean equal kinetic energy. For the same momentum, the lighter object must move much faster, and because $K=p^2/(2m)$, it carries more kinetic energy.

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Impulse

Impulse is the change in momentum caused by a force acting over time:

$$\vec{J} = \int_{t_i}^{t_f} \vec{F}_{\text{net}}\,dt = \Delta \vec{p}.$$

This result is called the impulse-momentum theorem. It says a force changes motion by accumulating over time, not just by being large at one instant.

Proof (Impulse-Momentum Theorem). Start with Newton’s second law in momentum form:

$$\vec{F}_{\text{net}}=\frac{d\vec{p}}{dt}.$$

Multiply by $dt$:

$$\vec{F}_{\text{net}}\,dt=d\vec{p}.$$

Integrate over the time interval of the interaction:

$$\int_{t_i}^{t_f}\vec{F}_{\text{net}}\,dt =\int_{\vec p_i}^{\vec p_f}d\vec p.$$

The right side is just the change in momentum:

$$\int_{t_i}^{t_f}\vec{F}_{\text{net}}\,dt =\vec p_f-\vec p_i =\Delta \vec p.$$

By definition, the integral of force over time is impulse, so $\vec J=\Delta\vec p$.

For a constant force,

$$\vec{J} = \vec{F}_{\text{net}}\Delta t.$$

On a force-time graph, impulse is the signed area under the curve. This is the time-domain analog of how work is the area under a force-position graph in Unit 3. During a collision, the peak force may be hard to model, but the impulse can often be found from the initial and final momenta, since $\vec{J}=\Delta\vec{p}$ does not care about the detailed shape of $\vec{F}(t)$.

It is also useful to define the average force over a collision:

$$\vec{F}_{\text{avg}} = \frac{\vec{J}}{\Delta t} = \frac{\Delta \vec{p}}{\Delta t}.$$

The average force is the constant force that would deliver the same impulse over the same time, even if the actual force varies.

Tip

Tip

A longer contact time (a “follow-through,” an airbag, bent knees on landing) reduces the peak force for a fixed change in momentum, thus effectively “softening” the impact.

Example. A $0.50\ \text{kg}$ ball moving at $4.0\ \text{m/s}$ to the right strikes a wall. The wall pushes back with a force that rises linearly from $0$ to a peak of $200\ \text{N}$ over $0.010\ \text{s}$, then falls linearly back to $0$ over the next $0.010\ \text{s}$. Find the ball’s velocity after contact and the average force.

The impulse is the area under the $F$-$t$ graph, which is a triangle of base $0.020\ \text{s}$ and height $200\ \text{N}$:

$$J = \tfrac{1}{2}(0.020)(200) = 2.0\ \text{N}\cdot\text{s}.$$

Take rightward as positive. The wall pushes left, so the impulse on the ball is $J = -2.0\ \text{N}\cdot\text{s}$. Using $J=\Delta p = m(v_f - v_i)$,

$$v_f = v_i + \frac{J}{m} = 4.0 + \frac{-2.0}{0.50} = 4.0 - 4.0 = 0\ \text{m/s}.$$

The ball is brought exactly to rest. The average force is

$$F_{\text{avg}} = \frac{J}{\Delta t} = \frac{-2.0}{0.020} = -100\ \text{N},$$

half the peak force, as expected for a triangular pulse. The instantaneous force reached $200\ \text{N}$, but the average force is what determines the net momentum change.

Example. A stream of identical balls, each of mass $m = 0.10\ \text{kg}$, flies horizontally at $v = 20\ \text{m/s}$ and strikes a wall. The balls hit at a rate of $n = 5$ balls per second and rebound straight back with the same speed (elastic bounce). Find the average force the wall exerts on the stream, and the force the stream exerts on the wall.

Each ball reverses its velocity, so the change in momentum of one ball is

$$\Delta p_{\text{one}} = m(-v) - m(v) = -2mv = -2(0.10)(20) = -4.0\ \text{kg}\cdot\text{m/s}.$$

In one second, $n = 5$ balls bounce, so the total momentum change delivered by the wall per second is

$$\frac{\Delta p_{\text{total}}}{\Delta t} = n\,\Delta p_{\text{one}} = 5(-4.0) = -20\ \text{N}.$$

The wall pushes back on the stream with an average force of $20\ \text{N}$ opposing the incoming motion. By Newton’s third law, the stream pushes on the wall with $20\ \text{N}$ in the direction of incoming travel. Note that if the balls instead stuck to the wall (no rebound), each $\Delta p_{\text{one}}$ would be only $-mv$, giving half the force: rebounding transfers twice the momentum of sticking. This same reasoning, written as $F = \dot{m}\,v$ for a continuous mass flow rate, handles water from a hose or gas from a thruster.

