Momentum and Impulse

Linear Momentum (CED Unit 5)

The linear momentum of an object is:

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

Momentum is a vector quantity with SI units of $\mathrm{kg \cdot \mathrm{m/s}}$.

Newton’s Second Law in Terms of Momentum

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

When mass is constant: $\vec{F}_{\mathrm{net}} = m\vec{a}$.

Why Momentum Is More Fundamental Than Force

$F = dp/dt$ is the general form of Newton’s second law. $F = ma$ is the special case when mass is Constant. For a rocket expelling fuel, a raindrop accumulating mass, or a relativistic particle, $F = ma$ fails but $F = dp/dt$ remains valid. Momentum is conserved; force is not. The momentum form of Newton’s second law connects directly to the most powerful conservation law in classical Mechanics.

Impulse (CED Unit 5)

Impulse is the change in momentum:

$$\vec\{J\} = \vec\{F\}_\{\mathrm\{net\} \Delta t = \Delta\vec\{p\} = m\vec\{v\}_f - m\vec\{v\}_i$$

For a variable force:

$$\vec\{J\} = \int_\{t_i\}^\{t_f\} \vec\{F\}(t)\, dt$$

Impulse-Momentum Theorem

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

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

Geometric Interpretation

On a force-vs-time graph, the impulse equals the area under the curve. For a constant force, this is A rectangle. For a variable force, the area can be computed by integration or approximated by Counting squares.

Average Force

The average force during an interaction of duration $\Delta t$ is:

$$F_\{\mathrm\{avg\} = \frac\{J\}\{\Delta t\} = \frac\{\Delta p\}\{\Delta t\}$$

This is the constant force that would produce the same impulse over the same time interval.

Center of Mass (CED Unit 5)

The center of mass of a system of particles is:

$$\vec\{r\}_\{\mathrm\{cm\} = \frac\{\sum m_i \vec\{r\}_i\}\{\sum m_i\} = \frac\{\sum m_i \vec\{r\}_i\}\{M\}$$

For a continuous object:

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

Motion of the Center of Mass

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

The center of mass of a system moves as if all external forces were applied to a single particle of Mass $M$ at the center of mass.

Center of Mass and Collisions

In any collision, the velocity of the center of mass does not change (since there are no external Forces during the brief collision):

$$\vec\{v\}_\{\mathrm\{cm\} = \frac\{m_1 \vec\{v\}_1 + m_2 \vec\{v\}_2\}\{m_1 + m_2\} = \mathrm\{constant$$

Center of Mass Frame

Transforming to the center of mass frame simplifies collision analysis. In this frame, the total Momentum is zero. Before the collision, the particles approach each other; after the collision, they Move apart. For an elastic collision, the speeds are unchanged but the directions may be different.

Rocket Propulsion (AP Physics C)

A rocket expels mass at a rate $\frac{dm}{dt} = -\alpha$ (where $\alpha \gt 0$) with exhaust Velocity $v_e$ relative to the rocket. By conservation of momentum:

$$M\frac\{dv\}\{dt\} = -v_e \frac\{dm\}\{dt\} = \alpha v_e$$

This is the Tsiolkovsky rocket equation. Integrating gives:

$$\Delta v = v_e \ln\frac\{m_0\}\{m_f\}$$

Where $m_0$ is the initial mass and $m_f$ is the final mass.

Why Rockets Work in a Vacuum

Rockets do not “push against” the air or the ground. They work by conservation of momentum: the Rocket expels mass backward, and the rocket moves forward. The expelled gas carries momentum in one Direction, and the rocket acquires equal and opposite momentum. This works equally well in a vacuum.

Intuition

Momentum is the conserved quantity of motion, and it’s conserved more often than energy. When two objects collide, internal forces are enormous but brief, so momentum is approximately conserved even when kinetic energy is not (as in inelastic collisions).

Impulse intuition: Impulse is the area under the force-time graph. A small force applied for a long time can produce the same velocity change as a large force applied briefly. This is why airbags work: they increase the collision time, reducing the peak force on your body.

Why momentum is more fundamental than force: $F = ma$ is a special case of $F = dp/dt$. For rockets (changing mass), raindrops (accumulating mass), or relativistic particles, only the momentum form works. Momentum is conserved; force is not.

Elastic vs inelastic collisions: In elastic collisions, both momentum and kinetic energy are conserved, think billiard balls. In inelastic collisions, momentum is conserved but KE is converted to heat, sound, or deformation. In perfectly inelastic collisions, the objects stick together and KE loss is maximized.

