OCR M3 2006 January — Question 4 10 marks

Exam BoardOCR
ModuleM3 (Mechanics 3)
Year2006
SessionJanuary
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicOblique and successive collisions
TypeOblique collision, find velocities/angles
DifficultyChallenging +1.2 This is a standard M3 oblique collision problem requiring resolution of velocities parallel and perpendicular to the line of centres, conservation of momentum along the line of centres, and Newton's restitution equation. While it involves multiple steps and careful bookkeeping of components, it follows a well-established procedure taught in M3 with no novel insight required. The equal radii and perpendicular initial velocity simplify the setup slightly. Harder than a basic C1 question but routine for M3 students who have practiced this topic.
Spec6.03l Newton's law: oblique impacts6.04e Rigid body equilibrium: coplanar forces

4 \includegraphics[max width=\textwidth, alt={}, center]{5bb3bd29-a2eb-4124-802c-fb17b68c50e4-2_283_711_1754_722} Two uniform smooth spheres \(A\) and \(B\), of equal radius, have masses 5 kg and 2 kg respectively. They are moving on a horizontal surface when they collide. Immediately before the collision, \(A\) has speed \(4 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and is moving perpendicular to the line of centres, and \(B\) has speed \(4 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) along the line of centres (see diagram). The coefficient of restitution is 0.75 . Find the speed and direction of motion of each sphere immediately after the collision.

Question 4:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Perpendicular velocity of A after impact \(= 4\)B1
\([5\times0] - 2\times4 = 5a + 2b\)M1, A1 For using conservation of momentum // l.o.c.
\(0.75\times4 = b - a\)M1, A1 Using N.E.L. // l.o.c.
M1For solving simultaneous equations
Speed of B is \(1\) ms\(^{-1}\); direction //l.o.c. and to the rightA1
\(v_A = \sqrt{4^2 + (-2)^2}\)M1 For method of finding speed of A
\(\tan(\text{angle}) = 4/2\)M1 For method of finding direction of A
Speed of A is \(4.47\) ms\(^{-1}\); direction is \(63.4°\) to l.o.c. and to the leftA1 [10]
## Question 4:

| Answer/Working | Marks | Guidance |
|---|---|---|
| Perpendicular velocity of A after impact $= 4$ | B1 | |
| $[5\times0] - 2\times4 = 5a + 2b$ | M1, A1 | For using conservation of momentum // l.o.c. |
| $0.75\times4 = b - a$ | M1, A1 | Using N.E.L. // l.o.c. |
| | M1 | For solving simultaneous equations |
| Speed of B is $1$ ms$^{-1}$; direction //l.o.c. and to the right | A1 | |
| $v_A = \sqrt{4^2 + (-2)^2}$ | M1 | For method of finding speed of A |
| $\tan(\text{angle}) = 4/2$ | M1 | For method of finding direction of A |
| Speed of A is $4.47$ ms$^{-1}$; direction is $63.4°$ to l.o.c. and to the left | A1 | [10] |

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\includegraphics[max width=\textwidth, alt={}, center]{5bb3bd29-a2eb-4124-802c-fb17b68c50e4-2_283_711_1754_722}

Two uniform smooth spheres $A$ and $B$, of equal radius, have masses 5 kg and 2 kg respectively. They are moving on a horizontal surface when they collide. Immediately before the collision, $A$ has speed $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and is moving perpendicular to the line of centres, and $B$ has speed $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ along the line of centres (see diagram). The coefficient of restitution is 0.75 . Find the speed and direction of motion of each sphere immediately after the collision.

\hfill \mbox{\textit{OCR M3 2006 Q4 [10]}}