| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2005 |
| Session | January |
| Marks | 7 |
| 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 application of conservation of momentum in one dimension with clearly defined masses and velocities. Part (a) requires a single equation setup, part (b) is interpretation of the sign, and part (c) is direct impulse calculation using change in momentum. All steps are routine M1 techniques with no problem-solving insight required. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03e Impulse: by a force6.03f Impulse-momentum: relation |
| Answer | Marks |
|---|---|
| 1 | 3 4 |
Question 1:
1 | 3 4
1 .5 kg 2.5 kg
2.5 v
(a) CLM: 1.5 x 3 – 2.5 x 4 = – 1.5 x 2.5 + 2.5 x v M1 A1
⇒ v = – 0.7 m s–1 so speed = 0.7 m s–1 A1
(3)
(b) Direction of Q unchanged A1√
(1)
(c) Impulse = 1.5 ( 3 + 2.5) M1
= 8.25, Ns A1, B1
(3)A particle $P$ of mass 1.5 kg is moving along a straight horizontal line with speed 3 m s$^{-1}$. Another particle $Q$ of mass 2.5 kg is moving, in the opposite direction, along the same straight line with speed 4 m s$^{-1}$. The particles collide. Immediately after the collision the direction of motion of $P$ is reversed and its speed is 2.5 m s$^{-1}$.
\begin{enumerate}[label=(\alph*)]
\item Calculate the speed of $Q$ immediately after the impact. [3]
\item State whether or not the direction of motion of $Q$ is changed by the collision. [1]
\item Calculate the magnitude of the impulse exerted by $Q$ on $P$, giving the units of your answer. [3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2005 Q1 [7]}}