Example 3.3.1.
A moving object may or may not have
- momentum.
- velocity.
- kinetic energy.
- potential energy.
Hint.
Think of what if the object is moving on the ground.
Solution.
d.
Section 3.3 Examples
Example 3.3.2.
When two or more objects collide their
- total momentum remains the same.
- total kinetic energy remains the same.
- momentum of each object never changes.
- kinetic energy of each object never changes.
Hint.
Read the conservation principles.
Solution.
a.
Example 3.3.3.
Two objects of same volume but different masses are bing dropped from the top of a building onto the ground. Which of the following quantities remains the same when they have fallen halfway down? Neglect air resistance.
- kinetic energy.
- potential energy.
- speed.
- acceleration due to gravity.
Hint.
Read section Subsubsection 2.1.1.1
Solution.
c. and d.
Example 3.3.4.
If a book of mass 2 kg is hold at 2m above the ground for 40 seconds. How much is the work done?
Hint.
Read section Section 3.1
Answer.
0.
Solution.
Force has not displaced the object, so distance =0. Hence
\begin{equation*}
W=fd=0.
\end{equation*}
Example 3.3.5.
Find the velocity of 400 g bullet if its kinetic energy is 50 kJ.
Answer.
500 m/s.
Solution.
Given: m=400g = 0.4 kg, KE = 50 kJ = \(50\times 1000\, J = 50000\,J, v=?\)
\begin{align*}
KE \amp =\frac{1}{2}mv^2\\
50000 \amp =\frac{1}{2}\times 0.4 \times v^2\\
v^2 \amp =\frac{50000\times 2}{0.4}\\
\therefore v \amp =\sqrt{250000} = 500
\end{align*}
Example 3.3.6.
Find the linear momentum of a 50 kg dog running at 5 m/s.
Solution.
Given: m = 50 kg, v = 5 \(m/s,\) p =?
\begin{align*}
p \amp =mv\\
p \amp =50\times 5\\
\therefore p \amp =250
\end{align*}
Answer.
\(250 \,kgm/s.\)
Example 3.3.7.
Find the recoil velocity of a rocket of mass 100 kg when it ejects 50 g of gas from its propulsion pistol at a speed of 50 m/s.
Hint.
Use conservation of momentum principle.
Solution.
Given:\(m_1=100 \,kg, m_2= 50 \,g=0.05\,kg, v_1= ?, v_2=50 \,m/s.\)
\begin{align*}
p_1\amp =p_2\\
or, m_1v_1\amp =m_2v_2\\
or, v_2\amp =\frac {m_2v_2}{m_1} \\
\therefore v_2\amp =\frac {0.05\times 50}{100} = 0.025\, m/s
\end{align*}
Example 3.3.8.
What is the formula for calculating potential energy?
- \(\displaystyle PE = \frac{1}{2}mv^2\)
- PE = mgh
- PE = Fd
- PE = \frac{P}{t}
Solution.
b.
Example 3.3.9.
What is the SI unit of energy?
- Joule
- Watt
- Newton
- Pascal
Solution.
a.
Example 3.3.10.
A box weighing 50 kg is lifted to a height of 3 meters. How much work was done against gravity? Assume that the acceleration due to gravity is \(9.8 \,m/s^2\text{.}\)
Solution.
The work done against gravity when lifting the box is equal to the change in gravitational potential energy. We can use the following formula to find the work done:
\begin{equation*}
W = F \times d
\end{equation*}
where force is equal to the weight of the box (mass x gravity), and distance is the height the box is lifted. Plugging in the given values, we get:
\begin{equation*}
W = 50 \,kg \times 9.8 \,m/s^2 = 490 \,N
\end{equation*}
\begin{equation*}
\therefore \quad W = 490 \,N \times 3 \,m = 1470 \,J
\end{equation*}
Therefore, the work done against gravity when lifting the box weighing 50 kg to a height of 3 meters is 1470 joules.
Example 3.3.11.
If a machine does 500 joules of work in 10 seconds, what is its power output?
- 50 W
- 100 W
- 200 W
- 500 W
Solution.
a.
Example 3.3.12.
If a force of 20 newtons is applied to an object and it moves 4 meters in the direction of the force, how much potential energy does it gain?
- 4 joules
- 16 joules
- 80 joules
- 320 joules
Solution.
c.
Example 3.3.13.
Which of the following statements is true about the law of conservation of energy?
- Energy can be created but not destroyed.
- Energy can be destroyed but not created.
- Energy cannot be created or destroyed, only transformed from one form to another.
- Energy can be created or destroyed, but only in certain circumstances.
Solution.
c.
