Questions SPS SM Mechanics

7. A health centre claims that the time a doctor spends with a patient can be modelled by a normal distribution with a mean of 10 minutes and a standard deviation of 4 minutes.

Using this model, find the probability that the time spent with a randomly selected patient is more than 15 minutes. Some patients complain that the mean time the doctor spends with a patient is more than 10 minutes. The receptionist takes a random sample of 20 patients and finds that the mean time the doctor spends with a patient is 11.5 minutes.
Stating your hypotheses clearly and using a $5 \%$ significance level, test whether or not there is evidence to support the patients' complaint. The health centre also claims that the time a dentist spends with a patient during a routine appointment, $T$ minutes, can be modelled by the normal distribution where $T \sim \mathrm {~N} \left( 5,3.5 ^ { 2 } \right)$
Using this model,
1. find the probability that a routine appointment with the dentist takes less than 2 minutes
2. find $\mathrm { P } ( T < 2 \mid T > 0 )$
3. hence explain why this normal distribution may not be a good model for $T$. The dentist believes that she cannot complete a routine appointment in less than 2 minutes.
  She suggests that the health centre should use a refined model only including values of $T > 2$

Find the median time for a routine appointment using this new model, giving your answer correct to one decimal place. Name: □ \section*{U8th A LEVEL Single Mathematics Assessment Mechanics $7 ^ { \text {th } }$ January 2021 } Instructions

Answer all the questions
Write your answer to each question in the space provided under each question. The question number(s) must be clearly shown.
Use black or blue ink. Pencil may be used for graphs and diagrams only.
You should clearly write your name at the top of this page and on any additional sheets that you use.
You are permitted to use a scientific or graphical calculator in this paper.
Final answers should be given to a degree of accuracy appropriate to the context.

Information

The total mark for this paper is 55 marks.
The marks for each question are shown in brackets ( ).
You are reminded of the need for clear presentation in your answers.
You should allow approximately 60 minutes for this section of the test

\section*{Formulae} \section*{A Level Mathematics A (H240)} \section*{Arithmetic series} $S _ { n } = \frac { 1 } { 2 } n ( a + l ) = \frac { 1 } { 2 } n \{ 2 a + ( n - 1 ) d \}$ \section*{Geometric series} $S _ { n } = \frac { a \left( 1 - r ^ { n } \right) } { 1 - r }$
$S _ { \infty } = \frac { a } { 1 - r }$ for $| r | < 1$ \section*{Binomial series} $( a + b ) ^ { n } = a ^ { n } + { } ^ { n } \mathrm { C } _ { 1 } a ^ { n - 1 } b + { } ^ { n } \mathrm { C } _ { 2 } a ^ { n - 2 } b ^ { 2 } + \ldots + { } ^ { n } \mathrm { C } _ { r } a ^ { n - r } b ^ { r } + \ldots + b ^ { n } \quad ( n \in \mathbb { N } )$, where ${ } ^ { n } \mathrm { C } _ { r } = { } _ { n } \mathrm { C } _ { r } = \binom { n } { r } = \frac { n ! } { r ! ( n - r ) ! }$
$( 1 + x ) ^ { n } = 1 + n x + \frac { n ( n - 1 ) } { 2 ! } x ^ { 2 } + \ldots + \frac { n ( n - 1 ) \ldots ( n - r + 1 ) } { r ! } x ^ { r } + \ldots \quad ( | x | < 1 , n \in \mathbb { R } )$ \section*{Differentiation}

$\mathrm { f } ( x )$	$\mathrm { f } ^ { \prime } ( x )$
$\tan k x$	$k \sec ^ { 2 } k x$
$\sec x$	$\sec x \tan x$
$\cot x$	$- \operatorname { cosec } ^ { 2 } x$
$\operatorname { cosec } x$	$- \operatorname { cosec } x \cot x$

