| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2013 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Coalescing particles collision |
| Difficulty | Standard +0.8 This M3 question requires multiple sophisticated steps: finding initial extension using equilibrium conditions, calculating Q's speed after free fall using energy/kinematics, applying conservation of momentum for an inelastic collision, then using energy conservation with elastic potential energy for the combined system. The algebraic manipulation to reach the specific quadratic form adds complexity. Significantly harder than typical A-level mechanics due to the combination of elastic strings, collisions, and energy methods in one extended problem. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings6.03b Conservation of momentum: 1D two particles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2.5g = 36.75\,e/3\) | M1 | \(P\) in equilibrium; Allow missing \(g\) |
| \(e = 2\) | A1 | |
| \(v^2 = 0^2 + 2g(3 + e)\) | M1 | May be implied by \(v^2 = 98\) |
| \(v = 7\sqrt{2}\) | A1 | |
| \(1 \times v = 3.5\,V\) | M1 | |
| Combined speed \(= 2\sqrt{2}\) (ms\(^{-1}\)) | A1 AG [6] | Convincing derivation, no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Change in PE is \(3.5gX\) | B1 | \(34.3X\) |
| Change in KE is \(0.5 \times 3.5 \times (2\sqrt{2})^2\) | B1 | \(14\) |
| Change in EE is \(\frac{36.75(X+2)^2}{2\times3} - \frac{36.75\times2^2}{2\times3}\) | M1, A1 | Allow sign errors / omission of 2; Allow '\(x\)' or '\(x+5\)' for '\(x+2\)'; 2 terms or difference; Allow sign errors; at least PE, KE, EE term |
| Use conservation of energy | M1 | \(\frac{36.75(X+2)^2}{2\times3} = \frac{36.75\times2^2}{2\times3} + 3.5gX + \frac{3.5}{2}V^2\) |
| \(35X^2 - 56X - 80 = 0\) | A1 AG [6] | Convincing derivation, no errors; may see \(36.75X^2 - 58.8X - 84 = 0\) |
# Question 5:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2.5g = 36.75\,e/3$ | M1 | $P$ in equilibrium; Allow missing $g$ |
| $e = 2$ | A1 | |
| $v^2 = 0^2 + 2g(3 + e)$ | M1 | May be implied by $v^2 = 98$ |
| $v = 7\sqrt{2}$ | A1 | |
| $1 \times v = 3.5\,V$ | M1 | |
| Combined speed $= 2\sqrt{2}$ (ms$^{-1}$) | A1 **AG** **[6]** | Convincing derivation, no errors |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Change in PE is $3.5gX$ | B1 | $34.3X$ |
| Change in KE is $0.5 \times 3.5 \times (2\sqrt{2})^2$ | B1 | $14$ |
| Change in EE is $\frac{36.75(X+2)^2}{2\times3} - \frac{36.75\times2^2}{2\times3}$ | M1, A1 | Allow sign errors / omission of 2; Allow '$x$' or '$x+5$' for '$x+2$'; 2 terms or difference; Allow sign errors; at least PE, KE, EE term |
| Use conservation of energy | M1 | $\frac{36.75(X+2)^2}{2\times3} = \frac{36.75\times2^2}{2\times3} + 3.5gX + \frac{3.5}{2}V^2$ |
| $35X^2 - 56X - 80 = 0$ | A1 **AG** **[6]** | Convincing derivation, no errors; may see $36.75X^2 - 58.8X - 84 = 0$ |
---5 A particle $P$, of mass 2.5 kg , is in equilibrium suspended from a fixed point $A$ by a light elastic string of natural length 3 m and modulus of elasticity 36.75 N . Another particle $Q$, of mass 1 kg , is released from rest at $A$ and falls freely until it reaches $P$ and becomes attached to it.\\
(i) Show that the speed of the combined particles, immediately after $Q$ becomes attached to $P$, is $2 \sqrt { 2 } \mathrm {~ms} ^ { - 1 }$.
The combined particles fall a further distance $X \mathrm {~m}$ before coming to instantaneous rest.\\
(ii) Find a quadratic equation satisfied by $X$, and show that it simplifies to $35 X ^ { 2 } - 56 X - 80 = 0$.
\hfill \mbox{\textit{OCR M3 2013 Q5 [12]}}