Question

$\text{A}.$ State the principle behind the working of a hydraulic machine and explain the working of hydraulic brakes. $\left[3\text{marks}\right]$


$\text{B}.\text{(i)}$ Write the equations of motion for a body that is being uniformly accelerated. $\left[1\text{mark}\right]$
$\text{(ii)}$ A car moving with a constant acceleration covers the $60\text{m}$ distance between two points
in $6\text{s}$. Its speed as it passes the second point was $15\text{m/s}.$
Find:(a) Its speed at the first point.
(b) Its acceleration.
(c) Its distance prior to the first point at which the body was at rest. $\left[2\text{marks}\right]$


$\text{C}.\text{(i)}$ State the universal law of gravitation.$\left[2\text{marks}\right]$
$\text{(ii)}$ Two identical blocks of mass $10\text{kg}$ are left $1\text{m}$ apart on a smooth frictionless surface. What is the resulting acceleration of the blocks? $\left[2\text{marks}\right]$

Answer 1

Solution:
$\text{A}.$ The principle behind the working of a hydraulic machine is Pascal's law.
Statement: The pressure exerted anywhere in a confined liquid is transmitted equally and undiminished in all directions throughout the liquid. $\left[1\text{mark}\right]$

The working of hydraulic brakes is as follows:

• When we apply a little force on the brake pedal with our foot, the master piston moves inside the master cylinder, and the resulting pressure is transmitted through the brake oil to act on a piston of the larger area.
• A large force thus acts on the piston, expanding the brake shoes against the brake lining.
• The friction between the brake lining and brake shoe is then responsible for retarding the motion of the vehicle.​ $\left[2\text{marks}\right]$

$\text{B}.$
$\text{(i)}$ The equation of uniformly accelerated motion are:

1. $v=u+at$
2. $\text{S}=ut+\frac{1}{2}a{t}^2$
3. $v^2-u^2=2a\text{S}$ $\left[1\text{mark}\right]$

$\text{(ii)}$ Given,
distance between the points, $\text{S}=60\text{m}$
velocity at the second point, $v=15\text{m/s}$
time taken, $\text{t}=6\text{s}$
from the first equation, $v=u+at$
$15=u+a\times 6$ ----$1$
from the third equation of motion, $v^2-u^2=2a\text{S}$
${15}^2-u^2=2a\times 60$-----$2$
solving $1$ and $2$ for $u$ and $a$,
we get (a) $u=5\text{m/s}$, (b) $a=\frac{5}{3}{\text{m/s}}^2$ $\left[1\text{mark}\right]$
using the third equation of motion, $v^2-u^2=2a\text{S}$
(c) the distance prior to the first point, $5^2-0^2=2\times \frac{5}{3}\times \text{S}$
$\text{S}=\frac{25\times 3}{2\times 5}=7.5\text{m}$ $\left[1\text{mark}\right]$

$\text{C}.$
$\text{(i)}$ Universal law of gravitation: Newton’s universal law of gravitation states that every particle attracts every other particle in the universe with a force that is directly proportional to the product of the masses and inversely proportional to the square of the distance between them. $\left[1\text{mark}\right]$
$\text{F}=\frac{\text{G}{m}_1{m}_2}{r^2}$,
Where G is the universal gravitational constant
$\text{G}=6.67\times {10}^{-11}{\text{N m}}^2/{\text{kg}}^2$
$m_1$ and $m_2$ are the massess of the bodies.
$r$ is the distance between the bodies. $\left[1\text{mark}\right]$

$\text{(ii)}$ When two identical blocks are left on a smooth horizontal surface, they will come together due to the action of gravitational attraction between the two bodies. $\left[1\text{mark}\right]$
The force of gravitation will be,
$\text{F}=\frac{\text{G}\times m\times m}{r^2}=\frac{6.67\times {10}^{-11}\times 10\times 10}{1^2}$
$=6.67\times {10}^{-9}\text{N}$
We know, $\text{F}=m\times a$
$\Rightarrow a=\frac{\text{F}}{m}=6.67\times {10}^{-11}{\text{m/s}}^2$ $\left[1\text{mark}\right]$