Center of mass: The center of mass of a system moves as if all external forces were applied to a single particle. Internal forces don’t affect the CM motion, this is why a fireworks shell explodes symmetrically in mid-air.

Common Pitfalls

1. Confusing momentum and kinetic energy. Both depend on mass and velocity, but momentum is a vector ($m\vec{v}$) while KE is a scalar ($\frac{1}{2}mv^2$).
2. Forgetting that momentum is a vector. In 2D collisions, you must conserve momentum in both the $x$ and $y$ directions separately.
3. Assuming all collisions are elastic. Most real collisions are inelastic. Only use the elastic collision formulas when KE is explicitly conserved.
4. Sign errors with velocity directions. Always define a positive direction and be consistent.
5. Using impulse incorrectly with variable forces. For non-constant forces, integrate or find the area under the $F$-vs-$t$ graph.
6. Forgetting that momentum is conserved only when the net external force is zero. During a collision, external forces (like gravity) are negligible compared to the large internal forces, so momentum is approximately conserved.
7. Computing the center of mass incorrectly for non-uniform objects. Use the integral form.

Practice Questions

1. A $0.50 \mathrm{ kg}$ ball hits a wall at $10 \mathrm{ m/s}$ and rebounds at $8.0 \mathrm{ m/s}$. If the contact time is $0.020 \mathrm{ s}$ Find the average force on the ball.

2. A $10 \mathrm{ g}$ bullet embeds in a $2.0 \mathrm{ kg}$ wooden block on a frictionless surface. The block then slides into a spring ($k = 200 \mathrm{ N/m}$) and compresses it $5.0 \mathrm{ cm}$. Find the initial speed of the bullet.

3. Two ice skaters push off each other. Skater A ($60 \mathrm{ kg}$) moves at $2.0 \mathrm{ m/s}$ and skater B ($40 \mathrm{ kg}$) moves at $3.0 \mathrm{ m/s}$. What was the initial momentum of the system?

4. A $3.0 \mathrm{ kg}$ object moving at $4.0 \mathrm{ m/s}$ collides elastically with a $1.0 \mathrm{ kg}$ object moving at $-6.0 \mathrm{ m/s}$. Find the final velocities.

5. Find the center of mass of a uniform rod of length $L$ and mass $M$.

6. A $2000 \mathrm{ kg}$ cannon fires a $50 \mathrm{ kg}$ shell horizontally at $400 \mathrm{ m/s}$. What is the recoil velocity of the cannon?

7. Prove that in a one-dimensional elastic collision with a stationary target, the relative velocity of approach equals the relative velocity of separation: $v_{1i} - v_{2i} = -(v_{1f} - v_{2f})$.

8. Two identical balls collide. One was moving at $5.0 \mathrm{ m/s}$ and the other was stationary. After the collision, one ball moves at $30^\circ$ above the original direction. Find the speeds and direction of both balls after the collision (assume elastic).

9. A force $F(t) = 6t^2 - 2t$ (in N, with $t$ in s) acts on a $3.0 \mathrm{ kg}$ object initially at rest. Find the impulse from $t = 0$ to $t = 4.0 \mathrm{ s}$ and the velocity at $t = 4.0 \mathrm{ s}$.

10. A flat disk of mass $M$ and radius $R$ has a small hole of mass $m$ removed at a distance $r$ from the center. Find the center of mass of the remaining object.

11. Impulse: Extended Analysis and Examples

Impulse as the Area Under an F-t Graph

When the force is given graphically, the impulse equals the area under the $F$-vs-$t$ curve. For a Linearly decreasing force (a triangle), the area is $\frac{1}{2} \times \mathrm{base \times \mathrm{height}}$.

Worked Example: Triangular Force Profile

A ball hits a wall with a force that varies linearly from 0 to $F_{\max} = 800 \mathrm{ N}$ over $\Delta t = 4 \mathrm{ ms}$ Then linearly back to 0 over another $4 \mathrm{ ms}$. Find the total Impulse and the average force.

The $F$-vs-$t$ graph is a triangle with base $8 \mathrm{ ms}$ and height $800 \mathrm{ N}$.

$$J = \frac{1}{2} \times 0.008 \times 800 = 3.2 \mathrm{ N s}$$ $$F_{\mathrm{avg}} = \frac{J}{\Delta t} = \frac{3.2}{0.008} = 400 \mathrm{ N}$$

The average force is half the peak force, which is always the case for a triangular force profile.

