1(a) Determine the SI base units of stress. Show your working.[2]
1(b) A beam is clamped so that the beam is horizontal.A mass of 500g is hung from end Q and the beam bends slightly.
The length of the beam from the edge of the clamp R to end Q is 60cm.The width b of the beam is 30mm and the thickness d of the beam is 5mm.The material of the beam has a Youngs modulus E.
The mass M is made to oscillate vertically.The time period T of the oscillation is 0.58s.
The period T is given by the expression
T=2π√(4Ml³/Ebd³)
(i)Determine E in GPa.[3]
(ii)The quantities used to determine E should be measured with accuracy and precision
1.Explain the difference between accuracy and precision.
accuracy_____
precision_____[2]
2.In a particular experiment the quantities l and T are measured with the same percentage uncertainty.State and explain which of these two quantities contributes more of the uncertainty in the value of E.[1]
Reveal answer
1(a)stress=force/area or kgms-²/m²
=kgm-¹s-²
1(b)(i)0.58=2π x [(4x0.5x0.6³)/(E x0.03x0.005³)]^½
E=[4π²x4x0.5x0.6³]/[(0.58)²x0.03x(0.005)³]
1.35x10^10Pa
14(13.5)GPa
1(b)(ii)1.accuracy is determined by the closeness of the value(s) or measurement(s) to the true value
precision is determined by the range of the values or measurements
1(b)(ii)2. l is (cubed so) 3 x (percentage/fractional) uncertainty and T is (square so) 2 x (percentage/fractional) and so l contributes more.
2(a)A cylinder is made from a material of density 2.7g/cm³.The cylinder has a diameter of 2.4cm and length 5.0cm
Show that the cylinder has weight 0.6N [3]
(b)The cylinder in (a) is hung from the end A of a non-uniform bar AB.The bar has a length of 50cm and has a weight 0.25N.The centre of gravity of the bar is 20cm from B.The bar is pivoted at P.The pivot is 12cm from B.
An object X is hung from the end B.The weight of X is adjusted until the bar is horizontal and in equilibrium.
(i)explain what is meant by the centre of gravity [1]
(ii)calculate the weight of X [3]
(c)The cylinder is now immersed in water.An upthrust acts on the cylinder and the bar is not in equilibrium.
(i)Explain the origin of the upthrust [2]
(ii)Explain why the weight of X must be reduced inorder to obtain equilibriun for AB [1]
Reveal answer
density=m/V
V=πd²L/4 or πr²L
weight=2.7x10³xπ(1.2x10-²)²x5.0x10-²x9.81=0.60N
2(b)(i)the point from where all the weight(of a body) seems to act
2(b)(ii)W x 12
(0.25x8)+(0.6x38)
W=(2+22.8)/12
2.1(2.07)N
2(c)(i)pressure changes with depth(in water)
or
pressure on bottom(of cylinder) different from pressure on top
2(c)(ii)anticlockwise moment reduced and reducing the weight of X reduces clockwise moment
or
anticlockwise moment reduced so clockwise moment now greater than(total) anticlockwise moment
3(a)Describe the doppler effect. [1]
3(b)A car travels with a constant velocity along a straight road.The car horn with a frequency of 400Hz is sounded continously.A stationary observer on the roadside hears the sound from the horn at a frequency of 360Hz.
The speed of sound is 340m/s
Determine the magnitude v and the direction,of the velocity of the car relative to the observer. [3]
Reveal answer
3(a)Observed frequency is different to source frequency when source moves relative to the observer
3(b) 360=(400x340)/(340+-v)
v=38(37.8)m/s
away(from the observer)
4(a)Explain how a satellite may be in a circular orbit around a planet. [2]
(b)The Earth and the Moon may be considered to be uniform spheres that are isolated in space.The earth has radius R and mean density p.The moon mass m is in a circular orbit about the earth with radius of orbit nR.
The moon makes one complete orbit of the Earth in timeT.
Show that the density p of the Earth is given by the expression
p=3πn³/GT². [4]
(c)The radius R of the Earth is 6.38x10³km and the distance between the centre of the Earth and the centre of the Moon is 3.8x10^5km.
The period T of the orbit of the moon about the Earth is 27.3 days.
Use the expression in (b) to calculate p.
Reveal answer
4(a)gravitational force (of attraction between satellite and planet) provides centripetal force (on satellite and planet).
(b)M=(4/3)x πR³p
w=2π/T or 2πnR/T
GM/(nR)²=nRw² or v²/nR
substitution clear to give p=3πn³/GT²
(c)n=(3.84x10^5)/(6.38x10³)=60.19 or 60.2
p=3π x 60.19³/[(6.67x10-¹)x(27.3x24x3600)²]
p=5.54x10³kg/m³