GRE Physics Companion: Collisions in One Dimension
For every isolated one-dimensional collision,
The second equation depends on collision type:
| elastic: | Ki = Kf, | (2)
|
| perfectly inelastic: | v1 = v2, | (3)
|
| general restitution: | v2 − v1 = e(u1 − u2). | (4) |
Figure 1. A compact strategy for one-dimensional collision problems. Momentum conservation is
the first equation; collision type supplies the second.
1 High-value GRE facts
- Momentum is conserved when external impulse is negligible.
- kinetic energy is conserved only in elastic collisions.
- Perfectly inelastic means the objects stick together.
- Equal masses exchange velocities in a one-dimensional elastic collision.
- For an elastic collision, separation speed equals approach speed.
- For a target initially at rest, use the standard elastic formulas only if the collision is
elastic.
- e = 1 is elastic and e = 0 is perfectly inelastic.
- In the center of mass frame, elastic collision velocities reverse.
- Equal-and-opposite collision impulses do not imply equal velocity changes for unequal
masses.
- Always keep the velocity signs.
Part I: Original GRE-style problems
Problem 1: perfectly inelastic velocity
A mass m moving at speed v sticks to an identical stationary mass. Their final speed
is
- 0
- v∕4
- v∕2
- v
- 2v
Problem 2: perfectly inelastic kinetic energy
For the collision in Problem 1, the final translational kinetic energy is what fraction of the initial
kinetic energy?
- 1∕4
- 1∕2
- 3∕4
- 1
- 2
Problem 3: equal-mass elastic collision
A mass m moving at +v collides elastically with an identical stationary mass. The first mass leaves
with velocity
- −v
- −v∕2
- 0
- v∕2
- v
Problem 4: elastic relative speed
For a one-dimensional elastic collision,
- v1 + v2 = u1 + u2 always
- v2 − v1 = u1 − u2
- v1 − v2 = u1 − u2
- v1 + v2 = 0
- both final velocities must be positive
Problem 5: restitution
A collision has coefficient of restitution e = 0.40 and approach relative speed 10 m∕s. The
separation relative speed is
- 2 m∕s
- 4 m∕s
- 6 m∕s
- 10 m∕s
- 25 m∕s
Problem 6: heavy wall limit
A Light ball collides elastically with a stationary wall of effectively infinite mass. The ball’s final
velocity is approximately
- 0
- +u∕2
- +u
- −u∕2
- −u
Problem 7: heavy projectile limit
A very heavy object moving at speed u strikes a very light stationary object elastically. The light
object’s final speed approaches
- 0
- u∕2
- u
- 2u
- 4u
Problem 8: kinetic energy and restitution
For fixed masses and initial relative speed, the fraction of relative kinetic energy remaining after
impact is
- e
- e2
- 1 − e
- 1 − e2
- 2e
Problem 9: collision impulse
During a collision, object 1 receives impulse −7 N s. External impulse is negligible. Object 2
receives
- −14 N s
- −7 N s
- 0
- +7 N s
- +14 N s
Problem 10: perfectly inelastic center of mass
Two objects stick together in an isolated collision. Their common final velocity equals
- the faster initial velocity
- the slower initial velocity
- the initial center of mass velocity
- zero in every case
- the arithmetic average of the initial speeds
Problem 11: target at rest, elastic
A mass m moving at speed u elastically strikes a stationary mass 3m. The first mass’s final
velocity is
- −u
- −u∕2
- 0
- u∕2
- u
Problem 12: identifying collision type
An isolated collision conserves momentum and leaves the two bodies moving together afterward.
The collision is
- necessarily elastic
- perfectly inelastic
- impossible
- superelastic
- a non-collision
Part II: Complete worked solutions
Solution 1
Momentum conservation gives
Thus
Answer: (C).
Solution 2
Initially,
Finally,
| Kf | = (2m) 2 | (8)
|
| = mv2. | (9) |
Therefore
Answer: (B).
Solution 3
Equal masses exchange velocities in a one-dimensional elastic collision.
The moving mass stops:
Answer: (C).
Solution 4
For elastic one-dimensional impact,
Answer: (B).
Solution 5
By definition,
Thus
Answer: (B).
Solution 6
For m2 ≫ m1 and u2 = 0,
Answer: (E).
Solution 7
For m1 ≫ m2,
Answer: (D).
Solution 8
Because relative speed is multiplied by e and kinetic energy depends on speed squared,
Answer: (B).
Solution 9
Internal collision impulses are equal and opposite:
Thus
Answer: (D).
Solution 10
When the objects stick, their common final velocity is
Answer: (C).
Solution 11
For an elastic collision with the target at rest,
With m1 = m and m2 = 3m,
Answer: (B).
Solution 12
If two bodies stick together after impact, the collision is perfectly inelastic.
Answer: (B).
2 GRE checklist
- Choose a positive direction and keep every velocity sign.
- Write momentum conservation first.
- Identify the collision type before choosing the second equation.
- Elastic means conserve kinetic energy.
- Perfectly inelastic means set v1 = v2.
- For restitution problems, use v2 − v1 = e(u1 − u2).
- Check equal-mass and heavy-mass limiting cases.
- Do not assume equal-and-opposite impulses produce equal speed changes.
References
References
[1] J. R. Taylor, Classical Mechanics, University Science Books, 2005.
[2] D. Kleppner and R. Kolenkow, An Introduction to Mechanics, 2nd ed., Cambridge
University Press, 2014.
[3] OpenStax, University Physics, Volume 1, Rice University, 2016.