Variable-Mass Systems

Variable-mass problems are usually momentum problems where the mass of the object you are tracking changes with time. The key move is to remember that momentum is a product:

$$\vec p=m\vec v.$$

So if $m$ changes,

$$\frac{d\vec p}{dt} =m\frac{d\vec v}{dt}+\vec v\frac{dm}{dt}$$

by the product rule. That extra $\vec v\,dm/dt$ term is the part that is easy to forget. It represents momentum changing because mass is being added or removed, even if the velocity of the object itself is not changing at that instant.

However, for an open system, $\vec F_{\text{ext}}=d(m\vec v)/dt$ is not automatically enough unless you are careful about the velocity of the entering or leaving mass. A good strategy is:

Problem-solving strategy

1. Decide what object/system you are tracking.
2. Write the momentum of that system as $m(t)\vec v(t)$.
3. Use the product rule: $d(m\vec v)/dt=m\,d\vec v/dt+\vec v\,dm/dt$.
4. Account for the momentum carried in or out by entering/leaving mass.
5. Keep signs consistent: mass entering has $dm/dt>0$ for your chosen system; mass leaving has $dm/dt<0$.

For example, if a cart collects falling sand that has no horizontal velocity before landing, there is no external horizontal force, but the cart’s horizontal momentum is spread over more mass. If a rocket ejects exhaust backward, the rocket gains forward momentum because the exhaust carries backward momentum away.

Example. A cart of initial mass $m_0$ moves frictionlessly at speed $v_0$. It passes under a hopper that drops sand vertically into the cart at constant rate $\lambda$, so the sand has zero horizontal velocity before landing. Find the cart’s speed $v(t)$ after time $t$.

Track the cart plus the sand already inside it. The mass is

$$m(t)=m_0+\lambda t.$$

There is no external horizontal force, and the incoming sand brings in zero horizontal momentum. Therefore the horizontal momentum of the cart-plus-collected-sand stays constant:

$$m(t)v(t)=m_0v_0.$$

So

$$v(t)=\frac{m_0v_0}{m_0+\lambda t}.$$

You can also see this from the product rule. Since horizontal momentum is constant,

$$\frac{d}{dt}(mv)=0.$$

Using the product rule,

$$m\frac{dv}{dt}+v\frac{dm}{dt}=0.$$

Here $dm/dt=\lambda$, so

$$\frac{dv}{dt}=-\frac{\lambda}{m}v.$$

The cart slows down not because an external horizontal force pulls it backward, but because it must share its horizontal momentum with newly added mass.

Example. Suppose a rocket in space (so assume there is no gravity or other effects) ejects fuel backward at constant relative speed $u$. If the rocket’s mass changes from $m_0$ to $m_f$, find its change in speed $\Delta v$.

Take the rocket’s forward direction as positive. At some instant, the rocket has mass $m$ and speed $v$. It ejects a small positive amount of fuel $dM$ backward relative to the rocket, so the rocket’s mass becomes $m-dM$ and its speed becomes $v+dv$. The exhaust moves at speed $v-u$ in the inertial frame.

With no external force, conserve momentum over this tiny interval:

$$mv=(m-dM)(v+dv)+dM(v-u).$$

Expand the right side:

$$mv=mv+m\,dv-v\,dM-dM\,dv+dM\,v-u\,dM.$$

The terms $-v\,dM$ and $dM\,v$ cancel. The product $dM\,dv$ is second-order small, so ignore it:

$$mv=mv+m\,dv-u\,dM.$$

Thus

$$m\,dv=u\,dM.$$

Since $dM=-dm$, where $dm$ is the change in rocket mass,

$$m\,dv=-u\,dm.$$

Separate variables:

$$dv=-u\frac{dm}{m}.$$

Integrate from initial mass $m_0$ to final mass $m_f$:

$$\Delta v =-u\int_{m_0}^{m_f}\frac{dm}{m} =u\ln\left(\frac{m_0}{m_f}\right).$$

So the ideal rocket equation is

$$\Delta v=u\ln\left(\frac{m_0}{m_f}\right).$$

The logarithm appears because each bit of fuel gives a larger speed gain later, when the rocket has less remaining mass.

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Conservation of Momentum

For a system of particles,

$$\frac{d\vec{P}_{\text{sys}}}{dt} = \sum \vec{F}_{\text{ext}}.$$

If the net external force on the system is zero, or if its impulse is negligible during the event,

$$\vec{P}_{i} = \vec{P}_{f}.$$

Internal forces cancel in pairs by Newton’s third law, so they cannot change the total momentum of the system. They can, however, redistribute momentum among the objects inside the system.

