Edexcel M2 2017 January — Question 6 5 marks

Exam BoardEdexcel
ModuleM2 (Mechanics 2)
Year2017
SessionJanuary
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicWork done and energy
TypeWork done against resistance - penetration into material
DifficultyStandard +0.3 This is a straightforward application of the work-energy principle across two stages (upward motion and penetration into ground). Students must calculate initial KE, account for gravitational PE changes, and equate total energy to work done by resistance force. While it requires careful bookkeeping of energy changes and unit conversion (cm to m), it follows a standard template with no conceptual surprises—slightly easier than average for M2.
Spec6.02i Conservation of energy: mechanical energy principle

6. A ball of mass 0.6 kg is projected vertically upwards with speed \(22.4 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) from a point which is 1.5 m above horizontal ground. The ball moves freely under gravity until it reaches the ground. The ground is soft and the ball sinks 2.5 cm into the ground before coming to rest. The ball is modelled as a particle and the ground is assumed to exert a constant resistive force of magnitude \(R\) newtons on the ball. Using the work-energy principle, find, to 3 significant figures, the value of \(R\).
(5)

Question 6:
AnswerMarks Guidance
Answer/WorkingMark Guidance
W.D. against \(R\): \(0.025R\)B1 Seen or implied
W.D. against \(R\) = KE Loss + PE LossM1 Work-energy equation. Need all relevant terms and no extras. Condone 2.5 for 0.025. 1.5 for 1.525 is M0
\(0.025R = \frac{1}{2} \times 0.6 \times 22.4^2 + 0.6g \times (1.5 + 0.025)\)A1 At most one error. Use of 2.5 for 0.025 is one error
(unsimplified equation)A1 Correct unsimplified. Follow their WD
\(R = 6380\) NA1 Q asks for 3 s.f. - do not accept 6400
Total: 5 marks
## Question 6:

| Answer/Working | Mark | Guidance |
|---|---|---|
| W.D. against $R$: $0.025R$ | B1 | Seen or implied |
| W.D. against $R$ = KE Loss + PE Loss | M1 | Work-energy equation. Need all relevant terms and no extras. Condone 2.5 for 0.025. 1.5 for 1.525 is M0 |
| $0.025R = \frac{1}{2} \times 0.6 \times 22.4^2 + 0.6g \times (1.5 + 0.025)$ | A1 | At most one error. Use of 2.5 for 0.025 is one error |
| (unsimplified equation) | A1 | Correct unsimplified. Follow their WD |
| $R = 6380$ N | A1 | Q asks for 3 s.f. - do not accept 6400 |

**Total: 5 marks**

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6. A ball of mass 0.6 kg is projected vertically upwards with speed $22.4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from a point which is 1.5 m above horizontal ground. The ball moves freely under gravity until it reaches the ground. The ground is soft and the ball sinks 2.5 cm into the ground before coming to rest. The ball is modelled as a particle and the ground is assumed to exert a constant resistive force of magnitude $R$ newtons on the ball. Using the work-energy principle, find, to 3 significant figures, the value of $R$.\\
(5)\\

\hfill \mbox{\textit{Edexcel M2 2017 Q6 [5]}}