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OCR M3 2007 January Q4

13 marks Standard +0.3

4 A particle $P$ of mass 0.2 kg is suspended from a fixed point $O$ by a light elastic string of natural length 0.7 m and modulus of elasticity $3.5 \mathrm {~N} . P$ is at the equilibrium position when it is projected vertically downwards with speed $1.6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. At time $t \mathrm {~s}$ after being set in motion $P$ is $x \mathrm {~m}$ below the equilibrium position and has velocity $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$.

Show that the equilibrium position of $P$ is 1.092 m below $O$.
Prove that $P$ moves with simple harmonic motion, and calculate the amplitude.
Calculate $x$ and $v$ when $t = 0.4$.

OCR M3 2007 January Q5

12 marks Standard +0.8

5 The pilot of a hot air balloon keeps it at a fixed altitude by dropping sand from the balloon. Each grain of sand has mass $m \mathrm {~kg}$ and is released from rest. When a grain has fallen a distance $x \mathrm {~m}$, it has speed $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Each grain falls vertically and the only forces acting on it are its weight and air resistance of magnitude $m k v ^ { 2 } \mathrm {~N}$, where $k$ is a positive constant.

Show that $\left( \frac { v } { g - k v ^ { 2 } } \right) \frac { \mathrm { d } v } { \mathrm {~d} x } = 1$.
Find $v ^ { 2 }$ in terms of $k , g$ and $x$. Hence show that, as $x$ becomes large, the limiting value of $v$ is $\sqrt { \frac { g } { k } }$.
Given that the altitude of the balloon is 300 m and that each grain strikes the ground at $90 \%$ of its limiting velocity, find $k$.

OCR M3 2007 January Q6

12 marks Challenging +1.2

6 \includegraphics[max width=\textwidth, alt={}, center]{f334f6e4-2a60-4647-8b37-e48937c85747-3_446_821_1007_664} Two uniform smooth spheres $A$ and $B$ of equal radius are moving on a horizontal surface when they collide. $A$ has mass 0.4 kg , and $B$ has mass $m \mathrm {~kg}$. Immediately before the collision, $A$ is moving with speed $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an acute angle $\theta$ to the line of centres, and $B$ is moving with speed $u \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at $30 ^ { \circ }$ to the line of centres. Immediately after the collision $A$ is moving with speed $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at $45 ^ { \circ }$ to the line of centres, and $B$ is moving with speed $3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ perpendicular to the line of centres (see diagram).

Find $u$.
Given that $\theta = 88.1 ^ { \circ }$ correct to 1 decimal place, calculate the approximate values of $v$ and $m$.
The coefficient of restitution is 0.75 . Show that the exact value of $\theta$ is a root of the equation $8 \sin \theta - 6 \cos \theta = 9 \cos 30 ^ { \circ }$.

OCR M3 2007 January Q7

13 marks Challenging +1.2

7 \includegraphics[max width=\textwidth, alt={}, center]{f334f6e4-2a60-4647-8b37-e48937c85747-4_721_691_269_726} The diagram shows a particle $P$ of mass 0.5 kg attached to the highest point $A$ of a fixed smooth sphere by a light elastic string. The sphere has centre $O$ and radius 1.2 m . The string has natural length 0.6 m and modulus of elasticity $6.86 \mathrm {~N} . P$ is released from rest at a point on the surface of the sphere where the acute angle $A O P$ is at least 0.5 radians.

(a) For the case angle $A O P = \alpha , P$ remains at rest. Show that $\sin \alpha = 2.8 \alpha - 1.4$.
(b) Use the iterative formula $$\alpha _ { n + 1 } = \frac { \sin \alpha _ { n } } { 2.8 } + 0.5 ,$$ with $\alpha _ { 1 } = 0.8$, to find $\alpha$ correct to 2 significant figures.
Given instead that angle $A O P = 0.5$ radians when $P$ is released, find the speed of $P$ when angle $A O P = 0.8$ radians, given that $P$ is at all times in contact with the surface of the sphere. State whether the speed of $P$ is increasing or decreasing when angle $A O P = 0.8$ radians.