Proof (Conservation of Momentum). For a system of particles, the total momentum is

$$\vec{P}_{\text{sys}}=\sum_i \vec{p}_i.$$

Differentiate:

$$\frac{d\vec{P}_{\text{sys}}}{dt}=\sum_i \frac{d\vec{p}_i}{dt}.$$

For each particle, Newton’s second law says

$$\frac{d\vec{p}_i}{dt}=\vec{F}_{i,\text{net}}.$$

The forces on all particles can be split into external forces and internal forces. Internal forces occur in equal-and-opposite pairs:

$$\vec{F}_{i\text{ on }j}=-\vec{F}_{j\text{ on }i}.$$

When summed over the whole system, those internal pairs cancel. Therefore

$$\frac{d\vec{P}_{\text{sys}}}{dt}=\sum \vec{F}_{\text{ext}}.$$

If $\sum \vec{F}_{\text{ext}}=0$, then $d\vec{P}_{\text{sys}}/dt=0$, so total momentum is constant:

$$\vec{P}_i=\vec{P}_f.$$

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Center of Mass

The center of mass simplifies the motion of a complex shape or system of particles. Its velocity determines the system’s total momentum, and its acceleration is set by the net external force, so we can study the overall motion without tracking each particle separately.

For discrete particles, the position $r$ (in whatever coordinate direction you define $r$ to be in) for an object’s center of mass (for one coordinate, e.g. x-coordinate or y-coordinate) is equal to

$$\vec{r}_{\text{cm}} = \frac{1}{M}\sum_i m_i\vec{r}_i,$$

where

$$M = \sum_i m_i.$$

The proof for the formula uses topics covered later on, so will not be covered here. For a continuous body,

$$\vec{r}_{\text{cm}} = \frac{1}{M}\int \vec{r}\,dm.$$

The center of mass moves as if all external force acted on the total mass:

$$\sum \vec{F}_{\text{ext}} = M\vec{a}_{\text{cm}}.$$

The total momentum of a system is

$$\vec{P}_{\text{sys}} = M\vec{v}_{\text{cm}}.$$

Proof (Center-of-Mass Motion). Start from the discrete center-of-mass definition:

$$\vec{r}_{\text{cm}}=\frac{1}{M}\sum_i m_i\vec{r}_i.$$

Differentiate once:

$$\vec{v}_{\text{cm}}=\frac{1}{M}\sum_i m_i\vec{v}_i.$$

Multiplying both sides by $M$ gives

$$M\vec{v}_{\text{cm}}=\sum_i m_i\vec{v}_i=\vec{P}_{\text{sys}}.$$

Differentiate again:

$$M\vec{a}_{\text{cm}}=\frac{d\vec{P}_{\text{sys}}}{dt}.$$

Using the momentum result above,

$$\frac{d\vec{P}_{\text{sys}}}{dt}=\sum \vec{F}_{\text{ext}},$$

so

$$\sum \vec{F}_{\text{ext}}=M\vec{a}_{\text{cm}}.$$

Example. Find the center of mass of a uniform right-triangular plate with legs along the axes: vertices at $(0,0)$, $(b,0)$, and $(0,h)$. It may be helpful to define a surface mass density $\sigma$ (mass per area).

Since the plate is uniform, $\sigma$ is constant all throughout the plate. Thus, the total mass is $M = \sigma \cdot \tfrac{1}{2}bh$ (mass/area times area). Slice the triangle into thin vertical strips of width $dx$ at position $x$. At that $x$, the hypotenuse runs from $(0,h)$ to $(b,0)$, so its height is

$$y(x) = h\left(1 - \frac{x}{b}\right).$$

The strip has area $y(x)\,dx$ and mass $dm = \sigma\, y(x)\,dx$. Then

$$x_{\text{cm}} = \frac{1}{M}\int_0^b x\,dm = \frac{\sigma}{M}\int_0^b x\,h\left(1-\frac{x}{b}\right)dx.$$

Evaluate the integral:

$$\int_0^b \left(x - \frac{x^2}{b}\right)dx = \frac{b^2}{2} - \frac{b^2}{3} = \frac{b^2}{6}.$$

So

$$x_{\text{cm}} = \frac{\sigma h}{M}\cdot\frac{b^2}{6} = \frac{\sigma h b^2/6}{\sigma bh/2} = \frac{b}{3}.$$

By symmetry of the argument (slicing horizontally), $y_{\text{cm}} = h/3$. The centroid of a uniform triangle sits one-third of the way in from each leg, at $(b/3,\,h/3)$.

Example. A thin rod of length $L$ lies along the $x$-axis from $x=0$ to $x=L$. Its linear mass density increases as $\lambda(x) = \lambda_0\,\dfrac{x}{L}$. Find its center of mass.

The mass element is $dm = \lambda(x)\,dx = \lambda_0\dfrac{x}{L}\,dx$. The total mass is

$$M = \int_0^L \lambda_0\frac{x}{L}\,dx = \frac{\lambda_0}{L}\cdot\frac{L^2}{2} = \frac{\lambda_0 L}{2}.$$

The center of mass is

$$x_{\text{cm}} = \frac{1}{M}\int_0^L x\,dm = \frac{1}{M}\int_0^L x\cdot\lambda_0\frac{x}{L}\,dx = \frac{\lambda_0}{ML}\int_0^L x^2\,dx = \frac{\lambda_0}{ML}\cdot\frac{L^3}{3}.$$

Substituting $M = \lambda_0 L/2$,

$$x_{\text{cm}} = \frac{\lambda_0 L^3/3}{(\lambda_0 L/2)L} = \frac{2L}{3}.$$

The center of mass sits at $2L/3$, shifted toward the heavy end, as expected. For a uniform rod the answer would have been $L/2$.