Quotient rule $y = \frac { u } { v } , \frac { \mathrm {~d} y } { \mathrm {~d} x } = \frac { v \frac { \mathrm {~d} u } { \mathrm {~d} x } - u \frac { \mathrm {~d} v } { \mathrm {~d} x } } { v ^ { 2 } }$ \section*{Differentiation from first principles} $\mathrm { f } ^ { \prime } ( x ) = \lim _ { h \rightarrow 0 } \frac { \mathrm { f } ( x + h ) - \mathrm { f } ( x ) } { h }$ \section*{Integration} $\int \frac { \mathrm { f } ^ { \prime } ( x ) } { \mathrm { f } ( x ) } \mathrm { d } x = \ln | \mathrm { f } ( x ) | + c$
$\int \mathrm { f } ^ { \prime } ( x ) ( \mathrm { f } ( x ) ) ^ { n } \mathrm {~d} x = \frac { 1 } { n + 1 } ( \mathrm { f } ( x ) ) ^ { n + 1 } + c$
Integration by parts $\int u \frac { \mathrm {~d} v } { \mathrm {~d} x } \mathrm {~d} x = u v - \int v \frac { \mathrm {~d} u } { \mathrm {~d} x } \mathrm {~d} x$ Small angle approximations
$\sin \theta \approx \theta , \cos \theta \approx 1 - \frac { 1 } { 2 } \theta ^ { 2 } , \tan \theta \approx \theta$ where $\theta$ is measured in radians \section*{Trigonometric identities} $\sin ( A \pm B ) = \sin A \cos B \pm \cos A \sin B$
$\cos ( A \pm B ) = \cos A \cos B \mp \sin A \sin B$
$\tan ( A \pm B ) = \frac { \tan A \pm \tan B } { 1 \mp \tan A \tan B } \quad \left( A \pm B \neq \left( k + \frac { 1 } { 2 } \right) \pi \right)$ \section*{Numerical methods} Trapezium rule: $\int _ { a } ^ { b } y \mathrm {~d} x \approx \frac { 1 } { 2 } h \left\{ \left( y _ { 0 } + y _ { n } \right) + 2 \left( y _ { 1 } + y _ { 2 } + \ldots + y _ { n - 1 } \right) \right\}$, where $h = \frac { b - a } { n }$
The Newton-Raphson iteration for solving $\mathrm { f } ( x ) = 0 : x _ { n + 1 } = x _ { n } - \frac { \mathrm { f } \left( x _ { n } \right) } { \mathrm { f } ^ { \prime } \left( x _ { n } \right) }$ \section*{Probability} $\mathrm { P } ( A \cup B ) = \mathrm { P } ( A ) + \mathrm { P } ( B ) - \mathrm { P } ( A \cap B )$
$\mathrm { P } ( A \cap B ) = \mathrm { P } ( A ) \mathrm { P } ( B \mid A ) = \mathrm { P } ( B ) \mathrm { P } ( A \mid B )$ or $\mathrm { P } ( A \mid B ) = \frac { \mathrm { P } ( A \cap B ) } { \mathrm { P } ( B ) }$ \section*{Standard deviation} $\sqrt { \frac { \sum ( x - \bar { x } ) ^ { 2 } } { n } } = \sqrt { \frac { \sum x ^ { 2 } } { n } - \bar { x } ^ { 2 } }$ or $\sqrt { \frac { \sum f ( x - \bar { x } ) ^ { 2 } } { \sum f } } = \sqrt { \frac { \sum f x ^ { 2 } } { \sum f } - \bar { x } ^ { 2 } }$ \section*{The binomial distribution} If $X \sim \mathbf { B } ( n , p )$ then $\mathbf { P } ( X = x ) = \binom { n } { x } p ^ { x } ( 1 - p ) ^ { n - x }$, Mean of $X$ is $n p$, Variance of $X$ is $n p ( 1 - p )$ \section*{Hypothesis test for the mean of a normal distribution} If $X \sim \mathrm {~N} \left( \mu , \sigma ^ { 2 } \right)$ then $\bar { X } \sim \mathrm {~N} \left( \mu , \frac { \sigma ^ { 2 } } { n } \right)$ and $\frac { \bar { X } - \mu } { \sigma / \sqrt { n } } \sim \mathrm {~N} ( 0,1 )$ \section*{Percentage points of the normal distribution} If $Z$ has a normal distribution with mean 0 and variance 1 then, for each value of $p$, the table gives the value of $z$ such that $\mathrm { P } ( Z \leqslant z ) = p$.

