Note:
Formulas to include:
- F=ma
- a=v-u/t
- surface area of a sphere
- PV = nRT
Q1
The following equation relates the force, F, to the mass, m, and acceleration, a, of an object:
F=ma
Calculate the force for each of the following objects.
1(a) A ball with mass 0.5 kg with an acceleration of 10 m/s^2
Answer type: simple
ANSWER: 5 (kgm/s^2)
Workings:
Force, F=ma=0.5\times10=5 kgm/s^2
2 marks
1(b) A person of mass 70 kg with an acceleration of 2.5 m/s^2
Answer type: simple
ANSWER: 175 (kgm/s^2)
Workings:
Force, F=ma=70\times 2.5=175 kgm/s^2
2 marks
1(c) A car of mass 1500 kg moving with an acceleration of 8 m/s^2
Answer type: simple
ANSWER: 12000 (kgm/s^2)
Workings:
Force, F=ma=1500\times 8=12000 kgm/s^2
2 marks
Q2
The following equation describes the surface area of a sphere
area = 4\pi r^2
Where r is the radius of the circle.
Calculate the surface area of a sphere with a radius of 4 cm
Use the value \pi = 3.14
Give your answer to 2 decimal places.
Answer type: simple
Answer: 201.06 (cm^2)
Workings:
\begin{aligned}area&=4\pi r^2 \\ &=4\times3.14 \times 4^2 \\ &=4\times3.14 \times 16 \\ &= 64\times3.14 = 201.06\end{aligned}Q3
The acceleration, a of a moving object may be calculated using the following equation:
a=\dfrac{v-u}{t}
Where v is the final speed in m/s,
u is the initial speed in m/s,
t is the time period is s
Calculate the acceleration of the following objects.
3(a) A person accelerating from 0 m/s to 4 m/s in a time of 1.6 s
Give your answer to 1 decimal place.
Answer type: simple
Answer: 2.5 (m/s^2)
Workings:
\begin{aligned}a &= \dfrac{v-u}{t} \\ &= \dfrac{4-o}{1.6} \\ &= \dfrac{4}{1.6} \\ &= 2.5 \text{m/s}^2\end{aligned}2 marks
3(b) An ice hockey puck, with an initial speed of 2.5 m/s and a final speed of 22.6.
The time period is 0.03 seconds.
Answer type: simple
Answer: 670 (m/s^2)
Workings:
\begin{aligned}a &= \dfrac{v-u}{t} \\ &= \dfrac{22.6-2.5}{0.03} \\ &= \dfrac{20.1}{0.03} \\ &= 670 \text{m/s}^2\end{aligned}2 marks
3(c) A train which accelerates from stationary to a speed of 40 m/s over the course of 1 minute.
Give your answer to 2 decimal places.
Answer type: simple
Answer: 0.66 (m/s^2)
Workings:
1 minute = 60 seconds
\begin{aligned}a &= \dfrac{v-u}{t} \\ &= \dfrac{40-0}{60} \\ &= \dfrac{40}{60} \\ &= 0.66 \text{ m/s}^2 (\text{2 d.p.})\end{aligned}Q4
The following formula is used to calculate the volume of a pyramid:
\text{Volume} = \dfrac{1}{3}\times A \times h
Where A is the area of the base,
h is the height of the pyramid
The Great Pyramid of Giza in Egypt is a square based pyramid.
The area of the base is approximately 53000 m^2
The height of the pyramid is 147 m.
Calculate the volume of the pyramid.
Type: simple
Answer: 2592100
Workings:
\begin{aligned}\text{Volume} &= \dfrac{1}{3}\times 53000 \times 147 &= \dfrac{53000 \times 147}{3} &= \dfrac{7776300}{3}=2592100 \text{ m}^3 \end{aligned}3 marks
Q5
The gravitational field strength, g, acting on an object is given by the following equation:
g=\dfrac{F}{m}
Where F is the force acting on the object in Newtons (N),
m is the mass of the object in kg.
Calculate the gravitational field strength acting on a 70 kg person if the force acting on them is 687 N
Give your answer to 2 decimal places.
Answer type: simple
Answer: 9.81 (m/s^2)
Workings:
g= \dfrac{F}{m} = \dfrac{687}{70}=9.81 m/s^2
2 marks
Q6
The following equation describes the volume of a gas under specific conditions:
V=\dfrac{nRT}{P}
Where V= is the volume of the gas in m^3,
n= the number of moles (a fixed amount) of gas,
R=8.31
T is the temperature of the gas in Kelvin (K),
P is the pressure of the gas in Pascals (Pa),
Calculate the volume of 2 moles a gas at a temperature of 298 K when the pressure is 100000 Pa
Give your answer to 2 decimal places.
Answer type: simple
ANSWER: 0.05 (m^3)
Workings:
The volume can be calculated by substituting in the values n=2, R=8.31 and P=100000 as follows:
\begin{aligned}V &= \dfrac{nRT}{P} \\ &= \dfrac{2\times8.31\times298}{100000} \\ &= 0.0495... \\ &= 0.05 (2 \text{d.p.})\end{aligned}3 marks