Example. A firework of mass $M$ is launched and, at the top of its arc, is momentarily moving horizontally at $v_0$ when it explodes into two equal pieces. One piece is observed to fall straight down with zero horizontal velocity immediately after the burst. Where does the other piece go, and where is the center of mass?

The explosion is internal, and over the brief burst gravity’s impulse is negligible, so horizontal momentum is conserved across the explosion. Before:

$$P_x = Mv_0.$$

After, piece 1 (mass $M/2$) has horizontal velocity $0$, so piece 2 (mass $M/2$) must carry all the horizontal momentum:

$$Mv_0 = \frac{M}{2}(0) + \frac{M}{2}v_{2x} \;\Rightarrow\; v_{2x} = 2v_0.$$

The second piece moves forward at twice the original speed. Crucially, the center of mass continues on the original parabolic trajectory as if no explosion happened, because the only external force is still gravity. The pieces fan out around that path; their CM lands exactly where the unexploded firework would have landed.

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Collisions

All collisions conserve momentum for an isolated system. Kinetic energy may or may not be conserved.

Elastic collisions

An elastic collision conserves both momentum and kinetic energy:

$$\vec{P}_i = \vec{P}_f, \qquad K_i = K_f.$$

For a one-dimensional elastic collision between masses $m_1$ and $m_2$, conservation of momentum and kinetic energy imply that the relative speed reverses:

$$v_{1i}-v_{2i}=-(v_{1f}-v_{2f}).$$

Solving with momentum conservation gives

$$v_{1f}=\frac{m_1-m_2}{m_1+m_2}v_{1i}+\frac{2m_2}{m_1+m_2}v_{2i},$$ $$v_{2f}=\frac{2m_1}{m_1+m_2}v_{1i}+\frac{m_2-m_1}{m_1+m_2}v_{2i}.$$

The proof for the final velocities is left to the reader as an exercise.

Proof (Relative Speed Reversal in a 1D Elastic Collision). Momentum conservation gives

$$m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}.$$

Rearrange:

$$m_1(v_{1i}-v_{1f})=m_2(v_{2f}-v_{2i}).$$

Kinetic energy conservation gives

$$\frac{1}{2}m_1v_{1i}^2+\frac{1}{2}m_2v_{2i}^2 = \frac{1}{2}m_1v_{1f}^2+\frac{1}{2}m_2v_{2f}^2.$$

Rearrange and factor:

$$m_1(v_{1i}^2-v_{1f}^2)=m_2(v_{2f}^2-v_{2i}^2),$$ $$m_1(v_{1i}-v_{1f})(v_{1i}+v_{1f}) = m_2(v_{2f}-v_{2i})(v_{2f}+v_{2i}).$$

Divide this equation by the rearranged momentum equation:

$$v_{1i}+v_{1f}=v_{2f}+v_{2i}.$$

Move terms:

$$v_{1i}-v_{2i}=-(v_{1f}-v_{2f}).$$

So the relative velocity after the collision is the negative of the relative velocity before the collision.

Problem-solving strategy

1. Conserve momentum in one dimension.
2. Use the relative-speed reversal $v_{1i}-v_{2i}=-(v_{1f}-v_{2f})$ instead of expanding kinetic energy.
3. Solve the two linear equations.

Useful shortcuts:

• Equal masses in 1D exchange velocities.
• If a light object elastically hits a much heavier stationary object (usually denoted by $m << M$), the light object rebounds with nearly the same speed.
• If a heavy object elastically hits a much lighter stationary object, the heavy object barely changes speed and the light object leaves at nearly twice the heavy object’s speed.

Example. A $3.0\ \text{kg}$ cart moving right at $5.0\ \text{m/s}$ elastically collides with a $1.0\ \text{kg}$ cart moving left at $2.0\ \text{m/s}$. Find both final velocities.

Use the 1D elastic formulas:

$$v_{1f}=\frac{m_1-m_2}{m_1+m_2}v_{1i}+\frac{2m_2}{m_1+m_2}v_{2i},$$ $$v_{2f}=\frac{2m_1}{m_1+m_2}v_{1i}+\frac{m_2-m_1}{m_1+m_2}v_{2i}.$$

With $m_1=3.0$, $m_2=1.0$, $v_{1i}=5.0$, and $v_{2i}=-2.0$,

$$v_{1f}=\frac{2}{4}(5.0)+\frac{2}{4}(-2.0)=2.5-1.0=1.5\ \text{m/s},$$ $$v_{2f}=\frac{6}{4}(5.0)+\frac{-2}{4}(-2.0)=7.5+1.0=8.5\ \text{m/s}.$$

The lighter cart shoots right quickly because it receives momentum and kinetic energy from the heavier incoming cart.

