Edexcel M4 — Question 1 13 marks

Exam BoardEdexcel
ModuleM4 (Mechanics 4)
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMomentum and Collisions 1
TypeParticle-barrier collision with angle
DifficultyChallenging +1.2 This is a standard M4 oblique impact question requiring resolution of velocities parallel and perpendicular to the plane, application of Newton's experimental law with coefficient of restitution, and calculation of kinetic energy loss. While it involves multiple steps and careful angle work, it follows a well-established method taught in M4 with no novel insight required—moderately above average difficulty for A-level.
Spec6.02d Mechanical energy: KE and PE concepts6.03k Newton's experimental law: direct impact

1. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{cf941854-3a33-4d9d-9fa0-ce9a63227599-03_457_638_233_598} \captionsetup{labelformat=empty} \caption{Figure 1}
\end{figure} A fixed smooth plane is inclined to the horizontal at an angle of \(45 ^ { \circ }\). A particle \(P\) is moving horizontally and strikes the plane. Immediately before the impact, \(P\) is moving in a vertical plane containing a line of greatest slope of the inclined plane. Immediately after the impact, \(P\) is moving in a direction which makes an angle of \(30 ^ { \circ }\) with the inclined plane, as shown in Figure 1. Find the fraction of the kinetic energy of \(P\) which is lost in the impact.

Question 1:
Part (a):
AnswerMarks Guidance
Working/AnswerMark Guidance
Equation of motion: \(0.5g - 1.5v = 0.5\frac{dv}{dt}\)M1 Applying Newton's second law with weight and resistance
\(4.9 - 1.5v = 0.5\frac{dv}{dt}\) or \(9.8 - 3v = \frac{dv}{dt}\)A1 Correct equation
Separating variables: \(\int \frac{dv}{9.8-3v} = \int dt\)M1 Separating variables correctly
\(-\frac{1}{3}\ln(9.8-3v) = t + c\)A1 Correct integration
Applying \(v=0, t=0\): \(c = -\frac{1}{3}\ln(9.8)\)M1 Using initial condition
\(-\frac{1}{3}\ln(9.8-3v) + \frac{1}{3}\ln(9.8) = t\)A1
\(\ln\left(\frac{9.8}{9.8-3v}\right) = 3t\)M1 Taking exponentials
\(3v = 9.8(1-e^{-3t})\)A1 Obtaining given result
Part (b):
AnswerMarks Guidance
Working/AnswerMark Guidance
\(v = \frac{9.8}{3}(1-e^{-3t})\), integrate to find distanceM1 Integrating \(v\) w.r.t. \(t\)
\(s = \frac{9.8}{3}\left[t + \frac{1}{3}e^{-3t}\right]_0^2\)A1 Correct integration
\(s = \frac{9.8}{3}\left[\left(2 + \frac{1}{3}e^{-6}\right) - \frac{1}{3}\right]\)M1 Applying limits \(t=0\) and \(t=2\)
\(s = \frac{9.8}{3}\left(\frac{5}{3} + \frac{1}{3}e^{-6}\right)\)A1 Correct expression
\(s \approx 5.44\) mA1 Correct numerical answer
# Question 1:

## Part (a):

| Working/Answer | Mark | Guidance |
|---|---|---|
| Equation of motion: $0.5g - 1.5v = 0.5\frac{dv}{dt}$ | M1 | Applying Newton's second law with weight and resistance |
| $4.9 - 1.5v = 0.5\frac{dv}{dt}$ or $9.8 - 3v = \frac{dv}{dt}$ | A1 | Correct equation |
| Separating variables: $\int \frac{dv}{9.8-3v} = \int dt$ | M1 | Separating variables correctly |
| $-\frac{1}{3}\ln(9.8-3v) = t + c$ | A1 | Correct integration |
| Applying $v=0, t=0$: $c = -\frac{1}{3}\ln(9.8)$ | M1 | Using initial condition |
| $-\frac{1}{3}\ln(9.8-3v) + \frac{1}{3}\ln(9.8) = t$ | A1 | |
| $\ln\left(\frac{9.8}{9.8-3v}\right) = 3t$ | M1 | Taking exponentials |
| $3v = 9.8(1-e^{-3t})$ | A1 | Obtaining given result |

## Part (b):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $v = \frac{9.8}{3}(1-e^{-3t})$, integrate to find distance | M1 | Integrating $v$ w.r.t. $t$ |
| $s = \frac{9.8}{3}\left[t + \frac{1}{3}e^{-3t}\right]_0^2$ | A1 | Correct integration |
| $s = \frac{9.8}{3}\left[\left(2 + \frac{1}{3}e^{-6}\right) - \frac{1}{3}\right]$ | M1 | Applying limits $t=0$ and $t=2$ |
| $s = \frac{9.8}{3}\left(\frac{5}{3} + \frac{1}{3}e^{-6}\right)$ | A1 | Correct expression |
| $s \approx 5.44$ m | A1 | Correct numerical answer |

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1.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{cf941854-3a33-4d9d-9fa0-ce9a63227599-03_457_638_233_598}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}

A fixed smooth plane is inclined to the horizontal at an angle of $45 ^ { \circ }$. A particle $P$ is moving horizontally and strikes the plane. Immediately before the impact, $P$ is moving in a vertical plane containing a line of greatest slope of the inclined plane. Immediately after the impact, $P$ is moving in a direction which makes an angle of $30 ^ { \circ }$ with the inclined plane, as shown in Figure 1.

Find the fraction of the kinetic energy of $P$ which is lost in the impact.\\

\hfill \mbox{\textit{Edexcel M4  Q1 [13]}}