| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Direct collision with direction reversal |
| Difficulty | Standard +0.3 This is a standard M2 collision problem requiring application of conservation of momentum and Newton's law of restitution. While it involves algebraic manipulation across three parts, the techniques are routine for this module—students must apply two standard equations, eliminate variables, and use physical constraints (direction reversal, coefficient bounds). The multi-part structure and algebraic complexity place it slightly above average, but it requires no novel insight beyond textbook methods. |
| Spec | 3.02i Projectile motion: constant acceleration model3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force6.03b Conservation of momentum: 1D two particles6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| (a) Diagram showing weight, tension, normal reaction, friction | B2 | |
| (b) \(M(C): I(7a \cos \alpha) = 0.8g(2a \cos \alpha)\) → \(T = 2(0.8g) \div 7 = 2.24\) N | M1 A1 M1 A1 | |
| (c) Resolve perp. to rod: \(R + T \cos \alpha = 0.8g \cos \alpha\) | M1 A1 | |
| \(R = 5.6 \cos 20° = 5.26\) N | M1 A1 | Total: 10 marks |
(a) Diagram showing weight, tension, normal reaction, friction | B2 |
(b) $M(C): I(7a \cos \alpha) = 0.8g(2a \cos \alpha)$ → $T = 2(0.8g) \div 7 = 2.24$ N | M1 A1 M1 A1 |
(c) Resolve perp. to rod: $R + T \cos \alpha = 0.8g \cos \alpha$ | M1 A1 |
$R = 5.6 \cos 20° = 5.26$ N | M1 A1 | **Total: 10 marks**\begin{enumerate}
\item A uniform rod $A B$, of mass 0.8 kg and length $10 a$, is supported at the end $A$ by a light inextensible vertical string and rests in limiting equilibrium on a rough fixed peg at $C$, where $A C = 7 a$.\\
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\item Two particles $A$ and $B$, of mass $m$ and $k m$ respectively, are moving in the same direction on a smooth horizontal surface. $A$ has speed $4 u$ and $B$ has speed $u$. The coefficient of restitution between $A$ and $B$ is $e \quad A$ collides directly with $B$, and in the collision the direction of $A$ 's motion is reversed. Immediately after the impact, $B$ has speed $2 u$.\\
(a) Show that the speed of $A$ immediately after the impact is $u ( 3 e - 2 )$.\\
(b) Deduce the range of possible values of $e$.\\
(c) Show that $4 < k \leq 5$.
\item A ball is projected from ground level with speed $34 \mathrm {~ms} ^ { - 1 }$ at an angle $\alpha$ above the horizontal, where $\tan \alpha = \frac { 8 } { 15 }$.\\
(a) Find the greatest height reached by the ball above ground level.
\end{enumerate}
While it is descending, the ball hits a horizontal ledge 6 metres above ground level.\\
(b) Find the horizontal distance travelled by the ball before it hits the ledge.\\
(c) Find the speed of the ball at the instant when it hits the ledge.
\hfill \mbox{\textit{Edexcel M2 Q6 [10]}}