Inelastic collisions

An inelastic collision is a collision that conserves momentum but not kinetic energy. Some mechanical energy becomes internal energy, deformation, heat, or sound. A perfectly inelastic collision is the special case where objects stick together after impact:

$$m_1\vec{v}_{1i}+m_2\vec{v}_{2i} = (m_1+m_2)\vec{v}_f.$$

The kinetic energy lost in a perfectly inelastic collision can be computed directly:

$$\Delta K = K_f - K_i = \frac{1}{2}(m_1+m_2)v_f^2 - \left(\frac{1}{2}m_1 v_{1i}^2 + \frac{1}{2}m_2 v_{2i}^2\right).$$

This loss is maximal among all collisions with the same initial momenta, because sticking together leaves the objects with the least possible kinetic energy consistent with conserved momentum (the energy of the center-of-mass motion alone).

For inelastic collisions, the most reliable shortcut is to solve for the center-of-mass velocity:

$$\vec{v}_{\text{cm}}=\frac{\vec{P}_{\text{tot}}}{M_{\text{tot}}}.$$

In a perfectly inelastic collision, the stuck-together object moves at exactly $\vec{v}_{\text{cm}}$. The kinetic energy after sticking is the kinetic energy of the center-of-mass motion; everything else has been converted into internal energy.

Example. A $0.20\ \text{kg}$ puck moving east at $6.0\ \text{m/s}$ sticks to a $0.30\ \text{kg}$ puck moving north at $4.0\ \text{m/s}$. Find the final velocity of the stuck pair and the kinetic energy lost.

Conserve momentum in components. The total mass is $0.50\ \text{kg}$. Initial momentum components:

$$P_x=(0.20)(6.0)=1.2\ \text{kg}\cdot\text{m/s},$$ $$P_y=(0.30)(4.0)=1.2\ \text{kg}\cdot\text{m/s}.$$

Thus the final velocity components are

$$v_{fx}=\frac{1.2}{0.50}=2.4\ \text{m/s},\qquad v_{fy}=\frac{1.2}{0.50}=2.4\ \text{m/s}.$$

The stuck pair moves northeast with speed

$$v_f=\sqrt{2.4^2+2.4^2}=3.4\ \text{m/s}.$$

Initial kinetic energy:

$$K_i=\tfrac12(0.20)(6.0)^2+\tfrac12(0.30)(4.0)^2=3.6+2.4=6.0\ \text{J}.$$

Final kinetic energy:

$$K_f=\tfrac12(0.50)(3.4)^2\approx2.9\ \text{J}.$$

So about $3.1\ \text{J}$ is lost to deformation, heat, and sound.

The Ballistic Pendulum

The ballistic pendulum is the classic problem that requires both momentum and energy, applied to different stages. The most standard example involves a bullet embedding itself in a hanging block which causes the block to swing up.

Problem-solving strategy

The trick is recognizing that the embedding (a perfectly inelastic collision) conserves momentum but loses energy, while the subsequent swing (no friction, no collision) conserves mechanical energy. Analyze each stage with the correct conserved quantity, then combine.

Example. A bullet of mass $m$ moving at speed $v$ embeds in a block of mass $M$ hanging at rest from a string. The block-plus-bullet then rises to a maximum height $h$. Find $v$ in terms of $m$, $M$, $h$, and $g$.

Stage 1 — collision (momentum conserved, energy not). The embedding is fast and perfectly inelastic. During it, momentum is conserved:

$$mv = (m+M)V,$$

where $V$ is the speed of the combined mass just after impact. Solving,

$$V = \frac{m}{m+M}\,v.$$

Do not set the bullet’s kinetic energy equal to anything here; most of it is lost to embedding.

Stage 2 — swing (energy conserved, momentum not). After impact, the combined mass rises. The string tension does no work, so mechanical energy is conserved during the swing (momentum is not conserved here, because gravity and tension are external):

$$\tfrac{1}{2}(m+M)V^2 = (m+M)gh.$$

Solving for $V$,

$$V = \sqrt{2gh}.$$

Combine. Set the two expressions for $V$ equal:

$$\frac{m}{m+M}\,v = \sqrt{2gh},$$

so

$$v = \frac{m+M}{m}\sqrt{2gh}.$$

The two stages must be analyzed separately with the correct conserved quantity for each. Mixing them (e.g. equating the bullet’s initial kinetic energy to the final potential energy) gives a wrong, larger answer because it ignores the energy lost in embedding.

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Momentum and Collisions in Two Dimensions

In two dimensions, conserve components separately:

$$\sum p_{x,i} = \sum p_{x,f}, \qquad \sum p_{y,i} = \sum p_{y,f}.$$

Angles enter through vector components. The momentum vector triangle is often more important than speed alone, because momentum depends on both mass and velocity. A useful sanity check: the total momentum vector before equals the total momentum vector after, so the “after” vectors must tip-to-tail close the same vector as the “before” vectors.

A very useful formula when dealing with 2D elastic collisions is the 90° separation rule, where unless the collision is head-on, the two objects move off at right angles.

