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HENRYTAIGO

Cambridge International AS and A Level

Name of student HENRYTAIGO Date
Adm. number Year/grade HenryTaigo Stream HenryTaigo
Subject Mechanics 2 (M2) Variant(s) P41, P42, P43
Start time Duration Stop time

Qtn No. 1 2 3 4 5 6 Total
Marks 8 7 8 8 8 9 48
Score

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Question 1 Code: 9709/51/M/J/15/5, Topic: -

A particle $P$ of mass $0.3 \mathrm{~kg}$ is attached to one end of a light elastic string of natural length $0.9 \mathrm{~m}$ and modulus of elasticity $18 \mathrm{~N}$. The other end of the string is attached to a fixed point $O$ which is $3 \mathrm{~m}$ above the ground.

$\text{(i)}$ Find the extension of the string when $P$ is in the equilibrium position. $[2]$

$P$ is projected vertically downwards from the equilibrium position with initial speed $6 \mathrm{~m} \mathrm{~s}^{-1}.$ At the instant when the tension in the string is $12 \mathrm{~N}$ the string breaks. $P$ continues to descend vertically.

$\text{(ii)}~~$ $\text{(a)}$ Calculate the height of $P$ above the ground at the instant when the string breaks. $[2]$

$\text{(b)}$ Find the speed of $P$ immediately before it strikes the ground. $[4]$

Question 2 Code: 9709/52/M/J/15/5, Topic: -

 

A uniform solid cube with edges of length $0.4 \mathrm{~m}$ rests in equilibrium on a rough plane inclined at an angle of $30^{\circ}$ to the horizontal. $A B C D$ is a cross-section through the centre of mass of the cube, with $A B$ along a line of greatest slope. $B$ lies below the level of $A$. One end of a light elastic string with modulus of elasticity $12 \mathrm{~N}$ and natural length $0.4 \mathrm{~m}$ is attached to $C$. The other end of the string is attached to a point below the level of $B$ on the same line of greatest slope, such that the string makes an angle of $30^{\circ}$ with the plane (see diagram). The cube is on the point of toppling. Find

$\text{(i)}$ the tension in the string, $[3]$

$\text{(ii)}$ the weight of the cube. $[4]$

Question 3 Code: 9709/53/M/J/15/5, Topic: -

 

A uniform triangular prism of weight $20 \mathrm{~N}$ rests on a horizontal table. $A B C$ is the cross-section through the centre of mass of the prism, where $B C=0.5 \mathrm{~m}, A B=0.4 \mathrm{~m}, A C=0.3 \mathrm{~m}$ and angle $B A C=90^{\circ}$. The vertical plane $A B C$ is perpendicular to the edge of the table. The point $D$ on $A C$ is at the edge of the table, and $A D=0.25 \mathrm{~m}$. One end of a light elastic string of natural length $0.6 \mathrm{~m}$ and modulus of elasticity $48 \mathrm{~N}$ is attached to $C$ and a particle of mass $2.5 \mathrm{~kg}$ is attached to the other end of the string. The particle is released from rest at $C$ and falls vertically (see diagram).

$\text{(i)}$ Show that the tension in the string is $60 \mathrm{~N}$ at the instant when the prism topples. $[3]$

$\text{(ii)}$ Calculate the speed of the particle at the instant when the prism topples. $[5]$

Question 4 Code: 9709/51/O/N/15/5, Topic: -

A particle $P$ of mass $0.2 \mathrm{~kg}$ is attached to one end of a light elastic string of natural length $0.75 \mathrm{~m}$ and modulus of elasticity $21 \mathrm{~N}$. The other end of the string is attached to a fixed point $A$ which is $0.8 \mathrm{~m}$ vertically above a smooth horizontal surface. $P$ rests in equilibrium on the surface.

$\text{(i)}$ Find the magnitude of the force exerted on $P$ by the surface. $[2]$

$P$ is now projected horizontally along the surface with speed $3 \mathrm{~m} \mathrm{~s}^{-1}$.

$\text{(ii)}$ Calculate the extension of the string at the instant when $P$ leaves the surface. $[3]$

$\text{(iii)}$ Hence find the speed of $P$ at the instant when it leaves the surface. $[3]$

Question 5 Code: 9709/52/O/N/15/5, Topic: -

Question 6 Code: 9709/53/O/N/15/5, Topic: -

A particle $P$ of mass $0.5 \mathrm{~kg}$ is projected vertically upwards from a point on a horizontal surface. A resisting force of magnitude $0.02 v^{2} \mathrm{~N}$ acts on $P$, where $v \mathrm{~m} \mathrm{~s}^{-1}$ is the upward velocity of $P$ when it is a height of $x \mathrm{~m}$ above the surface. The initial speed of $P$ is $8 \mathrm{~m} \mathrm{~s}^{-1}$.

$\text{(i)}$ Show that, while $P$ is moving upwards, $v \displaystyle\frac{\mathrm{d} v}{\mathrm{~d} x}=-10-0.04 v^{2}$. $[2]$

$\text{(ii)}$ Find the greatest height of $P$ above the surface. $[3]$

$\text{(iii)}$ Find the speed of $P$ immediately before it strikes the surface after descending. $[4]$

Worked solutions: P1, P3 & P6 (S1)

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