| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2010 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions |
| Type | Direct collision, find impulse magnitude |
| Difficulty | Moderate -0.8 This is a straightforward M1 momentum/impulse question requiring direct application of standard formulas. Part (a) uses impulse = change in momentum with given values, and part (b) applies conservation of momentum with one unknown. Both are routine textbook exercises with no problem-solving insight required, making it easier than average. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03e Impulse: by a force6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| \(I = 2 \times 12 - 2 \times 3 = 18\) (N s) | M1 A1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Solving to \(m = 1.5\) | M1 A1 DM1 A1 | (4) [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Solving to \(m = 1.5\) | M1 A1 DM1 A1 | (4) |
$I = 2 \times 12 - 2 \times 3 = 18$ (N s) | M1 A1 | (2)
**Part (b):**
LM: $2 \times 12 - 8m = 2 \times 3 + 4m$
Solving to $m = 1.5$ | M1 A1 DM1 A1 | (4) [6]
**Alternative to (b):**
$I = m(4-(-8)) = 18$
Solving to $m = 1.5$ | M1 A1 DM1 A1 | (4)A particle $A$ of mass 2 kg is moving along a straight horizontal line with speed 12 m s$^{-1}$. Another particle $B$ of mass $m$ kg is moving along the same straight line, in the opposite direction to $A$, with speed 8 m s$^{-1}$. The particles collide. The direction of motion of $A$ is unchanged by the collision. Immediately after the collision, $A$ is moving with speed 3 m s$^{-1}$ and $B$ is moving with speed 4 m s$^{-1}$. Find
\begin{enumerate}[label=(\alph*)]
\item the magnitude of the impulse exerted by $B$ on $A$ in the collision, [2]
\item the value of $m$. [4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2010 Q1 [6]}}