| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Session | Specimen |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Collision with mass ratio parameter |
| Difficulty | Standard +0.3 This is a standard M2 collision problem requiring systematic application of conservation of momentum, impulse-momentum theorem, and coefficient of restitution. While it has multiple parts and involves algebraic manipulation with a parameter m, each step follows routine procedures without requiring novel insight or complex problem-solving strategies. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03i Coefficient of restitution: e6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Change of momentum of A is \(0.24 \times 2\). Hence magnitude of impulse is 0.48 N s | M1, A1, A1 (3 marks) | For considering momentum of A; For correct expression for change in mom; For correct answer 0.48 |
| (ii) \(m v_B \geq 0.48\), \(v_B \geq 6\). Hence \(m \leq \frac{0.48}{6} = 0.08\) | M1, M1, A1 (3 marks) | For considering momentum of B; For using the inequality \(v_B \geq v_A\); For showing given answer correctly |
| (iii) \(m = 0.06 \Rightarrow v_B = 8\). Hence \(8 - 6 = e(8 - 0)\), i.e. \(e = \frac{1}{4}\) | B1, M1, A1 (3 marks) | For correct speed of B; For correct use of Newton's law; For correct answer \(\frac{1}{4}\) or equivalent |
| (iv) \(0.24 \times 4 - 0.06 \times 4 = 0.24u + 0.06b\), \(b - u = \frac{1}{4}(4 + 4)\). Hence speeds of A and B are 2 m s\(^{-1}\) and 4 m s\(^{-1}\) | B1, B1, M1, A1 (4 marks) | For a correct momentum equation; For a correct restitution equation; For solution of relevant simultaneous equns; For both answers correct |
(i) Change of momentum of A is $0.24 \times 2$. Hence magnitude of impulse is 0.48 N s | M1, A1, A1 (3 marks) | For considering momentum of A; For correct expression for change in mom; For correct answer 0.48
(ii) $m v_B \geq 0.48$, $v_B \geq 6$. Hence $m \leq \frac{0.48}{6} = 0.08$ | M1, M1, A1 (3 marks) | For considering momentum of B; For using the inequality $v_B \geq v_A$; For showing given answer correctly
(iii) $m = 0.06 \Rightarrow v_B = 8$. Hence $8 - 6 = e(8 - 0)$, i.e. $e = \frac{1}{4}$ | B1, M1, A1 (3 marks) | For correct speed of B; For correct use of Newton's law; For correct answer $\frac{1}{4}$ or equivalent
(iv) $0.24 \times 4 - 0.06 \times 4 = 0.24u + 0.06b$, $b - u = \frac{1}{4}(4 + 4)$. Hence speeds of A and B are 2 m s$^{-1}$ and 4 m s$^{-1}$ | B1, B1, M1, A1 (4 marks) | For a correct momentum equation; For a correct restitution equation; For solution of relevant simultaneous equns; For both answers correct8 Two uniform smooth spheres, $A$ and $B$, have the same radius. The mass of $A$ is 0.24 kg and the mass of $B$ is $m \mathrm {~kg}$. Sphere $A$ is travelling in a straight line on a horizontal table, with speed $8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, when it collides directly with sphere $B$, which is at rest. As a result of the collision, sphere $A$ continues in the same direction with a speed of $6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(i) Find the magnitude of the impulse exerted by $A$ on $B$.\\
(ii) Show that $m \leqslant 0.08$.
It is given that $m = 0.06$.\\
(iii) Find the coefficient of restitution between $A$ and $B$.
On another occasion $A$ and $B$ are travelling towards each other, each with speed $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, when they collide directly.\\
(iv) Find the speeds of $A$ and $B$ immediately after the collision.
\hfill \mbox{\textit{OCR M2 Q8 [13]}}