OCR MEI Further Mechanics Minor 2023 June — Question 3 11 marks

Exam BoardOCR MEI
ModuleFurther Mechanics Minor (Further Mechanics Minor)
Year2023
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMomentum and Collisions 1
TypeCollision followed by wall impact
DifficultyStandard +0.8 This is a multi-stage mechanics problem requiring conservation of momentum, coefficient of restitution, impulse calculations, and energy considerations across three distinct phases (collision, wall impact, friction). While the individual techniques are A-level standard, the problem requires careful tracking of velocities through multiple events and connecting friction work to kinetic energy loss, making it moderately challenging with extended reasoning.
Spec3.03r Friction: concept and vector form3.03t Coefficient of friction: F <= mu*R model6.03a Linear momentum: p = mv6.03b Conservation of momentum: 1D two particles6.03f Impulse-momentum: relation

3 The diagram shows two blocks P and Q of masses 0.5 kg and 2 kg respectively, on a horizontal surface. The points \(\mathrm { A } , \mathrm { B }\) and C lie on the surface in a straight line. There is a wall at C . The surface between B and C is smooth, and the surface between A and B is rough, such that the coefficient of friction between P and AB is \(\frac { 2 } { 3 }\). \includegraphics[max width=\textwidth, alt={}, center]{cad8805d-59f6-4ed2-81f4-9e8c749461f5-3_229_1271_1601_278} P is projected with a speed of \(6 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) directly towards Q , which is at rest. As a result of the collision between P and Q, P changes direction and subsequently comes to rest at A. You may assume that P only collides with Q once.
  1. Determine the coefficient of restitution between P and Q .
  2. Calculate the impulse exerted on P by Q during their collision. After colliding with P , Q strikes the wall, which is perpendicular to the direction of the motion of Q , and comes to rest exactly halfway between A and B . The collision between Q and the wall is perfectly elastic.
  3. Determine the coefficient of friction between Q and AB .

Question 3:
AnswerMarks Guidance
3(a) Let the speed of P when it reaches B be u m s-1.
Frictional force = 23  0 .5 gB1 3.3
Either 12  0 .5 u 2 − 23  0 .5 g  0 .6 = 0
AnswerMarks Guidance
Or a = − 23 g , so 0 2 = u 2 − 2  23 g  0 .6M1 3.4
against friction terms need to be
present.
Let the speed of Q be v m s-1 after collision with P.
AnswerMarks Guidance
COLM: 0 .5  6 = 0 .5  − 2 .8 + 2 vM1 3.3
u = 2 .8 and v = 2 .2A1 1.1
Coefficient of restitution between P and Q
AnswerMarks Guidance
= 2 .2 +6 2 .8M1 3.3
se p a p p
errors but not num/dem switch.
AnswerMarks Guidance
= 2 .2 +6 2 .8 = 56 ( = 0 .8 3 3 3 )A1 1.1
[6]
AnswerMarks Guidance
(b)0 .5 ( 2 .8 − ( − 6 ) ) = 4 .4 N s B1
towards A.B1 1.1
CB’
[2]
AnswerMarks Guidance
(c)Let the coefficient of friction between Q and AB be .
Speed of Q when it reaches B is 2.2 m s-1B1 3.4
122.22−2g0.3=0
Either
2
AnswerMarks Guidance
Or a=−g, so 0 2 2 .2 2 2 g 0 .3  = − M1 3.1b
11 24 17  = ( = 0 .8 2 3 1 2 9 )A1 1.1
[3]
Question 3:
3 | (a) | Let the speed of P when it reaches B be u m s-1.
Frictional force = 23  0 .5 g | B1 | 3.3
Either 12  0 .5 u 2 − 23  0 .5 g  0 .6 = 0
Or a = − 23 g , so 0 2 = u 2 − 2  23 g  0 .6 | M1 | 3.4 | If using WEP, both KE and WD
against friction terms need to be
present.
Let the speed of Q be v m s-1 after collision with P.
COLM: 0 .5  6 = 0 .5  − 2 .8 + 2 v | M1 | 3.3 | Three terms. Condone sign error.
u = 2 .8 and v = 2 .2 | A1 | 1.1 | or if both seen in (b) and (c)
Coefficient of restitution between P and Q
= 2 .2 +6 2 .8 | M1 | 3.3 | Attempt at v  v . Allow sign
se p a p p
errors but not num/dem switch.
= 2 .2 +6 2 .8 = 56 ( = 0 .8 3 3 3 ) | A1 | 1.1
[6]
(b) | 0 .5 ( 2 .8 − ( − 6 ) ) = 4 .4 N s | B1 | 3.3 | Use “their” value for 𝑢
towards A. | B1 | 1.1 | oe eg ‘towards B’, ‘in the direction
CB’
[2]
(c) | Let the coefficient of friction between Q and AB be .
Speed of Q when it reaches B is 2.2 m s-1 | B1 | 3.4 | Use “their” value for 𝑣
122.22−2g0.3=0
Either
2
Or a=−g, so 0 2 2 .2 2 2 g 0 .3  = −  | M1 | 3.1b
11 24 17  = ( = 0 .8 2 3 1 2 9 ) | A1 | 1.1 | cao
[3]
3 The diagram shows two blocks P and Q of masses 0.5 kg and 2 kg respectively, on a horizontal surface. The points $\mathrm { A } , \mathrm { B }$ and C lie on the surface in a straight line. There is a wall at C . The surface between B and C is smooth, and the surface between A and B is rough, such that the coefficient of friction between P and AB is $\frac { 2 } { 3 }$.\\
\includegraphics[max width=\textwidth, alt={}, center]{cad8805d-59f6-4ed2-81f4-9e8c749461f5-3_229_1271_1601_278}

P is projected with a speed of $6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ directly towards Q , which is at rest. As a result of the collision between P and Q, P changes direction and subsequently comes to rest at A. You may assume that P only collides with Q once.
\begin{enumerate}[label=(\alph*)]
\item Determine the coefficient of restitution between P and Q .
\item Calculate the impulse exerted on P by Q during their collision.

After colliding with P , Q strikes the wall, which is perpendicular to the direction of the motion of Q , and comes to rest exactly halfway between A and B . The collision between Q and the wall is perfectly elastic.
\item Determine the coefficient of friction between Q and AB .
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Mechanics Minor 2023 Q3 [11]}}