Question
(a) (i) Define the moment of a force about a point.
………………………………………………………………………………………………………………………….
……………………………………………………………………………………………………………………. [1]
(ii) Determine the SI base units of the moment of a force.
base units ………………………………………………… [1]
(b) A uniform rigid rod of length 2.4m is shown in Fig. 1.1.
A = ………………………………………… mm2
(ii) By taking moments about end A of the rod, calculate the tension T.
T = …………………………………………….. N
Answer/Explanation
Ans:
(a)(i)force × perpendicular distance (of line of action of force to the point)
(ii)units:\(kgms^{-2}m\)
\(=kgm^{2}s^{-2}\)
(b)\(W=\rho Vg or W=\rho ALg\)
\(A=5.2/(790\times 2.4\times 9.81\)
\((=2.8\times 10^{-4}(m^{2})\)
\(=2.8\times 10^{2}mm^{2}\)
(c)(i)\((component)=5.2\sin 56^{\circ}=4.3(N)\)
or
\(5.2\cos 34^{\circ}=4.3(N)\)
(ii)\((T\times 2.4)\) or \((4.3\times 1.2)\)or \((4.6\times 1.8)\)
\((T\times 2.4)+(4.3\times 1.2)=(4.6\times 1.8)\)
T=1.3N
Question
Air resistance is negligible.
(a) State the feature of the graph in Fig. 2.2 that shows the block has a constant acceleration.
…………………………………………………………………………………………………………………………… [1]
(b) Use Fig. 2.2 to determine the height of the block above the ground when the string breaks at
time t = 0.
height = …………………………………………….. m
(c) The block has a weight of 0.86N.
Calculate the difference in gravitational potential energy of the block between time t = 0 and
time t = 0.90s.
difference in gravitational potential energy = ……………………………………………… J [2]
(d) On Fig. 2.3, sketch a line to show the variation of the distance moved by the block with time t
from t = 0 to t = 0.20s. Numerical values of distance are not required.
Answer/Explanation
(a) coinstant gradient
(b) (displacement untill 0.20s=\(\frac{1}{2}\times 1.96\times 20=(0.196m)\)
or
(displacement after 0.20s=\(\frac{1}{2}\times 6.86\times 0.70=(2.401m)\)
height=2.401-0.196
=2.2m
(alternative methods are possible using equations of uniformly accelertated motion)
(c)\(\Delta E=mg\Delta h~ or ~W\Delta h\)
\(\Delta E=0.86\times 2.2\)
=1.9J
(d) curve line from the origin
gradient of curved line decreases and is zero at t=0.20s onl;y
(e) acceleration (of free fall) is unchanged/is not dpenedent on mass and (so) no effect
Question
loss of kinetic energy = ……………………………………………… J [2]
(ii) Determine the magnitude of the change in momentum of the ball during the collision.
change in momentum = …………………………………………… Ns [2]
(iii) Show that the magnitude of the average resultant force acting on the ball during the
collision is 4.2N.
(iv) Use the information in (iii) to calculate the magnitude of:
1. the average force of the floor on the ball during the collision
average force = …………………………………………………. N
2. the average force of the ball on the floor during the collision.
average force = …………………………………………………. N
Answer/Explanation
Ans:
\(E=\frac{1}{2}mv^{2} or \frac{1}{2}\times 0.062\times 3.8^{2} or \frac{1}{2}\times 0.062\times 1.7^{2}\)
loss of KE\(=\frac{1}{2}\times 0.062\times (3.8^{2}\times 1.7^{2})\)
=0.36J
(ii)p=mv or \(0.062\times 3.8 or 0.062\times 1.7\)
change in momentum \(=0.062\times (1.7+3.8)\)
=0.34Ns
(iii)average resultant force\(=\frac{0.34}{0.081}=4.2N\)
or
average resultant force\(=0.062\times (1.7+3.8)/0.081=4.2N\)
(iv)average force\(=4.2(0.062\times 9.81)\)
=4.8N
average force=4.8N
Question
(a) Define, for a wire:
(i) stress
………………………………………………………………………………………………………………………….
……………………………………………………………………………………………………………………. [1]
(ii) strain.
………………………………………………………………………………………………………………………….
