Work, Energy and Power · Lesson 15 of 16
Oblique Collisions and Multi-Stage Energy Problems
“Resolve two-dimensional collisions and combine collision laws with earlier energy methods.”
• Resolve momentum into components. • Derive the equal-mass right-angle result. • Separate swings, impacts and resource calculations into stages.
Momentum must balance in each direction
Take the incident velocity along +x and the target initially at rest. After an oblique collision, suppose the projectile leaves above the x-axis at angle θ1 and the target below it at angle θ2. Balance x and y components separately. The y components must cancel because initial y momentum is zero. Changing the directions in a diagram changes the component signs, so draw the outgoing arrows before writing equations.
There are four unknowns: two speeds and two angles. Momentum supplies two equations. If the collision is elastic, kinetic-energy conservation supplies a third. One more piece of information, such as a measured outgoing angle or the collision geometry and interaction model, is generally needed. Conservation laws alone do not always determine the entire scattering pattern.
Equal masses: why a right angle appears
Expanding the square gives u² = v1²+v2²+2v1·v2. Elasticity removes the first three terms, leaving v1·v2 = 0. If both outgoing velocities are nonzero, they are perpendicular. This special result requires equal masses, one initially at rest, and elastic translational motion without relevant spin or friction. It does not describe the degenerate head-on case in which one final velocity is zero, since a zero vector has no direction.
Problem
Two equal smooth balls collide elastically, with the target initially stationary. The target leaves at 37° below the incident direction. Find the other angle and speeds in terms of incident speed u.
- 1.The nonzero outgoing directions are perpendicular, so θ1+37°=90° and θ1=53°.
- 2.From the right-angle velocity triangle, v1 = u cos53° = u sin37° ≈ 0.602u and v2 = u cos37° ≈ 0.799u.
- 3.Their squared speed fractions add to one, and their y momentum components cancel.
Separate a pendulum swing from the impact
Problem
A bob is released from 30° to vertical and strikes an equal mass initially at rest on a table. The impact is horizontal, head-on and elastic. How high does the incident bob rise afterwards?
- 1.Before contact, energy conservation gives u² = 2gL(1−cos30°).
- 2.During the brief impact, conserve horizontal momentum and kinetic energy. Equal masses exchange velocities: the incident bob stops and the target leaves at u.
- 3.The incident bob is then at its lowest point with zero velocity. In the ideal geometry it does not rise after the collision.
- 4.The release angle determines the transferred speed but does not alter the zero rebound height for this special equal-mass head-on elastic case.
Three balls: momentum alone is insufficient
Problem
Three identical balls lie in a line. One arrives at speed V while two touch at rest. Which proposed output is compatible with a lossless head-on transfer?
- 1.Initial momentum is mV and kinetic energy is ½mV².
- 2.Proposal (i), two balls moving at V/2 and the first stopped, has the same momentum but only half the initial kinetic energy.
- 3.Proposal (ii), one ball leaving at V with the other two at rest, has both the original momentum and kinetic energy. It is the compatible elastic-transfer outcome.
- 4.Proposal (iii), all three moving at V/3, has the same momentum but only one third of the initial kinetic energy. It corresponds to a loss of relative-motion energy.
- 5.Momentum conservation alone would accept all three proposals, so the elastic energy condition is essential.
Scattering need not require contact
A comet deflected by the Sun and a charged particle deflected by a nucleus interact at a distance. Such encounters are called scattering. The force law, masses, incoming speeds and geometry determine the outgoing directions. For isolated conservative interactions, momentum of the full system and total energy still supply useful constraints even when no surfaces touch. A fixed massive source approximation can hide the source’s tiny recoil; include it for exact momentum accounting.
Check your understanding
Quiz
Two-dimensional momentum conservation gives
Explanation: Balance x and y separately.
For equal-mass elastic glancing impact with one at rest, nonzero final velocities are
Explanation: The dot product vanishes.
Does elasticity plus planar momentum always fix four final unknowns?
Explanation: Two momentum equations plus one energy equation leave one degree of freedom.
An equal-mass elastic head-on impact at a pendulum’s bottom makes the incident bob
Explanation: It transfers its velocity to the initially stationary equal mass.
Three equal balls all leaving together at V/3 preserve
Explanation: Final K is one third of initial K.
Key Takeaways
• Resolve vector momentum before applying collision energy conditions. • The right-angle result has specific assumptions. • Use separate equations for each physical stage of a problem.