Proof (equal-mass 2D elastic collision: 90° separation). A moving object of mass $m$ elastically strikes an identical mass $m$ at rest. Show that, unless the collision is head-on, the two objects move off at right angles.

Momentum conservation (the masses cancel):

$$\vec{v}_{1i} = \vec{v}_{1f} + \vec{v}_{2f}.$$

Kinetic energy conservation (factors of $\tfrac{1}{2}m$ cancel):

$$v_{1i}^2 = v_{1f}^2 + v_{2f}^2.$$

Take the dot product of the momentum equation with itself:

$$v_{1i}^2 = \vec{v}_{1i}\cdot\vec{v}_{1i} = (\vec{v}_{1f}+\vec{v}_{2f})\cdot(\vec{v}_{1f}+\vec{v}_{2f}) = v_{1f}^2 + v_{2f}^2 + 2\,\vec{v}_{1f}\cdot\vec{v}_{2f}.$$

Comparing with the energy equation forces

$$2\,\vec{v}_{1f}\cdot\vec{v}_{2f} = 0.$$

If both objects move ($v_{1f},v_{2f}\neq 0$), the dot product vanishing means the final velocities are perpendicular: the objects separate at $90^\circ$. This is the familiar billiards result for equal-mass balls; it fails if the masses differ or the collision is inelastic. Treat it as the two-dimensional cousin of the equal-mass velocity-exchange rule from elastic collisions.

Example. A $0.20\ \text{kg}$ puck moving east at $5.0\ \text{m/s}$ strikes a stationary $0.30\ \text{kg}$ puck. After the collision the $0.20\ \text{kg}$ puck moves at $3.0\ \text{m/s}$ at $37^\circ$ north of east. Find the velocity (magnitude and direction) of the $0.30\ \text{kg}$ puck.

Conserve momentum in each direction. Initial momentum is entirely along $x$ (east): $p_x = (0.20)(5.0) = 1.0\ \text{kg}\cdot\text{m/s}$, $p_y = 0$.

The $0.20\ \text{kg}$ puck afterward has components

$$p_{1fx} = (0.20)(3.0)\cos 37^\circ = (0.20)(3.0)(0.799) = 0.479,$$ $$p_{1fy} = (0.20)(3.0)\sin 37^\circ = (0.20)(3.0)(0.602) = 0.361.$$

For the $0.30\ \text{kg}$ puck, conservation gives

$$p_{2fx} = 1.0 - 0.479 = 0.521, \qquad p_{2fy} = 0 - 0.361 = -0.361.$$

Its velocity components are $v_{2fx} = 0.521/0.30 = 1.74\ \text{m/s}$ and $v_{2fy} = -0.361/0.30 = -1.20\ \text{m/s}$. The speed is

$$v_{2f} = \sqrt{1.74^2 + 1.20^2} = \sqrt{3.03 + 1.44} = 2.11\ \text{m/s},$$

at an angle below the east axis of

$$\theta = \tan^{-1}\!\left(\frac{1.20}{1.74}\right) = 34.6^\circ \text{ south of east}.$$

The struck puck recoils to the opposite side, balancing the $y$-momentum that the first puck gained.

Example. Three identical smooth disks lie on a frictionless table. Two disks are initially at rest and touching. A third disk is launched with speed $v$ directly toward the midpoint of the two stationary disks, so all three disks collide simultaneously and elastically. Find the final velocity of the originally moving disk.

By symmetry, the originally moving disk continues along the same axis after the collision. Let its final velocity along the original direction be $v_f$, where a negative value means it rebounds backward. Let each of the two originally stationary disks leave with speed $u$ along the line of centers. Those directions make $30^\circ$ with the original motion, so each contributes $u\cos30^\circ$ of forward momentum.

Momentum along the original direction gives

$$mv=mv_f+2mu\cos30^\circ.$$

Cancel $m$ and use $2\cos30^\circ=\sqrt{3}$:

$$v=v_f+\sqrt{3}u.$$

Energy is conserved because the collision is perfectly elastic:

$$\frac{1}{2}mv^2=\frac{1}{2}mv_f^2+2\left(\frac{1}{2}mu^2\right),$$

so

$$v^2=v_f^2+2u^2.$$

From the momentum equation, $u=(v-v_f)/\sqrt{3}$. Substitute into energy:

$$v^2=v_f^2+\frac{2}{3}(v-v_f)^2.$$

Solving gives two mathematical roots. One is $v_f=v$, the no-collision case, so the physical collision root is

$$v_f=-\frac{v}{5}.$$

The originally moving disk rebounds with speed $v/5$ opposite its initial direction.