……………………………………………………………………………………………………………………. [1]
(b) (i) A school experiment is performed on a metal wire to determine the Young modulus of
the metal. A force is applied to one end of the wire which is fixed at the other end. The
variation of the force F with extension x of the wire is shown in Fig. 4.1.
(iv) Each student in the class performs the experiment in (b)(i). The teacher describes the
values of the Young modulus calculated by the students as having high accuracy and
low precision.
Explain what is meant by low precision.
………………………………………………………………………………………………………………………….
……………………………………………………………………………………………………………………. [1]
Answer/Explanation
Ans:
(a)(i)stress =force / cross-sectional area
(ii)strain = extension / original length
b(i) E = FL / Ax
= GL / A
(ii)straight line from origin above the original line
line ends at point (4 small squares,\(F_{1}\))
(iii)1. shaded area below the graph line and between the two vertical dashed lines
2.Remove the force/F/\(F_{2}\) and the wire goes back to original length/zero extension
(iv)values have a large range
Question
A progressive wave Y passes a point P. The variation with time t of the displacement x for the
wave at P is shown in Fig. 5.1.
ratio = …………………………………………………
Answer/Explanation
Ans
(a)\(v=\frac{\lambda }{T}\)
or \(v=f\lambda\) and \(f=\frac{1}{T}\)
\(v=8.0\times 10^{-2}/40\)
\(=0.20ms^{-1}\)
(b)\(I\alpha A^{2}\)
ratio\(=\frac{2^{2}}{4^{2}}\)
=0.25
Question
(a) Describe the conditions required for two waves to be able to form a stationary wave.
…………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………… [2]
(b) A stationary wave on a string has nodes and antinodes. The distance between a node and an adjacent antinode is 6.0cm.
(i) State what is meant by a node.
……………………………………………………………………………………………………………………. [1]
(ii) Calculate the wavelength of the two waves forming the stationary wave.
wavelength = …………………………………………… cm [1]
(iii) State the phase difference between the particles at two adjacent antinodes of the stationary wave.
phase difference = ………………………………………………. °
Answer/Explanation
Ans:
(a)the waves (of the same type) move in opposite directions and overlap
the waves have the same (speed and) frequency/wavelength
(b)(i)zero amplitude
(ii)distance\(=6.0\times 4\)
=24cm
(iii)\(180^{\circ}\)
Question
(a) Define the ohm.
…………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………… [1]
(b) A uniform wire has resistance 3.2Ω. The wire has length 2.5m and is made from metal of
resistivity 460nΩm.
Calculate the cross-sectional area of the wire.
(ii) State an expression for E in terms of I, R and r.
E = ………………………………………………… [1]
(iii) The resistance R of the variable resistor is changed so that it is equal to r.
Determine an expression, in terms of only E and r, for the power P dissipated in the
variable resistor.
P = …………………………………………………
Answer/Explanation
Ans:
(a)\(\frac{volt}{ampere}\)
(b)\(R=\frac{\rho L}{A}\)
\(A=460\times 10^{-9}\times 2.5/3.2\)
\(=3.6\times 10^{-7}m^{2}\)
(c)(i)enery is dissipated in the internal resistance/r
(ii)E=IR+Ir or E=I(R+r)
(iii)\(P=I^{2}R \)or \(P=I^{2}r\)
\(I=\frac{E}{2r}\)
(so)\(P=\frac{E^{2}}{4r}\)
Question
(a) State a similarity and a difference between a down quark and a down antiquark.
similarity: …………………………………………………………………………………………………………………..
difference: …………………………………………………………………………………………………………………
Answer/Explanation
Ans:
(a)similarity:same/equal mass
or
same/equal (magnitude of) charge
or
both fundamental(particles)
diffirence:opposite sign of charge
or
one is matter and the other is antimatter
(b)(i)munber of protons=13 and number of neutrons=12
(ii)charge\(=13\times 1.60\times 10^{-19}(C)=2.1\times 10^{-18}(C)\)
(c)force\(=11\times 10^{3}\times 2.1\times 10^{-18}\)
work done\(=11\times 10^{3}\times 2.1\times 10^{-18}\times 0.04\)
\(=9.2\times 10^{-16}J\)