Edexcel M4 2018 June — Question 2 8 marks

Exam BoardEdexcel
ModuleM4 (Mechanics 4)
Year2018
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicOblique and successive collisions
TypeSphere rebounds off fixed wall obliquely
DifficultyStandard +0.8 This M4 mechanics question requires setting up velocity components before/after collision, applying the coefficient of restitution to the normal component, using the kinetic energy condition to form an equation, and solving a trigonometric equation. It combines multiple concepts (restitution, energy, trigonometry) in a non-routine way that requires careful algebraic manipulation, making it moderately harder than average A-level questions.
Spec6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact
2. A small ball \(B\), moving on a smooth horizontal plane, collides with a fixed smooth vertical wall. Immediately before the collision the angle between the direction of motion of \(B\) and the wall is \(\alpha\). The coefficient of restitution between \(B\) and the wall is \(\frac { 3 } { 4 }\). The kinetic energy of \(B\) immediately after the collision is \(60 \%\) of its kinetic energy immediately before the collision. Find, in degrees, the size of angle \(\alpha\).
AnswerMarks Guidance
Velocity before & after: parallel to wall: \(u\) and \(u\)B1
Perpendicular to the wall: \(v\) and \(\frac{3}{4}v\)B1 Allow with \(ev\)
Kinetic energy: \(\frac{1}{2}m(\frac{9}{16}v^2 + u^2) = 0.6 \times \frac{1}{2}m(v^2 + u^2)\)M1A2
\(\frac{90}{16}v^2 + 10u^2 = 6v^2 + 6u^2\)
\(4u^2 = \frac{6}{16}v^2\), \(u^2 = \frac{3}{32}v^2\)
\(\tan\alpha = \frac{v}{u} = \sqrt{\frac{32}{3}}\)M1A1
\(\alpha = 73°\) (or better 72.976...)A1
Question 2 Alt
AnswerMarks
Velocity before & after: parallel to wall: \(u\cos\alpha\) and \(u\cos\alpha\)B1
Perpendicular to the wall: \(u\sin\alpha\) and \(\frac{3}{4}u\sin\alpha\)B1
Kinetic energy: \(\frac{1}{2}m(\frac{9}{16}(u\sin\alpha)^2 + (u\cos\alpha)^2) = 0.6 \times \frac{1}{2}m((u\sin\alpha)^2 + (u\cos\alpha)^2)\)M1A2
\(\frac{9}{16}\sin^2\alpha + \cos^2\alpha = \frac{3}{5} = \frac{9}{16} + \frac{7}{16}\cos^2\alpha\)M1
\(\cos^2\alpha = \frac{3}{35}\), \(\alpha = \cos^{-1}\sqrt{\frac{3}{35}} = 73.0°\) (1.27 radians)A1, A1 
| Velocity before & after: parallel to wall: $u$ and $u$ | B1 | |
| Perpendicular to the wall: $v$ and $\frac{3}{4}v$ | B1 | Allow with $ev$ |
| Kinetic energy: $\frac{1}{2}m(\frac{9}{16}v^2 + u^2) = 0.6 \times \frac{1}{2}m(v^2 + u^2)$ | M1A2 | |
| $\frac{90}{16}v^2 + 10u^2 = 6v^2 + 6u^2$ | | |
| $4u^2 = \frac{6}{16}v^2$, $u^2 = \frac{3}{32}v^2$ | | |
| $\tan\alpha = \frac{v}{u} = \sqrt{\frac{32}{3}}$ | M1A1 | |
| $\alpha = 73°$ (or better 72.976...) | A1 | |

## Question 2 Alt
| Velocity before & after: parallel to wall: $u\cos\alpha$ and $u\cos\alpha$ | B1 | |
| Perpendicular to the wall: $u\sin\alpha$ and $\frac{3}{4}u\sin\alpha$ | B1 | |
| Kinetic energy: $\frac{1}{2}m(\frac{9}{16}(u\sin\alpha)^2 + (u\cos\alpha)^2) = 0.6 \times \frac{1}{2}m((u\sin\alpha)^2 + (u\cos\alpha)^2)$ | M1A2 | |
| $\frac{9}{16}\sin^2\alpha + \cos^2\alpha = \frac{3}{5} = \frac{9}{16} + \frac{7}{16}\cos^2\alpha$ | M1 | |
| $\cos^2\alpha = \frac{3}{35}$, $\alpha = \cos^{-1}\sqrt{\frac{3}{35}} = 73.0°$ (1.27 radians) | A1, A1 | |
2. A small ball $B$, moving on a smooth horizontal plane, collides with a fixed smooth vertical wall. Immediately before the collision the angle between the direction of motion of $B$ and the wall is $\alpha$. The coefficient of restitution between $B$ and the wall is $\frac { 3 } { 4 }$. The kinetic energy of $B$ immediately after the collision is $60 \%$ of its kinetic energy immediately before the collision.

Find, in degrees, the size of angle $\alpha$.\\

\hfill \mbox{\textit{Edexcel M4 2018 Q2 [8]}}
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