OCR M3 2008 January Q1

6 marks Moderate -0.8

1 A smooth horizontal surface lies in the $x - y$ plane. A particle $P$ of mass 0.5 kg is moving on the surface with speed $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in the $x$-direction when it is struck by a horizontal blow whose impulse has components - 3.5 N s and 2.4 N s in the $x$-direction and $y$-direction respectively.

Find the components in the $x$-direction and the $y$-direction of the velocity of $P$ immediately after the blow. Hence show that the speed of $P$ immediately after the blow is $5.2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. $P$ is struck by a second horizontal blow whose impulse is $\mathbf { I }$.
Given that $P$ 's direction of motion immediately after this blow is parallel to the $x$-axis, write down the component of $\mathbf { I }$ in the $y$-direction.

OCR M3 2008 January Q2

9 marks Standard +0.8

2 \includegraphics[max width=\textwidth, alt={}, center]{7e0f600a-18f1-458b-8549-27fca592b19c-2_515_1065_861_541} Two uniform rods $A B$ and $B C$, each of length 2 m , are freely jointed at $B$. The weights of the rods are $W \mathrm {~N}$ and 50 N respectively. The end $A$ of $A B$ is hinged at a fixed point. The rods $A B$ and $B C$ make angles $\tan ^ { - 1 } \left( \frac { 3 } { 4 } \right)$ and $\beta$ respectively with the downward vertical, and are held in equilibrium in a vertical plane by a horizontal force of magnitude 75 N acting at $C$ (see diagram).

By taking moments about $B$ for $B C$, show that $\tan \beta = 3$.
Write down the horizontal and vertical components of the force acting on $A B$ at $B$.
Find the value of $W$.

OCR M3 2008 January Q3

9 marks Standard +0.8

3 \includegraphics[max width=\textwidth, alt={}, center]{7e0f600a-18f1-458b-8549-27fca592b19c-3_419_921_267_612} Two uniform smooth spheres $A$ and $B$, of equal radius, have masses 6 kg and 3 kg respectively. They are moving on a horizontal surface when they collide. Immediately before the collision the velocity of $A$ has components $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ along the line of centres towards $B$, and $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ perpendicular to the line of centres. $B$ is moving with speed $8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ along the line of centres towards $A$ (see diagram). The coefficient of restitution between the spheres is $e$.

Find, in terms of $e$, the component of the velocity of $A$ along the line of centres immediately after the collision.
Given that the speeds of $A$ and $B$ are the same immediately after the collision, and that $3 e ^ { 2 } = 1$, find $v$.

OCR M3 2008 January Q4

10 marks Standard +0.8

4 A particle of mass $m \mathrm {~kg}$ is released from rest at a fixed point $O$ and falls vertically. The particle is subject to an upward resisting force of magnitude $0.49 m v \mathrm {~N}$ where $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ is the velocity of the particle when it has fallen a distance of $x \mathrm {~m}$ from $O$.

Write down a differential equation for the motion of the particle, and show that the equation can be written as $\left( \frac { 20 } { 20 - v } - 1 \right) \frac { \mathrm { d } v } { \mathrm {~d} x } = 0.49$.
Hence find an expression for $x$ in terms of $v$.

OCR M3 2008 January Q5

11 marks Standard +0.3

5 A particle $P$ of mass $m \mathrm {~kg}$ is attached to one end of a light elastic string of natural length 1.2 m and modulus of elasticity 0.75 mgN . The other end of the string is attached to a fixed point $O$ of a smooth plane inclined at $30 ^ { \circ }$ to the horizontal. $P$ is released from rest at $O$ and moves down the plane.

Show that the maximum speed of $P$ is reached when the extension of the string is 0.8 m .
Find the maximum speed of $P$.
Find the maximum displacement of $P$ from $O$.

