Allow a jump in velocity while keeping position continuous.
Syllabus 4.2
Notes and key methods
An idealised short collision may be represented by an instantaneous jump in velocity. The position remains continuous. The ordinary acceleration at the jump is not defined by a finite slope, although velocities on either side can be specified.
For displacement, integrate the regular portions of the velocity graph separately. A vertical segment at one time has zero width and adds no displacement. On each side of an idealised event, use that stage’s own initial velocity and duration.
Check the conditions
Choose and retain a positive direction; distinguish signed quantities from magnitudes.
Use SI units and \(\displaystyle g= \frac{10m}{s^{2}}\) wherever gravity is relevant.
Apply only the model and time interval stated in the question.
Three worked examples
Example 1 · Idealised instantaneous velocity changes
A particle moves right at \(\displaystyle \frac{6m}{s}\) for 3 s. An idealised instantaneous collision reverses its velocity to \(\displaystyle - \frac{5m}{s}\), after which it moves uniformly for 6 s. The collision has zero duration in this model and position remains continuous. Positive is right.
Find the signed velocity change at the collision, in m/s.
Find the total displacement over both uniform stages, in m.
Find the total distance travelled over both uniform stages, in m.
Find the average velocity over the full modelled interval, in m/s.
Show the reasoning
(a)
Use final minus initial signed velocity, not a difference of speed magnitudes.
Subtract the pre-collision velocity from the post-collision velocity: \(\displaystyle - \frac{11m}{s}\).
Answer: \(-11\).
(b)
Add both signed stage displacements; the zero-duration event adds none.
The signed contributions 18 and -30 sum to -12 m.
Answer: \(-12\).
(c)
Add speed times duration for each regular stage.
The two positive path lengths sum to 48 m.
Answer: \(48\).
(d)
Use elapsed time from both stages and the signed displacement.
Divide the signed displacement by 9 s, giving \(\displaystyle - \frac{4}{3}\) m/s.
Answer: \(\displaystyle - \frac{4}{3}\).
Example 2 · Idealised instantaneous velocity changes
A particle moves right at \(\displaystyle \frac{7m}{s}\) for 4 s. An idealised instantaneous collision reverses its velocity to \(\displaystyle - \frac{6m}{s}\), after which it moves uniformly for 7 s. The collision has zero duration in this model and position remains continuous. Positive is right.
Find the signed velocity change at the collision, in m/s.
Find the total displacement over both uniform stages, in m.
Find the total distance travelled over both uniform stages, in m.
Find the average velocity over the full modelled interval, in m/s.
Show the reasoning
(a)
Use final minus initial signed velocity, not a difference of speed magnitudes.
Subtract the pre-collision velocity from the post-collision velocity: \(\displaystyle - \frac{13m}{s}\).
Answer: \(-13\).
(b)
Add both signed stage displacements; the zero-duration event adds none.
The signed contributions 28 and -42 sum to -14 m.
Answer: \(-14\).
(c)
Add speed times duration for each regular stage.
The two positive path lengths sum to 70 m.
Answer: \(70\).
(d)
Use elapsed time from both stages and the signed displacement.
Divide the signed displacement by 11 s, giving \(\displaystyle - \frac{14}{11}\) m/s.
Answer: \(\displaystyle - \frac{14}{11}\).
Example 3 · Idealised instantaneous velocity changes
A particle moves right at \(\displaystyle \frac{8m}{s}\) for 5 s. An idealised instantaneous collision reverses its velocity to \(\displaystyle - \frac{7m}{s}\), after which it moves uniformly for 3 s. The collision has zero duration in this model and position remains continuous. Positive is right.
Find the signed velocity change at the collision, in m/s.
Find the total displacement over both uniform stages, in m.
Find the total distance travelled over both uniform stages, in m.
Find the average velocity over the full modelled interval, in m/s.
Show the reasoning
(a)
Use final minus initial signed velocity, not a difference of speed magnitudes.
Subtract the pre-collision velocity from the post-collision velocity: \(\displaystyle - \frac{15m}{s}\).
Answer: \(-15\).
(b)
Add both signed stage displacements; the zero-duration event adds none.
The signed contributions 40 and -21 sum to 19 m.
Answer: \(19\).
(c)
Add speed times duration for each regular stage.
The two positive path lengths sum to 61 m.
Answer: \(61\).
(d)
Use elapsed time from both stages and the signed displacement.
Divide the signed displacement by 8 s, giving \(\displaystyle \frac{19}{8}\) m/s.
Answer: \(\displaystyle \frac{19}{8}\).
Common mistakes
A jump in velocity is not a jump in position.
Do not assign the vertical graph segment an extra distance or a finite constant acceleration.