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The Zero-Momentum (Center-of-Mass) Frame

For some problems it helps to work in the center-of-mass frame, the reference frame moving with $\vec{v}_{\text{cm}}$. In this frame the total momentum is zero by construction:

$$\vec{P}'_{\text{sys}} = M(\vec{v}_{\text{cm}} - \vec{v}_{\text{cm}}) = 0.$$

Since the total momentum is zero, the objects always have equal and opposite momenta in this frame, both before and after a collision. An elastic collision in the CM frame simply reverses each object’s velocity; an inelastic collision brings them to rest in this frame, which makes the maximum-energy-loss statement obvious. The lab-frame (the stationary frame) results then follow by adding $\vec{v}_{\text{cm}}$ back.

Problem-solving strategy

1. Compute $v_{\text{cm}}$.
2. Subtract it from every velocity to enter the CM frame.
3. Apply the collision rule there.
4. Add $v_{\text{cm}}$ back to return to the lab frame.

Example. A $3.0\ \text{kg}$ cart moving at $5.0\ \text{m/s}$ hits a $1.0\ \text{kg}$ cart moving at $-2.0\ \text{m/s}$ elastically. Solve using the center-of-mass frame.

The center-of-mass velocity is

$$v_{\text{cm}}=\frac{(3.0)(5.0)+(1.0)(-2.0)}{4.0}=\frac{13}{4}=3.25\ \text{m/s}.$$

In the CM frame,

$$v'_{1i}=5.0-3.25=1.75\ \text{m/s},$$ $$v'_{2i}=-2.0-3.25=-5.25\ \text{m/s}.$$

For a 1D elastic collision in the CM frame, velocities reverse:

$$v'_{1f}=-1.75\ \text{m/s},\qquad v'_{2f}=5.25\ \text{m/s}.$$

Add $v_{\text{cm}}$ back:

$$v_{1f}= -1.75+3.25=1.5\ \text{m/s},$$ $$v_{2f}=5.25+3.25=8.5\ \text{m/s}.$$

Example. A $2.0\ \text{kg}$ cart moving right at $7.0\ \text{m/s}$ collides with a $3.0\ \text{kg}$ cart moving left at $3.0\ \text{m/s}$. The carts stick together. Use the center-of-mass frame to find how much kinetic energy is lost in the collision.

First find the center-of-mass velocity:

$$v_{\text{cm}} =\frac{(2.0)(7.0)+(3.0)(-3.0)}{2.0+3.0} =\frac{14-9}{5.0} =1.0\ \text{m/s}.$$

Now switch to the CM frame by subtracting $v_{\text{cm}}$:

$$v'_{1i}=7.0-1.0=6.0\ \text{m/s},$$ $$v'_{2i}=-3.0-1.0=-4.0\ \text{m/s}.$$

Because the carts stick together, they are at rest in the CM frame after the collision. Therefore all kinetic energy that existed in the CM frame is lost to deformation, heat, and sound:

$$K'_{\text{initial}} =\frac{1}{2}(2.0)(6.0)^2+\frac{1}{2}(3.0)(4.0)^2 =36+24 =60\ \text{J}.$$

So the collision loses

$$60\ \text{J}.$$

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Key equations

Idea	Equation
Momentum	$\vec{p} = m\vec{v}$
Newton’s second law (general)	$\vec{F}_{\text{net}} = d\vec{p}/dt$
Impulse	$\vec{J} = \int \vec{F}\,dt = \Delta\vec{p}$
Average force	$\vec{F}_{\text{avg}} = \Delta\vec{p}/\Delta t$
Variable-mass product rule	$\dfrac{d}{dt}(m\vec v)=m\dfrac{d\vec v}{dt}+\vec v\dfrac{dm}{dt}$
Ideal rocket equation	$\Delta v=u\ln\left(\dfrac{m_0}{m_f}\right)$
Conservation of momentum	$\vec{P}_i = \vec{P}_f$ (no external impulse)
Center of mass (discrete)	$\vec{r}_{\text{cm}} = \tfrac{1}{M}\sum_i m_i\vec{r}_i$
Center of mass (continuous)	$\vec{r}_{\text{cm}} = \tfrac{1}{M}\int \vec{r}\,dm$
System momentum	$\vec{P}_{\text{sys}} = M\vec{v}_{\text{cm}}$
CM-frame energy lost when objects stick	$\Delta E_{\text{lost}}=K'_{\text{initial}}$
1D elastic, relative speed	$v_{1i}-v_{2i} = -(v_{1f}-v_{2f})$
1D elastic final velocities	$v_{1f} = \tfrac{m_1-m_2}{m_1+m_2}v_{1i} + \tfrac{2m_2}{m_1+m_2}v_{2i}$
Perfectly inelastic collision	$m_1\vec{v}_{1i} + m_2\vec{v}_{2i} = (m_1+m_2)\vec{v}_f$

Practice

Multiple Choice

1. A net force on a particle varies as $F(t)=F_0(1-t/T)$ from $t=0$ to $t=T$. The impulse is

(A) $F_0T$

(B) $F_0T/2$

(C) $F_0/T$

(D) zero

2. A ball of mass $0.20\ \text{kg}$ hits a wall moving to the right at $15\ \text{m/s}$ and rebounds to the left at $10\ \text{m/s}$. If the contact time is $0.050\ \text{s}$, the magnitude of the average force exerted by the wall is

(A) $20\ \text{N}$

(B) $60\ \text{N}$

(C) $100\ \text{N}$

(D) $250\ \text{N}$

3. A system of particles has total mass $M$. Which equation remains true even if the particles collide inelastically with each other?