OCR M3 2008 January Q6

12 marks Standard +0.8

6 \includegraphics[max width=\textwidth, alt={}, center]{7e0f600a-18f1-458b-8549-27fca592b19c-4_497_524_276_804} A particle $P$ of mass 0.4 kg is attached to one end of a light inextensible string of length 2 m . The other end of the string is attached to a fixed point $O$. With the string taut the particle is travelling in a circular path in a vertical plane. The angle between the string and the downward vertical is $\theta ^ { \circ }$ (see diagram). When $\theta = 0$ the speed of $P$ is $7 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.

At the instant when the string is horizontal, find the speed of $P$ and the tension in the string.
At the instant when the string becomes slack, find the value of $\theta$.

OCR M3 2008 January Q7

15 marks Challenging +1.2

7 A particle $P$, of mass $m \mathrm {~kg}$, is attached to one end of a light elastic string of natural length 3.2 m and modulus of elasticity $4 m g \mathrm {~N}$. The other end of the string is attached to a fixed point $A$. The particle is released from rest at a point 4.8 m vertically below $A$. At time $t \mathrm {~s}$ after $P$ 's release $P$ is ( $4 + x ) \mathrm { m }$ below $A$.

Show that $4 \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } = - 49 x$. $P$ 's motion is simple harmonic.
Write down the amplitude of $P$ 's motion and show that the string becomes slack instantaneously at intervals of approximately 1.8 s . A particle $Q$ is attached to one end of a light inextensible string of length $L \mathrm {~m}$. The other end of the string is attached to a fixed point $B$. The particle is released from rest with the string taut and inclined at a small angle with the downward vertical. At time $t \mathrm {~s}$ after $Q$ 's release $B Q$ makes an angle of $\theta$ radians with the downward vertical.
Show that $\frac { \mathrm { d } ^ { 2 } \theta } { \mathrm {~d} t ^ { 2 } } \approx - \frac { g } { L } \theta$. The period of the simple harmonic motion to which $Q$ 's motion approximates is the same as the period of $P$ 's motion.
Given that $\theta = 0.08$ when $t = 0$, find the speed of $Q$ when $t = 0.25$.

OCR M3 2009 January Q1

8 marks Standard +0.3

1 \includegraphics[max width=\textwidth, alt={}, center]{14403602-94a6-4441-a673-65f9b98180e5-2_385_741_269_701} A particle $P$ of mass 0.5 kg is moving in a straight line with speed $6.3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. An impulse of magnitude 2.6 N s applied to $P$ deflects its direction of motion through an angle $\theta$, and reduces its speed to $2.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ (see diagram). By considering an impulse-momentum triangle, or otherwise,

show that $\cos \theta = 0.6$,
find the angle that the impulse makes with the original direction of motion of $P$.

OCR M3 2009 January Q2

8 marks Standard +0.3

2 \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{14403602-94a6-4441-a673-65f9b98180e5-2_501_752_1133_356} \captionsetup{labelformat=empty} \caption{Fig. 1}

\end{figure} \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{14403602-94a6-4441-a673-65f9b98180e5-2_519_558_1183_1231} \captionsetup{labelformat=empty} \caption{Fig. 2}

\end{figure} Two uniform rods $A B$ and $B C$, of weights 70 N and 110 N respectively, are freely jointed at $B$. The rods are in equilibrium in a vertical plane with $A$ and $C$ at the same horizontal level and $A C = 2 \mathrm {~m}$. The $\operatorname { rod } A B$ is freely jointed to a fixed point at $A$ and the rod $B C$ is freely jointed to a fixed point at $C$. The horizontal distance between $B$ and $A$ is 4 m and $B$ is 4 m below $A C$; angle $B A C$ is obtuse (see Fig. 1). The force exerted on the $\operatorname { rod } A B$ at $B$, by the $\operatorname { rod } B C$, has horizontal and vertical components as shown in Fig. 2.

By taking moments about $A$ for the $\operatorname { rod } A B$ find the value of $X - Y$.
By taking moments about $C$ for the rod $B C$ show that $2 X - 3 Y + 165 = 0$.
Find the magnitude of the force acting between $A B$ and $B C$ at $B$.

