Laws of Motion · Lesson 3 of 14

Impulse and Collisions

“Use momentum changes and force-time areas to describe brief interactions and distinguish collision-angle conventions.”

Learning Objectives

• Relate impulse to change in momentum. • Calculate average force when contact time is known. • Resolve rebound impulses into normal and tangential components. • State the angle between velocity vectors before using a deflection formula.

A brief force can produce a finite change

During a bat-ball collision or a rebound from a wall, force may vary sharply over a short contact time. The detailed force history can be difficult to measure, while initial and final velocities are much easier to determine. Impulse is the accumulated effect of that force. The impulse of the resultant force equals the change in momentum, regardless of whether the force is constant.

Impulse-momentum theoremLaTeX

J_{net} = \int_{t_{i}}^{t_{f}} F_{net} d t = p_{f} - p_{i}

Impulse has units N s = kg m/s and dimensions [M L T⁻¹]. For constant mass, Δp = m(v_f − v_i).

Average forceLaTeX

F_{av} = \frac{Δ p}{Δ t}, J = F_{av} Δ t

This is a time average. Force-time graph area gives impulse component by component.

For the same stopping momentum change, increasing the stopping time reduces the average force magnitude when the force direction is fixed. A catcher moves the hands backward with the ball to lengthen the slowing interval. Soft packaging and cushioning work by increasing the stopping duration and distance. A large impulsive force is still an ordinary force; it does not need a separate law of mechanics.

Impulse as area under a force-time graph— The triangular force pulse has area ½ × 0.020 × 400 = 4 N s. Its time-average force over the pulse is 200 N.

Reversing direction changes the sign

Returning a ball toward the bowler

Problem
A 0.15 kg ball approaches a bat at 12 m/s and is sent straight back at the same speed. Find impulse on the ball.

1.Choose the outgoing direction toward the bowler positive. Initial velocity is −12 m/s and final velocity is +12 m/s.
2.Change in momentum is 0.15[12 − (−12)] = +3.6 kg m/s.
3.Impulse is 3.6 N s toward the bowler. Unchanged speed did not mean unchanged momentum.
4.If contact lasts 0.010 s, the average resultant force is 360 N in that direction. Without a contact time, impulse alone does not determine average force.

Two head-on rebounds

Problem
Two 0.05 kg balls approach in opposite directions at 6 m/s and each reverses its velocity without changing speed. Find their impulses.

1.For a ball initially moving in +x, Δp = 0.05(−6 − 6) = −0.60 kg m/s.
2.The other ball receives +0.60 kg m/s. The impulses are equal and opposite.
3.Each impulse follows from that ball’s own initial and final velocities; do not add both balls’ momentum changes when asked for one ball’s impulse.

A smooth wall changes the normal component

For an ideal smooth stationary wall and a rebound with unchanged speed, the component parallel to the wall is unchanged while the normal component reverses. The impulse on the ball is normal to the wall. The wall’s force need not point opposite the full incoming velocity. A wall with significant tangential friction would require a different model.

Oblique reflection from a smooth wallLaTeX

∣ J ∣ = 2 m u cos α

α is the incoming angle with the wall’s normal; the normal velocity changes from +u cosα to −u cosα. Tangential impulse is zero in this model.

Compare normal and oblique hits

Problem
Identical balls strike a smooth wall at the same speed: one normally and one at 30° to the normal. Compare impulse magnitudes.

1.The normal hit has impulse magnitude 2mu. The oblique hit has 2mu cos30°.
2.The ratio is 1/cos30° = 2/√3 = 1.155, about 1.2.
3.Both impulses on the balls point along the wall normal away from the wall. The impulses exerted by the balls on the wall have the opposite direction.

Be precise about a deflection angle

Equal-speed change through angle δLaTeX

∣Δ p ∣ = m u^{2} + u^{2} - 2 u^{2} cos δ = 2 m u sin \frac{δ}{2}

δ is the angle between the incoming and outgoing velocity vectors drawn with a common tail. The impulse vector is m(v_f − v_i).

A verbal “45° deflection” is ambiguous unless its reference direction is shown. A 45° turn between velocity vectors gives a different impulse from an outgoing path that makes 45° with the direction directly back toward the source. In the latter case, the velocity-vector angle is 135°. Draw the two velocity arrows before choosing the angle.

A 54 km/h ball and two interpretations

Problem
A ball of mass 0.15 kg keeps speed 54 km/h = 15 m/s after a hit. Compare a 45° velocity turn with a 135° velocity turn.

1.Here 2mu = 2 × 0.15 × 15 = 4.5 N s.
2.For δ = 45°, impulse magnitude is 4.5 sin22.5° = 1.722 N s.
3.For δ = 135°, it is 4.5 sin67.5° = 4.5 cos22.5° = 4.157 N s, about 4.2 N s. This is the interpretation used by the source answer appendix.
4.The impulse points along the vector difference, not generally along the bisector of the two same-tail velocity arrows. It can be viewed as the bisector of v_f and −v_i when their magnitudes are equal.

Net impulse and a single contact impulse

The theorem gives net impulse. If a bat or wall force is extremely large for a very short time, gravity’s impulse mgΔt may be negligible by comparison, so contact impulse is approximately the momentum change. If the duration is not brief or another force is significant, include its impulse. An ideal instantaneous jump is a useful approximation; a real interaction takes a finite time.

Quick check

A ball reverses from +5 m/s to −5 m/s with unchanged mass. What is its velocity change?

Key Takeaways

• Impulse equals momentum change and the area under a force-time graph. • Use signed velocities or full vectors for reversals. • Longer stopping time can reduce average force for the same momentum change. • A smooth wall changes only the normal velocity component. • Define the angle between velocity vectors before using a deflection formula.

Previous · Lesson 2

Momentum and Newton’s Second Law

Next · Lesson 4

Newton’s Third Law and System Boundaries