Question Answer (5 Marks)

Question 15 Marks

A body of $2\ kg$ falls from rest. What will be its kinetic energy during the fall at the end of $2 s$ ? (Assume $g =10\ m / s ^2$ )

Answer

Mass of body $=2 kg$
Initial velocity $u =0$
Time taken $=2 s$
Acceleration due to gravity, $g=10 m / s ^2$
Final velocity v
Using first equation of motion
$v=u+g t$
$=0+10 \times 2=20 m / s$
$KE=\frac{1}{2} m \times v^2$
$=\frac{1}{2} \times 2 \times 20^2=400 J$

View full question & answer→

Question 25 Marks

State and explain the law of conservation of energy with an example.

Answer

Law of conservation of energy: Law of conservation of energy states that whenever energy changes from one form to another form, the total amount of energy remains constant. Energy can never be created nor destroyed, it transforms from one form to another.
Example: When electrical energy is converted into light energy in an electric bulb, then some energy is wasted as heat during conversion but the total energy remains the same.

View full question & answer→

Question 35 Marks

A ball of mass $200\ g$ falls from a height of $5$ metres. What is its kinetic energy when it just reaches the ground? $(8=$ $9.8 m / s$ ).

Answer

Mass $=200 g=0.2 kg$
Height $=5 m$
Initial velocity $u =0$
Acceleration due to gravity, $g=9.8 m / s ^2$
Final velocity, $v$
Using third equation of motion,
$v^2-u^2=2 g s$
$v ^2-0=2 \times 9.8 \times 5$
$v^2=98 \ldots$ (i)
Kinetic energy $=\frac{1}{2} m \times v ^2$
Put the value of $v^2$ from equ. (i)
Kinetic energy $=\frac{1}{2} \times 0.2 \times 98=9.8 J$

View full question & answer→

Question 45 Marks

Two objects having equal masses are moving with uniform velocities of 2m/s and 6m/s respectively. Calculate the ratio of their kinetic energies.

Answer

Let masses of the two objects be m.
$\text{v}_1 = 2\text{m}/\text{s}$
$\text{v}_2 = 6\text{m}/\text{s}$
$\text{KE}_1=\frac{1}{2}\text{m}\times\text{v}_1^2$
$=\frac{1}{2}\text{m}\times2^2$
$\text{KE}_2=\frac{1}{2}\text{m}\times\text{v}_2^2$
$=\frac{1}{2}\text{m}\times6^2$
$\text{Ratio}=\frac{\text{KE}_1}{\text{KE}_2}$
$=\frac{2^2}{6^2}=\frac{4}{36}=\frac{1}{9}$

View full question & answer→

Question 55 Marks

A bullet of mass $15\ g$ has a speed of $400\ m/s$. What is its kinetic energy? If the bullet strikes a thick target and is brought to rest in $2\ cm$, calculate the average net force acting on the bullet. What happens to the kinetic energy originally in the bullet?

Answer

Mass, $m =15 g=0.015 kg$
Velocity, $v=400 m / s$
$KE=\frac{1}{2} mv^2$
$=\frac{1}{2} \times 0.015 \times 400^2=1200 J$
Distance covered before stopping, $s =2 cm=0.02 cm$
Here, initial velocity, $u=400 m / s$
$\text { Final velocity }=0$
$v^2-u^2=2 as$
$0^2-400^2=2 \times a \times 0.02$
$a=-4 \times 10^6 m / s^2 \text { (Negative sign shows retardation) }$
$F=ma$
$=0.015 \times-4 \times 10^6=6 \times 10^4 N$
Kinetic energy of the bullet is mainly converted into heat energy (by friction).

View full question & answer→

Question 65 Marks

Explain the transformation of energy in the following cases:

A ball thrown upwards.
A stone dropped from the roof of a building.

Answer

When a ball is thrown upwards its kinetic energy gradually converts into potential energy and potential energy becomes maximum at the maximum height attained by the ball.
When a stone is dropped from the roof of the building its potential energy gradually converts into kinetic energy and kinetic energy becomes maximum when the stone is just above the ground.

View full question & answer→

Question 75 Marks

Calculate the work done by the brakes of a car of mass $1000\ kg$ when its speed is reduced from $20\ m/s$ to $10\ m/s$?

Answer

Mass of car $=1000 kg$
Initial velocity $u =20 m / s$
Final velocity $v =10 m / s$
Retardation $=a$
Distance covered $=s$
Using third equation of motion
$v^2-u^2=2 a s$
$10^2-20^2=2 a s$
$\text { as }=-\frac{300}{2} \ldots \text { (i) }$
$\text { as }=-150 \ldots \text { (i) }$
Work done $W=F \times s$
But $F=m \times a$
So, $W=m \times a \times s$
Put the value of 'as' from equation (i)
$W=1000 \times-150=-150000=-150 kJ$
Negative sign implies that force of brakes acts opposite to the direction of motion.

View full question & answer→

Question 85 Marks

The following data was obtained for a body of mass $1\ kg$ dropped from a height of 5 metres:

	Distance above ground	Velocity
i.	$5\ m$	$0\ m/s$
ii.	$3.2\ m$	$6\ m/s$
iii.	$0\ m$	$10\ m/s$

Show by calculations that the above data verifies the law of conservation of energy (Neglect air resistance). $(g = 10m/s^2)$.

Answer

$\text { Mass of body }=1 kg$
$g=10 m / s^2$
$PE=m \times g \times h$
$=1 \times 10 \times h$
$=10 \times h$
$KE=\frac{1}{2} mv^2=\frac{1}{2} \times 1 \times v^2=\frac{v^2}{2}$
$\text { At } h=5 m$
$v=0 m / s$
$PE=10 \times 5=50 J$
$KE=0 J$
$\text { Total energy }=PE+KE$
$\text { At } h=3.2 m$
$v=m / s$
$PE=10 \times 3.2=32 J$
$KE=\frac{6^2}{2}=18 J$
Total energy $= PE + KE$
$32+18=50$
At $h =0 m$
$v=10 m / s$
$PE=0 J$
$KE=\frac{10^2}{2}=50 J$
Total energy $= PE + KE =50 J$
The total energy in all three cases is constant. This proves the law of conservation of energy.

View full question & answer→

Question 95 Marks

On a level road, a scooterist applies brakes to slow down from a speed of $10\ m/s$ to $5\ m/s$. If the mass of the scooterist and the scooter be $150\ kg$, calculate the work done by the brakes. (Neglect air resistance and friction)

Answer

Mass of scooter + scooterist $= 150\ kg$
Initial velocity $u = 10\ m/s$
Final velocity $v = 5\ m/s$
Retardation $= a$
Distance covered $= s$
Using third equation of motion
$v^2 - u^2 = 2\ as$
$5^2 - 10^2 = 2\ as$
$\text{as} =-\frac{75}{2}\ ...(\text{i})$
Work done $W = F \times s$
But $F = m \times a$
So, $W= m \times a \times s$
Put the value of $‘as’$ from eq(i)
$\text{W} = 150\times\Big(-\frac{75}{2}\Big) = -5625 \text{J}$
Neagtive sign implies that force of brakes acts opposite to the direction of motion.

View full question & answer→

Science Question Answer (5 Marks)

Test yourself on this topic