MYP · 04 · extended response
A collision with an external impulse
Momentum · Original GioPhysics question with a detailed, mark-by-mark answer guide.
- Demand
- discriminating
- Marks
- 6
- Topics
- 3
- Answer
- Complete
External force during collision
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Force east / N
Horizontal axis (x): Time / s
Read the exact data as a table
| Time / s | Force east / N |
|---|---|
| 0 | 0 |
| 0.1 | -10 |
| 0.2 | 0 |
These are the supplied values, not a worked answer. Lines join the supplied values; they are not a fitted model.
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. A 2.0 kg cart moving east at 3.0 m s⁻¹ collides and sticks to a 1.0 kg cart moving west at 1.0 m s⁻¹. During the 0.20 s interaction the combined system also experiences the external force shown below. Take east as positive. Supplied data (Time / s; Force east / N): (0, 0), (0.1, -10), (0.2, 0).
- (a)
Calculate the final velocity, explaining why momentum is not conserved for the two-cart system alone.
3 marks - (b)
Compare initial and final kinetic energies. Explain why their difference cannot all be assigned to deformation from the data given.
3 marks
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(a)
Calculate the final velocity, explaining why momentum is not conserved for the two-cart system alone.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
Initial signed momentum is 2(3)+1(−1) = 5 kg m s⁻¹.
- 2
External impulse is the triangular signed area, ½(0.20)(−10) = −1.0 N s; final momentum is therefore 4.0 kg m s⁻¹.
- 3
The stuck mass is 3.0 kg, so v = 4/3 m s⁻¹ east. Internal collision forces cancel in the system sum, but the stated external impulse does not.
(b)
Compare initial and final kinetic energies. Explain why their difference cannot all be assigned to deformation from the data given.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
Initial kinetic energy is ½(2)(3²)+½(1)(1²) = 9.5 J; final is ½(3)(4/3)² = 8/3 J.
- 2
The reduction is 9.5−8/3 = 41/6 J. The collision may transfer energy to sound, internal energy and deformation.
- 3
The external force may also do work, which requires displacement information. Impulse alone does not specify work, so the energy loss cannot be attributed entirely to collision deformation.
Complete marking guidance
Use these worked steps alongside the part-by-part answers. They explain the method and mark allocation; they do not add extra marks.
- Initial signed momentum is 2(3)+1(−1) = 5 kg m s⁻¹.
- External impulse is the triangular signed area, ½(0.20)(−10) = −1.0 N s; final momentum is therefore 4.0 kg m s⁻¹.
- The stuck mass is 3.0 kg, so v = 4/3 m s⁻¹ east. Internal collision forces cancel in the system sum, but the stated external impulse does not.
- Initial kinetic energy is ½(2)(3²)+½(1)(1²) = 9.5 J; final is ½(3)(4/3)² = 8/3 J.
- The reduction is 9.5−8/3 = 41/6 J. The collision may transfer energy to sound, internal energy and deformation.
- The external force may also do work, which requires displacement information. Impulse alone does not specify work, so the energy loss cannot be attributed entirely to collision deformation.
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