$p$	0.75	0.90	0.95	0.975	0.99	0.995	0.9975	.0 .999	0.9995
$z$	0.674	1.282	1.645	1.960	2.326	2.576	2.807	3.090	3.291

\section*{Kinematics} Motion in a straight line
$v = u + a t$
$s = u t + \frac { 1 } { 2 } a t ^ { 2 }$
$s = \frac { 1 } { 2 } ( u + v ) t$
$v ^ { 2 } = u ^ { 2 } + 2 a s$
$s = v t - \frac { 1 } { 2 } a t ^ { 2 }$ Motion in two dimensions
$\mathbf { v } = \mathbf { u } + \mathbf { a } t$
$\mathbf { s } = \mathbf { u } t + \frac { 1 } { 2 } \mathbf { a } t ^ { 2 }$
$\mathbf { s } = \frac { 1 } { 2 } ( \mathbf { u } + \mathbf { v } ) \boldsymbol { t }$
$\mathbf { s } = \mathbf { v } t - \frac { 1 } { 2 } \mathbf { a } t ^ { 2 }$
[0pt] [BLANK PAGE]
1. A vehicle is driven at a constant speed of $12 \mathrm {~ms} ^ { - 1 }$ along a straight horizontal road. Only one of the statements below is correct. Identify the correct statement.
Tick ( $\checkmark$ ) one box. The vehicle is accelerating □ The vehicle's driving force exceeds the total force resisting its motion □ The resultant force acting on the vehicle is zero □ The resultant force acting on the vehicle is dependent on its mass □
2. A number of forces act on a particle such that the resultant force is $\binom { 6 } { - 3 } \mathrm {~N}$
One of the forces acting on the particle is $\binom { 8 } { - 5 } \mathrm {~N}$
Calculate the total of the other forces acting on the particle.
Circle your answer.
[0pt] [1 mark] $$\binom { 2 } { - 2 } \mathrm {~N} \quad \binom { 14 } { - 8 } \mathrm {~N} \quad \binom { - 2 } { 2 } \mathrm {~N} \quad \binom { - 14 } { 8 } \mathrm {~N}$$ 3. A rough plane is inclined to the horizontal at an angle $\alpha$, where $\tan \alpha = \frac { 3 } { 4 }$
A brick $P$ of mass $m$ is placed on the plane.
The coefficient of friction between $P$ and the plane is $\mu$
Brick $P$ is in equilibrium and on the point of sliding down the plane.
Brick $P$ is modelled as a particle.
Using the model,

