Change in Momentum and Impulse

Learning Goals

After this lesson, you should be able to:

• calculate vector momentum change from initial and final velocities;
• apply the impulse–momentum theorem;
• find average net force or interaction time;
• determine impulse from signed force–time graph area;
• determine net force from momentum–time graph slope;
• analyze direction changes in one and two dimensions;
• explain how increasing collision time changes average force; and
• design an experiment comparing measured impulse with momentum change.

1. Momentum Change Is a Vector Difference

Always subtract the initial momentum vector from the final momentum vector:

For constant mass,

The magnitude of \(\Delta\vec p\) is not generally \(p_f-p_i\). When direction changes, components must be subtracted before calculating the magnitude.

2. Newton's Second Law in Momentum Form

The average net external force on an object or system equals its momentum change per time:

The force and momentum-change vectors point in the same direction. A large momentum change can result from a large force, a long interaction, or both.

3. Impulse–Momentum Theorem

The impulse delivered by the net external force during an interval is

Equal units express the theorem's physical meaning: impulse is the transfer that changes momentum. It is a vector, not merely the product of two positive magnitudes.

5. Worked Example: Reversing Direction

Question: A \(0.150\,\mathrm{kg}\) ball initially moves right at \(20.0\,\mathrm{m/s}\) and rebounds left at \(30.0\,\mathrm{m/s}\). Take right as positive. Find the impulse on the ball.

Answer: The impulse is \(7.50\,\mathrm{N\,s}\) to the left. Adding speed magnitudes without first assigning velocity signs often produces the wrong direction.

6. Average Force and Interaction Time

Rearrange the theorem to match the requested quantity:

The second expression must use components along the same axis. Dividing a vector magnitude by an unrelated signed force component is not valid.

7. Worked Example: Find Average Force

Question: A \(0.20\,\mathrm{kg}\) ball changes velocity from \(-12\,\mathrm{m/s}\) to \(+8.0\,\mathrm{m/s}\) in \(0.040\,\mathrm s\). Find the average net force.

Answer: The average net force is \(100\,\mathrm N\) in the positive direction. It is the net force, not automatically the force from one named agent.

8. Variable Force and Signed Graph Area

When force varies, divide the graph into familiar shapes and add their signed areas:

Area above the time axis contributes positive impulse; area below contributes negative impulse. A rectangle of height \(F_{\mathrm{avg}}\) with the same width and signed area represents an equivalent average force.

9. Worked Example: Triangular Force Pulse

Question: A force rises linearly from zero to \(600\,\mathrm N\) and returns linearly to zero over a total interval of \(8.0\,\mathrm{ms}\). Find the impulse.

Answer: The positive impulse is \(2.4\,\mathrm{N\,s}\), so the object's momentum changes by \(+2.4\,\mathrm{kg\,m/s}\). The peak force should not be multiplied by the entire interval.

10. Reading Momentum–Time Graphs

A negative momentum value and a negative slope mean different things: the value gives motion direction, while the slope gives net-force direction.

11. Two-Dimensional Impulse

Apply the theorem independently to perpendicular components:

The impulse vector can point in neither the initial nor final velocity direction because it represents the vector needed to turn \(\vec p_i\) into \(\vec p_f\).

12. Worked Example: Turning in Two Dimensions

Question: A \(0.25\,\mathrm{kg}\) puck changes velocity from \((4.0\hat i)\,\mathrm{m/s}\) to \((6.0\hat j)\,\mathrm{m/s}\). Find the impulse.

Answer: The impulse is \((-1.0\hat i+1.5\hat j)\,\mathrm{N\,s}\), directed \(56.3^\circ\) above the negative \(x\)-axis.

13. Same Impulse, Different Average Force

For a fixed momentum change, increasing the stopping time reduces the required average-force magnitude:

Example: A momentum change of \(1500\,\mathrm{kg\,m/s}\) occurring in \(0.030\,\mathrm s\) requires \(50\,\mathrm{kN}\) on average. Extending it to \(0.150\,\mathrm s\) reduces the average to \(10\,\mathrm{kN}\).

Airbags, padding, helmets, and “giving” while catching increase interaction time. They do not eliminate the impulse required for the same initial and final momentum.

14. Investigation: Compare Force Area with \(\Delta p\)

Use a low-friction cart, force sensor, motion sensor or photogates, balance, and elastic or magnetic bumper.

1. Measure the cart's total mass and define a positive track direction.
2. Record force and velocity throughout one collision.
3. Calculate \(p_i=mv_i\), \(p_f=mv_f\), and \(\Delta p=p_f-p_i\).
4. Estimate impulse from signed trapezoidal areas under the force–time data.
5. Compare \(J\) with \(\Delta p\) using measurement uncertainties.
6. Repeat with softer padding and compare peak force, duration, area, and momentum change.
7. Discuss sensor zero, sampling rate, force calibration, and timing synchronization.

15. Common Traps and Key Takeaways

• Calculate \(\vec p_f-\vec p_i\), not a sum of speed magnitudes.
• Impulse is a vector in the same direction as momentum change and average net force.
• Use net external force when analyzing the momentum of a system.
• For variable force, use signed graph area, not peak force times total time.
• The slope of \(p(t)\) is force; the area under \(F(t)\) is impulse.
• Longer stopping time lowers average force for the same impulse but does not reduce the required momentum change.
• In two dimensions, solve components before magnitude and direction.