(A) $\sum\vec F_{\text{ext}}=M\vec a_{\text{cm}}$

(B) $\sum\vec F_{\text{int}}=M\vec a_{\text{cm}}$

(C) $K_i=K_f$

(D) $\vec r_{\text{cm}}=\vec 0$

4. A projectile explodes at the top of its path into two fragments of masses $m$ and $3m$. If the smaller fragment stops immediately after the explosion, the speed of the larger fragment immediately after is

(A) $v/3$

(B) $v$

(C) $4v/3$

(D) $3v$

5. Two skaters push off from rest on frictionless ice. One has three times the mass of the other. If no external horizontal force acts, the heavier skater’s kinetic energy is

(A) one-ninth the lighter skater’s kinetic energy

(B) one-third the lighter skater’s kinetic energy

(C) equal to the lighter skater’s kinetic energy

(D) three times the lighter skater’s kinetic energy

6. A force on a mass $m$ is $F(t)=F_0t/T$ from $t=0$ to $T$ and then $F(t)=F_0(2-t/T)$ from $t=T$ to $2T$. If the mass starts from rest, its speed at $t=2T$ is

(A) $F_0T/m$

(B) $F_0T/(2m)$

(C) $2F_0T/m$

(D) $\sqrt{F_0T/m}$

7. A stationary object explodes into three equal masses. Two pieces leave at speed $v$ with angle $120^\circ$ between their velocities. The third piece leaves with speed

(A) $0$

(B) $v$

(C) $\sqrt{3}v$

(D) $2v$

8. A mass $m$ moving right with speed $v$ collides elastically in one dimension with an initially stationary mass $3m$. After the collision, the velocity of the mass $m$ is

(A) $-v/2$

(B) $-v/3$

(C) $v/3$

(D) $v/2$

9. A mass $m$ with speed $5\ \text{m/s}$ elastically collides head-on with a mass $3m$ initially moving toward it at $1\ \text{m/s}$. The final velocity of the mass $m$ is

(A) $-4\ \text{m/s}$

(B) $-2\ \text{m/s}$

(C) $1\ \text{m/s}$

(D) $5\ \text{m/s}$

10. A cart moves to the right at $4\ \text{m/s}$ while sand leaks out vertically downward at rate $2\ \text{kg/s}$ relative to the ground. Ignoring external horizontal forces, the horizontal acceleration of the remaining cart-sand system is

(A) zero

(B) $2\ \text{m/s}^2$ to the right

(C) $2\ \text{m/s}^2$ to the left

(D) impossible to determine without the cart mass

11. A cart of initial mass $M$ and speed $v_0$ collects rain falling vertically at rate $\lambda$. Neglect horizontal external forces. Its speed after time $t$ is

(A) $v_0$

(B) $\dfrac{Mv_0}{M+\lambda t}$

(C) $v_0+\lambda t/M$

(D) $\dfrac{(M+\lambda t)v_0}{M}$

12. A rocket expels fuel backward at speed $u$ relative to the rocket. With no external force, the rocket’s speed change as its mass decreases from $M_i$ to $M_f$ is

(A) $u\ln(M_i/M_f)$

(B) $u\ln(M_f/M_i)$

(C) $u(M_i-M_f)$

(D) $u(M_f/M_i)$

FRQ

1. A cart of initial mass $M$ moves on a frictionless horizontal track with speed $v_0$. Sand falls vertically into the cart at constant rate $\lambda$.

   $(A)$ Derive the cart’s speed as a function of time.

   $(B)$ Determine the horizontal force the cart exerts on newly collected sand.

   $(C)$ Determine the rate at which mechanical energy is lost.

   $(D)$ Explain why horizontal momentum is conserved even though kinetic energy is not.

2. A block of mass $m$ moving with speed $v_0$ collides with and sticks to a block of mass $2m$ attached to a spring of constant $k$ on a frictionless track.

   $(A)$ Find the speed of the combined blocks just after the collision.

   $(B)$ Determine the maximum compression of the spring.

   $(C)$ Find the fraction of the initial kinetic energy lost in the collision.

   $(D)$ Describe how the answer changes if the collision is elastic instead.

3. A projectile of mass $3m$ moving horizontally at speed $v_0$ explodes into three fragments of equal mass. One fragment moves straight upward at speed $v_0$, and a second moves at angle $30^\circ$ below the original direction with speed $2v_0$.

   $(A)$ Determine the velocity components of the third fragment.

   $(B)$ Determine the speed of the third fragment.

   $(C)$ Compare the total kinetic energy before and after the explosion.

   $(D)$ Explain what supplied the change in kinetic energy.