Worked Example: Variable Force from Calculus

A force $F(t) = 6t^2 - 2t$ (in N) acts on a $3 \mathrm{ kg}$ object initially at rest. Find the Impulse and velocity at $t = 4 \mathrm{ s}$.

$$J = \int_0^4 (6t^2 - 2t)\, dt = \left[2t^3 - t^2\right]_0^4 = 128 - 16 = 112 \mathrm{ N s}$$ $$v = \frac{J}{m} = \frac{112}{3} = 37.3 \mathrm{ m/s}$$

12. Collisions: Extended Analysis

Worked Example: Two-Dimensional Collision

A $2 \mathrm{ kg}$ ball moving at $5 \mathrm{ m/s}$ strikes a stationary $3 \mathrm{ kg}$ ball. After the collision, the $2 \mathrm{ kg}$ ball moves at $30^{\circ}$ above the original direction at $3 \mathrm{ m/s}$. Find the velocity of the $3 \mathrm{ kg}$ ball.

Conservation of $x$-momentum:

$$2(5) + 0 = 2(3\cos 30^{\circ}) + 3v_{2x}$$ $$10 = 5.196 + 3v_{2x}$$ $$v_{2x} = \frac{4.804}{3} = 1.601 \mathrm{ m/s}$$

Conservation of $y$-momentum:

$$0 = 2(3\sin 30^{\circ}) + 3v_{2y}$$ $$0 = 3 + 3v_{2y}$$ $$v_{2y} = -1.0 \mathrm{ m/s}$$

Speed of $3 \mathrm{ kg}$ ball:

$$v_2 = \sqrt{1.601^2 + (-1.0)^2} = \sqrt{2.563 + 1} = \sqrt{3.563} = 1.89 \mathrm{ m/s}$$

Direction: $\theta = \arctan\left(\frac{-1.0}{1.601}\right) = -32.0^{\circ}$ (below the original Direction).

Worked Example: Ballistic Pendulum

A $10 \mathrm{ g}$ bullet travelling at $400 \mathrm{ m/s}$ embeds itself in a $2 \mathrm{ kg}$ Wooden block hanging from a string. Find the maximum height the block reaches.

Phase 1: Collision (conservation of momentum, perfectly inelastic).

$$mv = (m + M)V$$ $$0.01 \times 400 = (0.01 + 2)V$$ $$V = \frac{4}{2.01} = 1.99 \mathrm{ m/s}$$

Phase 2: Swing (conservation of energy).

$$\frac{1}{2}(m + M)V^2 = (m + M)gh$$ $$h = \frac{V^2}{2g} = \frac{1.99^2}{2 \times 9.8} = \frac{3.96}{19.6} = 0.202 \mathrm{ m}$$

The key insight is that momentum is conserved during the collision (very short time, large internal Forces), but mechanical energy is not (the collision is inelastic). After the collision, energy is Conserved during the upward swing (only gravity does work).

13. Center of Mass: Extended Examples

Worked Example: CM of a Non-Uniform Rod

A thin rod of length $L$ and mass $M$ has linear mass density $\lambda(x) = \lambda_0 x/L$ Where $x$ is the distance from one end. Find the center of mass.

$$\bar{x} = \frac{\int_0^L x\lambda(x)\, dx}{\int_0^L \lambda(x)\, dx} = \frac{\int_0^L x \cdot \frac{\lambda_0 x}{L}\, dx}{\int_0^L \frac{\lambda_0 x}{L}\, dx}$$ $$= \frac{\frac{\lambda_0}{L}\int_0^L x^2\, dx}{\frac{\lambda_0}{L}\int_0^L x\, dx} = \frac{\frac{L^3}{3}}{\frac{L^2}{2}} = \frac{2L}{3}$$

The center of mass is at $\frac{2}{3}L$ from the lighter end (not at the midpoint), because more Mass is concentrated toward the heavier end.

Worked Example: CM of an L-Shaped Object

Two uniform rods, each of mass $M$ and length $L$ Are joined at right angles to form an L shape. Find the center of mass.

Take the origin at the corner. Rod 1 lies along the positive $x$-axis with CM at $(L/2, 0)$. Rod 2 Lies along the positive $y$-axis with CM at $(0, L/2)$.

$$\bar{x} = \frac{M \cdot L/2 + M \cdot 0}{2M} = \frac{L}{4}$$ $$\bar{y} = \frac{M \cdot 0 + M \cdot L/2}{2M} = \frac{L}{4}$$

The center of mass is at $(L/4, L/4)$ from the corner.