OCR M3 2009 January Q3

9 marks Standard +0.8

3 \includegraphics[max width=\textwidth, alt={}, center]{14403602-94a6-4441-a673-65f9b98180e5-3_387_181_274_982} $A$ and $B$ are fixed points with $B$ at a distance of 1.8 m vertically below $A$. One end of a light elastic string of natural length 0.6 m and modulus of elasticity 24 N is attached to $A$, and one end of an identical elastic string is attached to $B$. A particle $P$ of weight 12 N is attached to the other ends of the strings (see diagram).

Verify that $P$ is in equilibrium when it is at a distance of 1.05 m vertically below $A$. $P$ is released from rest at the point 1.2 m vertically below $A$ and begins to move.
Show that, when $P$ is $x \mathrm {~m}$ below its equilibrium position, the tensions in $P A$ and $P B$ are $( 18 + 40 x ) \mathrm { N }$ and $( 6 - 40 x ) \mathrm { N }$ respectively.
Show that $P$ moves with simple harmonic motion of period 0.777 s , correct to 3 significant figures.
Find the speed with which $P$ passes through the equilibrium position. \includegraphics[max width=\textwidth, alt={}, center]{14403602-94a6-4441-a673-65f9b98180e5-3_540_655_1564_744} One end of a light inextensible string of length 0.5 m is attached to a fixed point $O$. A particle $P$ of mass 0.2 kg is attached to the other end of the string. With the string taut and horizontal, $P$ is projected with a velocity of $3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ vertically downward. $P$ begins to move in a vertical circle with centre $O$. While the string remains taut the angular displacement of $O P$ is $\theta$ radians from its initial position, and the speed of $P$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ (see diagram).

OCR M3 2009 January Q5

10 marks Challenging +1.2

5 \includegraphics[max width=\textwidth, alt={}, center]{14403602-94a6-4441-a673-65f9b98180e5-4_369_953_269_596} Two smooth uniform spheres $A$ and $B$, of equal radius, have masses 3 kg and 4 kg respectively. They are moving on a horizontal surface, each with speed $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, when they collide. The directions of motion of $A$ and $B$ make angles $\alpha$ and $\beta$ respectively with the line of centres of the spheres, where $\sin \alpha = \cos \beta = 0.6$ (see diagram). The coefficient of restitution between the spheres is 0.75 . Find the angle that the velocity of $A$ makes, immediately after impact, with the line of centres of the spheres.
[0pt] [10]

OCR M3 2009 January Q6

12 marks Standard +0.8

6 A stone of mass 0.125 kg falls freely under gravity, from rest, until it has travelled a distance of 10 m . The stone then continues to fall in a medium which exerts an upward resisting force of $0.025 v \mathrm {~N}$, where $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ is the speed of the stone $t \mathrm {~s}$ after the instant that it enters the resisting medium.

Show by integration that $v = 49 - 35 \mathrm { e } ^ { - 0.2 t }$.
Find how far the stone travels during the first 3 seconds in the medium.

OCR M3 2009 January Q7

15 marks Standard +0.3

7 A particle of mass 0.8 kg is attached to one end of a light elastic string of natural length 2 m and modulus of elasticity 20 N . The other end of the string is attached to a fixed point $O$. The particle is held at rest at $O$ and then released. When the extension of the string is $x \mathrm {~m}$, the particle is moving with speed $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$.

By considering energy show that $v ^ { 2 } = 39.2 + 19.6 x - 12.5 x ^ { 2 }$.
Hence find
1. the maximum extension of the string,
2. the maximum speed of the particle,
3. the maximum magnitude of the acceleration of the particle.

By taking moments about $B$ for $A B$, show that $W = \sqrt { 2 } T$.
Find the value of $\tan \alpha$.

OCR M3 2010 January Q4

11 marks Challenging +1.2

4 A particle $P$ of mass 0.2 kg travels in a straight line on a horizontal surface. It passes through a point $O$ on the surface with speed $2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. A resistive force of magnitude $0.2 \left( v + v ^ { 2 } \right) \mathrm { N }$ acts on $P$ in the direction opposite to its motion, where $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ is the speed of $P$ when it is at a distance $x \mathrm {~m}$ from $O$.

