Edexcel IAL Maths M3.2 Elastic Strings and Springs

[Maximum number: 9]

Figure 1

A light elastic spring has natural length l and modulus of elasticity $\lambda$ .
One end of the spring is attached to a point A on a smooth plane.
The plane is inclined at angle $\theta$ to the horizontal, where $\tan\theta=\frac5{12}$ .
A particle P of mass m is attached to the other end of the spring.
Initially P is held at the point B on the plane, where AB is a line of greatest slope of the plane.
The point B is lower than A and AB=2l, as shown in Figure 1.
The particle is released from rest at B and first comes to instantaneous rest at the point C on AB, where AC=0.7l.

(a)

Use the principle of conservation of mechanical energy to show that

$\lambda=\frac{100}{91}mg.$

[ 5 ]

Change in GPE
$=mg\cdot1.3l\sin\theta=0.5lmg$ $\mathrm{EPE}=\frac{\lambda l^2}{2l}\quad\text{or}\quad \frac{\lambda(0.3l)^2}{2l}$
Energy equation from B to A:
$\frac{\lambda l^2}{2l}-\frac{\lambda(0.3l)^2}{2l}=0.5lmg$ $\lambda=mg\frac1{1-0.09}=\frac{100}{91}mg$
A1*

(b)

Find the acceleration of P when it is released from rest at B.

[ 4 ]

Equation of motion:
$T-mg\sin\theta=ma$ $\frac{\lambda l}{l}-mg\sin\theta=ma$ $\frac{100}{91}mg-\frac5{13}mg=ma$ $a=\frac57g$

[Maximum number: 4]

The fixed point A is vertically above the fixed point B, with A B=3 l

A light elastic string has natural length l and modulus of elasticity 4 m g One end of the string is attached to A and the other end is attached to a particle P of mass m

A second light elastic string also has natural length l and modulus of elasticity 4 m g One end of this string is attached to P and the other end is attached to B.

Initially P rests in equilibrium at the point E, where AEB is a vertical straight line.

Show that $A E=\frac{13}{8} l$

mg+4mg\left(\frac{2l-AE}{l}\right)=4mg\left(\frac{AE-l}{l}\right)

13mg=\frac{8mg}{l}AE

AE=\frac{13l}{8}

A1*

AE=\frac{13}{8}l

[Maximum number: 15]

A particle P of mass 1.2 kg is attached to the midpoint of a light elastic string of naturaly length 0.5 m and modulus of elasticity $\lambda$ newtons.

The fixed points A and B are 0.8 m apart on a horizontal ceiling. One end of the string is attached to A and the other end of the string is attached to B.

Initially P is held at rest at the midpoint M of the line AB and the tension in the string is 30 N.

(a)

Show that $\lambda=50$

The particle is now held at rest at the point C, where C is 0.3 m vertically below M. The particle is released from rest.

[ 3 ]

At M, the length of the string is $0.8\text{ m}$ , so the extension is
$0.8-0.5=0.3\text{ m}.$

Using Hooke's law:
$30=\lambda\frac{0.3}{0.5}$ $\lambda=50$

Marking guidance for this part:
Obtain the given value for $\lambda$ from fully correct working.

(b)

Find the magnitude of the initial acceleration of P

[ 6 ]

At C, each half of the string has length $0.5\text{ m}$ , so the extension used is $0.5\text{ m}$ .

$T=\frac{50\times0.5}{0.5}=50$ $2T\cos\theta-1.2g=1.2a,\qquad \cos\theta=\frac35$ $100\times\frac35-1.2\times9.8=1.2a$ $a=40.2\text{ m s}^{-2}$
Marking guidance for this part:
B1 for the correct extension, seen explicitly or used.
M1 for using Hooke's law with $\lambda=50$ and the extension.
A1 for the correct tension equation, following through the extension.
A1 for a correct vertical equation of motion with T resolved.
DM1 for solving the equation of motion.
A1 for 40 or $40.2\text{ m s}^{-2}$ , positive.

(c)

Find the speed of P at the instant immediately before it hits the ceiling.

Figure for Question (c) — Edexcel A-Level Mathematics A2

[ 6 ]

$\text{EPE at }C=\frac12\times50\times\frac{0.5^2}{0.5}$

Energy from C to the ceiling:
$1.2g\times0.3+\frac12\times1.2v^2 =\frac12\times50\times\frac{0.5^2}{0.5} -\frac12\times50\times\frac{0.3^2}{0.5}$ $v^2=\frac1{0.6} \left(25\times\frac{0.5^2}{0.5}-25\times\frac{0.3^2}{0.5}-1.2g\times0.3\right) =7.452$ $v=2.730\ldots=2.7\text{ or }2.73\text{ m s}^{-1}$

B1ft M1 A1 A1 DM1 A1

Marking guidance for this part:
B1ft for either correct elastic potential energy term, following through the extension.
M1 for an energy equation from C to the ceiling with PE, KE and two EPE terms.
A1 for a correct elastic potential energy form.
A1 for a completely correct equation.
DM1 for solving for $v^2$ or v.
A1 for 2.7 or $2.73\text{ m s}^{-1}$ .

[Maximum number: 4]

Figure 1

A light elastic spring has natural length l and modulus of elasticity $\lambda$ .
One end of the spring is attached to a point A on a smooth plane.
The plane is inclined at angle $\theta$ to the horizontal, where $\tan\theta=\frac5{12}$ .
A particle P of mass m is attached to the other end of the spring.
Initially P is held at the point B on the plane, where AB is a line of greatest slope of the plane.
The point B is lower than A and AB=2l, as shown in Figure 1.
The particle is released from rest at B and first comes to instantaneous rest at the point C on AB, where AC=0.7l.

Find the acceleration of P when it is released from rest at B.

Equation of motion:
$T-mg\sin\theta=ma$ $\frac{\lambda l}{l}-mg\sin\theta=ma$ $\frac{100}{91}mg-\frac5{13}mg=ma$ $a=\frac57g$

[Maximum number: 5]

Figure 1

A light elastic spring has natural length l and modulus of elasticity $\lambda$ .
One end of the spring is attached to a point A on a smooth plane.
The plane is inclined at angle $\theta$ to the horizontal, where $\tan\theta=\frac5{12}$ .
A particle P of mass m is attached to the other end of the spring.
Initially P is held at the point B on the plane, where AB is a line of greatest slope of the plane.
The point B is lower than A and AB=2l, as shown in Figure 1.
The particle is released from rest at B and first comes to instantaneous rest at the point C on AB, where AC=0.7l.

Use the principle of conservation of mechanical energy to show that

$\lambda=\frac{100}{91}mg.$

Change in GPE
$=mg\cdot1.3l\sin\theta=0.5lmg$ $\mathrm{EPE}=\frac{\lambda l^2}{2l}\quad\text{or}\quad \frac{\lambda(0.3l)^2}{2l}$
Energy equation from B to A:
$\frac{\lambda l^2}{2l}-\frac{\lambda(0.3l)^2}{2l}=0.5lmg$ $\lambda=mg\frac1{1-0.09}=\frac{100}{91}mg$
A1*

Edexcel A-Level Mathematics A2 M3.2 Elastic Strings and Springs Questions

Exam points

Covered syllabus points

Covered syllabus points

Question 1

Question (a)

Question (b)

Question 2

Question 3

Question (a)

Question (b)

Question (c)

Question 4

Question 5