| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2005 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions |
| Type | Direct collision, find impulse magnitude |
| Difficulty | Moderate -0.3 This is a standard M1 momentum conservation problem with straightforward application of formulas. Students must set up conservation of momentum with given constraints and solve simultaneous equations, then calculate impulse using the change in momentum formula. While it requires careful sign convention and algebraic manipulation, it follows a well-practiced template with no novel insight required, making it slightly easier than average. |
| Spec | 6.03a Linear momentum: p = mv6.03b Conservation of momentum: 1D two particles6.03e Impulse: by a force6.03f Impulse-momentum: relation |
| Answer | Marks |
|---|---|
| 2 | 8 2 CLM: 0.6 x 8 – 0.2 x 2 = 0.6 x v + 0.2 x w |
| Answer | Marks |
|---|---|
| = 2.16 Ns | M1 A1 |
Question 2:
2 | 8 2 CLM: 0.6 x 8 – 0.2 x 2 = 0.6 x v + 0.2 x w
v w Using w = 2v to form equn in v/w only
Solve to get v = 4.4 m s–1
(b) Impulse on B = 0.2(2 + 8.8)
= 2.16 Ns | M1 A1
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(3)Two small steel balls $A$ and $B$ have mass 0.6 kg and 0.2 kg respectively. They are moving towards each other in opposite directions on a smooth horizontal table when they collide directly. Immediately before the collision, the speed of $A$ is $8 \text{ m s}^{-1}$ and the speed of $B$ is $2 \text{ m s}^{-1}$. Immediately after the collision, the direction of motion of $A$ is unchanged and the speed of $B$ is twice the speed of $A$. Find
\begin{enumerate}[label=(\alph*)]
\item the speed of $A$ immediately after the collision, [5]
\item the magnitude of the impulse exerted on $B$ in the collision. [3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2005 Q2 [8]}}