Show that $\frac { 1 } { 1 + v } \frac { \mathrm {~d} v } { \mathrm {~d} x } = - 1$.
By solving the differential equation in part (i) show that $\frac { - \mathrm { e } ^ { x } } { 3 - \mathrm { e } ^ { x } } \frac { \mathrm {~d} x } { \mathrm {~d} t } = - 1$, where $t$ s is the time taken for $P$ to travel $x \mathrm {~m}$ from $O$.
Hence find the value of $t$ when $x = 1$.

OCR M3 2010 January Q5

11 marks Standard +0.8

5 A light elastic string of natural length 1.6 m has modulus of elasticity 120 N . One end of the string is attached to a fixed point $O$ and the other end is attached to a particle $P$ of weight 1.5 N . The particle is released from rest at the point $A$, which is 2.1 m vertically below $O$. It comes instantaneously to rest at $B$, which is vertically above $O$.

Verify that the distance $A B$ is 4 m .
Find the maximum speed of $P$ during its upward motion from $A$ to $B$. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{08760a55-da6c-41f2-a88a-289ecc227f69-4_351_442_303_479} \captionsetup{labelformat=empty} \caption{Fig. 1}
\end{figure} \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{08760a55-da6c-41f2-a88a-289ecc227f69-4_394_648_260_1018} \captionsetup{labelformat=empty} \caption{Fig. 2}
\end{figure} A light inextensible string of length $0.8 \pi \mathrm {~m}$ has particles $P$ and $Q$, of masses 0.4 kg and 0.58 kg respectively, attached to its ends. The string passes over a smooth horizontal cylinder of radius 0.8 m , which is fixed with its axis horizontal and passing through a fixed point $O$. The string is held at rest in a vertical plane perpendicular to the axis of the cylinder, with $P$ and $Q$ at opposite ends of the horizontal diameter of the cylinder through $O$ (see Fig. 1). The string is released and $Q$ begins to descend. When $O P$ has rotated through $\theta$ radians, with $P$ remaining in contact with the cylinder, the speed of each particle is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ (see Fig. 2).

OCR M3 2010 January Q7

14 marks Challenging +1.2

7 A particle $P$ of mass 0.5 kg is attached to one end of each of two identical light elastic strings of natural length 1.6 m and modulus of elasticity 19.6 N . The other ends of the strings are attached to fixed points $A$ and $B$ on a line of greatest slope of a smooth plane inclined at $30 ^ { \circ }$ to the horizontal. The distance $A B$ is 4.8 m and $A$ is higher than $B$.

Find the distance $A P$ for which $P$ is in equilibrium on the line $A B$. $P$ is released from rest at a point on $A B$ where both strings are taut. The strings remain taut during the subsequent motion of $P$ and $t$ seconds after release the distance $A P$ is $( 2.5 + x ) \mathrm { m }$.
Use Newton's second law to obtain an equation of the form $\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } = k x$. State the property of the constant $k$ for which the equation indicates that $P$ 's motion is simple harmonic, and find the period of this motion.
Given that $x = 0.5$ when $t = 0$, find the values of $x$ for which the speed of $P$ is $2.8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. www.ocr.org.uk after the live examination series.
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OCR M3 2011 January Q1

6 marks Standard +0.3

1 \includegraphics[max width=\textwidth, alt={}, center]{67af8d98-85af-42b1-9e7f-c6380a1f8a3f-2_476_583_258_781} A ball of mass 0.5 kg is moving with speed $22 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in a straight line when it is struck by a bat. The impulse exerted by the bat has magnitude 15 N s and the ball is deflected through an angle of $90 ^ { \circ }$ (see diagram). Find

the direction of the impulse,
the speed of the ball immediately after it is struck.

OCR M3 2011 January Q2

6 marks Standard +0.3

2 A particle of mass 0.4 kg is attached to a fixed point $O$ by a light inextensible string of length 0.5 m . The particle is projected horizontally with speed $6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from the point 0.5 m vertically below $O$. The particle moves in a complete circle. Find the tension in the string when

the string is horizontal,
the particle is vertically above $O$.

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