14. Rocket Propulsion: Extended Analysis (AP Physics C)

Worked Example: Rocket Velocity

A rocket has initial mass $m_0 = 50000 \mathrm{ kg}$ and final mass $m_f = 10000 \mathrm{ kg}$. The Exhaust velocity is $v_e = 3000 \mathrm{ m/s}$. Find the velocity gained.

$$\Delta v = v_e \ln\frac{m_0}{m_f} = 3000 \times \ln\left(\frac{50000}{10000}\right) = 3000 \times \ln 5 = 3000 \times 1.609 = 4828 \mathrm{ m/s}$$

Why Multi-Stage Rockets Are More Efficient

For a single-stage rocket, the mass ratio $m_0/m_f$ is limited by the structural mass of the tank and engine. A multi-stage rocket discards the empty first-stage tank and engine, reducing the mass that must be accelerated in subsequent stages. This dramatically improves the mass ratio for each Stage.

For a two-stage rocket with equal exhaust velocities and equal structural fractions, the total Velocity gain is:

$$\Delta v_{\mathrm{total}} = v_e \ln\frac{m_0}{m_1} + v_e \ln\frac{m_1}{m_f} = v_e \ln\frac{m_0}{m_f}$$

This is the same as a single-stage rocket with the same mass ratio, but the structural mass is much less because each stage only needs to be strong enough for its own fuel load, not the total fuel Load.

15. Summary Table: Collision Types

16. Practice Questions (Additional)

1. A $0.05 \mathrm{ kg}$ tennis ball hits a racket at $20 \mathrm{ m/s}$ and rebounds at $25 \mathrm{ m/s}$ at an angle of $30^{\circ}$ from the original direction. If the contact time is $5 \mathrm{ ms}$ Find the average force exerted by the racket.

2. A $5 \mathrm{ kg}$ object moving at $8 \mathrm{ m/s}$ collides with a $3 \mathrm{ kg}$ object moving at $4 \mathrm{ m/s}$ in the same direction. After the collision, the $5 \mathrm{ kg}$ object moves at $5 \mathrm{ m/s}$ in the same direction. Find the velocity of the $3 \mathrm{ kg}$ object and the coefficient of restitution.

3. A force $F(t) = 12t - 3t^2$ (in N) acts on a $4 \mathrm{ kg}$ object initially moving at $2 \mathrm{ m/s}$. Find (a) the time when the object is at rest, (b) the impulse from $t = 0$ to that time, and (c) the impulse from $t = 0$ to $t = 6 \mathrm{ s}$.

4. Two ice skaters of masses $50 \mathrm{ kg}$ and $70 \mathrm{ kg}$ stand facing each other on frictionless ice. They push off each other, and the $50 \mathrm{ kg}$ skater moves at $3 \mathrm{ m/s}$. Find the velocity of the $70 \mathrm{ kg}$ skater and the total kinetic energy of the system.

5. A rocket burns fuel at a rate of $100 \mathrm{ kg/s}$ with an exhaust velocity of $2500 \mathrm{ m/s}$. The initial mass is $20000 \mathrm{ kg}$. Find (a) the thrust and (b) the velocity after $60 \mathrm{ s}$ of burning, assuming the rocket starts from rest in deep space (no gravity).

Extended Worked Examples

Example 16: Glancing Collision with Energy Analysis

A $4 \mathrm{ kg}$ object moving at $6 \mathrm{ m/s}$ collides with a stationary $6 \mathrm{ kg}$ Object on a frictionless surface. After the collision, the $6 \mathrm{ kg}$ object moves at $3 \mathrm{ m/s}$ at $45^\circ$ to the original direction. Determine the final velocity of the $4 \mathrm{ kg}$ object and classify the collision.