find, in terms of $m$ and $g$, the magnitude of the normal reaction of the plane on brick $P$
show that $\mu = \frac { 3 } { 4 }$ For parts (c) and (d), you are not required to do any further calculations.
Brick $P$ is now removed from the plane and a much heavier brick $Q$ is placed on the plane. The coefficient of friction between $Q$ and the plane is also $\frac { 3 } { 4 }$
Explain briefly why brick $Q$ will remain at rest on the plane. Brick $Q$ is now projected with speed $0.5 \mathrm {~ms} ^ { - 1 }$ down a line of greatest slope of the plane.
Brick $Q$ is modelled as a particle.
Using the model,
describe the motion of brick $Q$, giving a reason for your answer.
4. A particle $P$ moves with acceleration $( 4 \mathbf { i } - 5 \mathbf { j } ) \mathrm { ms } ^ { - 2 }$
At time $t = 0 , P$ is moving with velocity $( - 2 \mathbf { i } + 2 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$
Find the velocity of $P$ at time $t = 2$ seconds. At time $t = 0 , P$ passes through the origin $O$.
At time $t = T$ seconds, where $T > 0$, the particle $P$ passes through the point $A$.
The position vector of $A$ is $( \lambda \mathbf { i } - 4.5 \mathbf { j } ) \mathrm { m }$ relative to $O$, where $\lambda$ is a constant.
Find the value of $T$.
Hence find the value of $\lambda$
5.
1. At time $t$ seconds, where $t \geqslant 0$, a particle $P$ moves so that its acceleration $\mathbf { a } \mathrm { ms } ^ { - 2 }$ is given by $$\mathbf { a } = ( 1 - 4 t ) \mathbf { i } + \left( 3 - t ^ { 2 } \right) \mathbf { j }$$ At the instant when $t = 0$, the velocity of $P$ is $36 \mathbf { i } \mathrm {~m} \mathrm {~s} ^ { - 1 }$
Find the velocity of $P$ when $t = 4$
Find the value of $t$ at the instant when $P$ is moving in a direction perpendicular to $\mathbf { i }$
(ii) At time $t$ seconds, where $t \geqslant 0$, a particle $Q$ moves so that its position vector $\mathbf { r }$ metres, relative to a fixed origin $O$, is given by $$\mathbf { r } = \left( t ^ { 2 } - t \right) \mathbf { i } + 3 t \mathbf { j }$$ Find the value of $t$ at the instant when the speed of $Q$ is $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$
6. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{d1809ec7-dccf-446b-8a1d-f04a4252ebec-28_529_993_374_529} \captionsetup{labelformat=empty} \caption{Figure 2}
\end{figure} A small ball is projected with speed $U \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from a point $O$ at the top of a vertical cliff.
The point $O$ is 25 m vertically above the point $N$ which is on horizontal ground.
The ball is projected at an angle of $45 ^ { \circ }$ above the horizontal.
The ball hits the ground at a point $A$, where $A N = 100 \mathrm {~m}$, as shown in Figure 2 .
The motion of the ball is modelled as that of a particle moving freely under gravity.
Using this initial model,
show that $U = 28$
find the greatest height of the ball above the horizontal ground $N A$. In a refinement to the model of the motion of the ball from $O$ to $A$, the effect of air resistance is included. This refined model is used to find a new value of $U$.
How would this new value of $U$ compare with 28 , the value given in part (a)?
State one further refinement to the model that would make the model more realistic.
7. Block $A$, of mass 0.2 kg , lies at rest on a rough plane.
The plane is inclined at an angle $\theta$ to the horizontal, such that $\tan \theta = \frac { 7 } { 24 }$
A light inextensible string is attached to $A$ and runs parallel to the line of greatest slope until it passes over a smooth fixed pulley at the top of the slope. The other end of this string is attached to particle $B$, of mass 2 kg , which is held at rest so that the string is taut, as shown in the diagram below.
\includegraphics[max width=\textwidth, alt={}, center]{d1809ec7-dccf-446b-8a1d-f04a4252ebec-32_424_1070_815_486}
$\quad B$ is released from rest so that it begins to move vertically downwards with an acceleration of $\frac { 543 } { 625 } \mathrm {~g} \mathrm {~m} \mathrm {~s} ^ { - 2 }$ Show that the coefficient of friction between $A$ and the surface of the inclined plane is 0.17
In this question use $g = 9.81 \mathrm {~m} \mathrm {~s} ^ { - 2 }$ When $A$ reaches a speed of $0.5 \mathrm {~ms} ^ { - 1 }$ the string breaks.
1. Find the distance travelled by $A$ after the string breaks until first coming to rest.
(ii) State an assumption that could affect the validity of your answer to part (b)(i).

\(d\)	10	20	30	40	50
\(\mathrm { P } ( D = d )\)	\(\frac { k } { 10 }\)	\(\frac { k } { 20 }\)	\(\frac { k } { 30 }\)	\(\frac { k } { 40 }\)	\(\frac { k } { 50 }\)

\(\mathrm { f } ( x )\)	\(\mathrm { f } ^ { \prime } ( x )\)
\(\tan k x\)	\(k \sec ^ { 2 } k x\)
\(\sec x\)	\(\sec x \tan x\)
\(\cot x\)	\(- \operatorname { cosec } ^ { 2 } x\)
\(\operatorname { cosec } x\)	\(- \operatorname { cosec } x \cot x\)

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\(p\)	0.75	0.90	0.95	0.975	0.99	0.995	0.9975	.0 .999	0.9995
\(z\)	0.674	1.282	1.645	1.960	2.326	2.576	2.807	3.090	3.291