Step 1: Conservation of momentum (x-direction)

$$m_1 u_1 = m_1 v_{1x} + m_2 v_{2x}$$ $$4 \times 6 = 4v_{1x} + 6 \times 3\cos 45^\circ$$ $$24 = 4v_{1x} + 12.728$$ $$v_{1x} = \frac{24 - 12.728}{4} = 2.818 \mathrm{ m/s}$$

Step 2: Conservation of momentum (y-direction)

$$0 = m_1 v_{1y} + m_2 v_{2y}$$ $$0 = 4v_{1y} + 6 \times 3\sin 45^\circ$$ $$v_{1y} = \frac{-12.728}{4} = -3.182 \mathrm{ m/s}$$

Step 3: Final velocity of the $4 \mathrm{ kg}$ object

$$|v_1| = \sqrt{2.818^2 + (-3.182)^2} = \sqrt{7.941 + 10.125} = \sqrt{18.066} = 4.251 \mathrm{ m/s}$$ $$\theta = \arctan\left(\frac{-3.182}{2.818}\right) = -48.5^\circ$$

Step 4: Classify the collision

$$KE_i = \frac{1}{2}(4)(6^2) = 72 \mathrm{ J}$$ $$KE_f = \frac{1}{2}(4)(4.251)^2 + \frac{1}{2}(6)(3^2) = \frac{1}{2}(4)(18.07) + \frac{1}{2}(6)(9) = 36.14 + 27 = 63.14 \mathrm{ J}$$

$KE_f \lt KE_i$, so this is an inelastic collision (but not perfectly inelastic since the objects did not stick together). Energy lost $= 72 - 63.14 = 8.86 \mathrm{ J}$.

Example 17: Impulse and Force-Time Graph Analysis

A $2 \mathrm{ kg}$ object initially at rest is subjected to a force described by:

$$F(t) = \begin{cases} 20t & 0 \le t \le 2 \\ 40 - 10t & 2 \lt t \le 4 \\ 0 & t \gt 4 \end{cases}$$

Where $F$ is in newtons and $t$ in seconds. Find the velocity at $t = 4 \mathrm{ s}$ and the average Force.

Step 1: Calculate total impulse

$$J = \int_0^2 20t \, dt + \int_2^4 (40 - 10t) \, dt$$ $$J = \left[ 10t^2 \right]_0^2 + \left[ 40t - 5t^2 \right]_2^4$$ $$J = (40 - 0) + (160 - 80 - 80 + 20) = 40 + 20 = 60 \mathrm{ N\cdot\mathrm{s}}$$

Step 2: Final velocity

$$J = mv - mu = 2v - 0 \implies v = 30 \mathrm{ m/s}$$

Step 3: Average force

$$F_{\mathrm{avg}} = \frac{J}{\Delta t} = \frac{60}{4} = 15 \mathrm{ N}$$

Example 18: Perfectly Inelastic Collision with Rotation

A $0.1 \mathrm{ kg}$ bullet travelling at $400 \mathrm{ m/s}$ embeds itself in a wooden block of Mass $1.9 \mathrm{ kg}$ suspended by a $2 \mathrm{ m}$ string. Find the maximum height the block Rises.

Step 1: Momentum conservation (perfectly inelastic collision)

$$mv = (m + M)V$$ $$0.1 \times 400 = (0.1 + 1.9)V$$ $$40 = 2V \implies V = 20 \mathrm{ m/s}$$

Step 2: Energy conservation (pendulum swing)

$$\frac{1}{2}(m + M)V^2 = (m + M)gh$$ $$h = \frac{V^2}{2g} = \frac{400}{2 \times 9.8} = 20.4 \mathrm{ m}$$

Step 3: Check feasibility

The string length is only $2 \mathrm{ m}$ So the block cannot rise $20.4 \mathrm{ m}$. This means The block swings past the horizontal. Let us check the velocity at the top of the swing:

At the top of a vertical circle, the block must have enough speed to maintain tension:

$$V_{\mathrm{top}^2} = V^2 - 4gL = 400 - 4 \times 9.8 \times 2 = 400 - 78.4 = 321.6$$ $$V_{\mathrm{top}} = 17.9 \mathrm{ m/s}$$

Since $V_{\mathrm{top} \gt 0}$ The block completes full circles. The string remains taut at the top if:

T + (m+M)g = \frac{(m+M)V_{\mathrm{top}^2}{L} $$T = 2 \times \frac{321.6}{2} - 2 \times 9.8 = 321.6 - 19.6 = 302 \mathrm{ N}$$

The tension is positive, so the block does indeed complete full vertical circles.

Cross-References

• Kinematics: Momentum conservation applies during collisions, kinematics describes the motion before and after.
• Work, Energy, and Power: Energy is conserved in elastic collisions; momentum is always conserved.
• Newton’s Laws: Impulse is the integral of force over time, Newton’s second law in momentum form.
• AP Calculus, Integrals: Impulse is the integral of force over time, connecting the impulse-momentum theorem to integral calculus.