Q: A ball is thrown vertically upwards with a velocity of \${20 ms^{-1}}\$ from the top of a multi storey building.The height of the point where the ball is thrown 25 m from the ground.How long will it be before the ball hits the ground ? Take \${g= 10 ms^{-2}}\$.

Time to reach maximum height can be obtained from \$v=u+at\$\$0=20+(-10)t\$\$t=2 s\$\$s=ut+0.5at^2=20(2)+0.5(-10)(2)^2= 20 m\$Thus, total distance for maximum height is 45 m\$s=ut+0.5at^2\$\$45=0+0.5(10)(t')^2\$\$t'=3 s\$Total time= 3+2= 5s

Q: A farmer moves along the boundary of a square field of side \$10\$ m in \$40\$ sec What will be magnitude of displacement of the farmer at the end of \$2\$ minutes \$20\$ seconds from his initial position?

Given side of square \$=10 m\$, thus perimeter \$P=40 m\$Time taken to cover the boundary of 40 m =40 sThus in 1 second, the farmer covers a distance of 1 mNow distance covered by the farmer in 2 min 20 seconds = \$1\times 140=140 m\$Now the total number of rotation the farmer makes to cover a distance of 140 meters =\$\dfrac{total\quad distance }{perimeter}\$\$=3.5\$ At this point, the farmer is at a point say B from the origin OThus the displacement \$s=\sqrt{{10}^{2}+{10}^{2}}\$ from Pythagoras theorem.s=\$10\sqrt{2}=14.14 m\$

Q: The position \$x\$ of a particle varies with time as \$x = at^{2} - bt^{3}\$. The acceleration of particle is zero at time \$T\$ equal to

\$x=at^2-bt^3\$\$\Rightarrow\$ Velocity \$=\dfrac{dx}{dt}=2at-3bt^2\$Acceleration\$=\dfrac{d^2x}{dt^2}=\dfrac{d}{dt}(2at-3b^2)=2a-6bt\$or \$a=2a-6bt\$Making \$a=0\$\$\Rightarrow 2a=6bt\$\$\Rightarrow t=dfrac{2a}{6b}=\dfrac{a}{3b}\$.

Q: A stone is thrown in vertically upward direction with a velocity of \$5\ ms^{-1}\$. If the acceleration of the stone during its motion is \$10\ ms^{-2}\$ in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?

Given, initial velocity, \$u=5ms^{−1}\$Final velocity, \$v=0\$Since, u is upward & a is downward, it is a retarded motion. ∴\$a=−10ms^{−2}\$Height attained by stone, \$s=?\$Time take to attain height, \$ t=?\$(i) Using the relation, \$v^2−u^2=2as\$, we have\$s=v^2−u^2/2a\$\$=(0)^2−(5)^2/2×(−10)=1.25m\$(ii) Using the relation, \$v=u+at\$\$0=5+(−10)t\$ or\$t=5/10=0.5s\$

Q: An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.

Distance covered by the satellite in \$24\$ hours\$s = 2\pi r\$ \$=2 \times 3.14 \times 42250 = 265464.58\ km\$Therefore speed of satellite,\$v=\dfrac{\text{distance travelled}}{\text{time taken}} = \dfrac{265464.58}{24 \times 60 \times 60} = 3.07\ km\ s^{-1}\$

Q: A racing car has uniform acceleration of \$4\ ms^{-2}\$. What distance will it cover in \$10\ s\$ after start?

Given: Initial velocity is: \$u=0\$Acceleration is: \$ a=4\ ms^{2} \$Time is: \$ t=10\ s\$Using second equation of motion:\$s=ut+\cfrac{1}{2}at^{2}\$\$s=0 \times 10+\cfrac{1}{2} \times 4 \times 10^{2}\$ \$=0+200=200\ m\$

Question 1

A ball is thrown vertically upwards with a velocity of ${20 ms^{-1}}$ from the top of a multi storey building.The height of the point where the ball is thrown 25 m from the ground.How long will it be before the ball hits the ground ? Take ${g= 10 ms^{-2}}$.

Accepted Answer

Time to reach maximum height can be obtained from $v=u+at$$0=20+(-10)t$$t=2 s$$s=ut+0.5at^2=20(2)+0.5(-10)(2)^2= 20 m$Thus, total distance for maximum height is 45 m$s=ut+0.5at^2$$45=0+0.5(10)(t')^2$$t'=3 s$Total time= 3+2= 5s

Question 2

An athlete completes one round of a circular track of diameter 200 m in 40 s. If he continues to run, what will be the distance covered and the displacement respectively, at the end of 2 minutes 20 s?

Accepted Answer

The diameter of the circular track is given as 200 m. That is, 2r=200 m.From this radius is calculated as Radius, r = 100 m.In 40 sec the athlete complete one round. So, in 2 mins and 20 secs, that is, 140 sec the athlete will complete = 140 / 40 = 3.5 (three and a half) rounds.One round is considered as the circumference of the circular track.The distance covered in 140 sec = $2\pi r\times 3.5=2\times 3.14\times 100\times 3.5=2200\ m$.For each complete round the displacement is zero. Therefore for 3 complete rounds, the displacement will be zero.At the end of his motion, the athlete will be in the diametrically opposite position. That is, displacement = diameter = 200 m.Hence, the distance covered is 2200 m and the displacement is 200 m.

Question 3

A farmer moves along the boundary of a square field of side $10$ m in $40$ sec What will be magnitude of displacement  of the farmer at the end of $2$ minutes $20$ seconds from his initial position?

Accepted Answer

Given side of square $=10 m$, thus perimeter $P=40 m$Time taken to cover the boundary of 40 m =40 sThus in 1 second, the farmer covers a distance of 1 mNow distance covered by the farmer in 2 min 20 seconds = $1\times 140=140 m$Now the total number of rotation the farmer makes to cover a distance of 140 meters =$\dfrac{total\quad distance }{perimeter}$$=3.5$ At this point, the farmer is at a point say B from the origin OThus the displacement $s=\sqrt{{10}^{2}+{10}^{2}}$ from Pythagoras theorem.s=$10\sqrt{2}=14.14 m$

Question 4

A ball is released from the top of a tower of height h metres. It takes T seconds to reach the ground. What is the position of the ball in ${ T }/{ 3 }$ seconds?

Accepted Answer

the acceleration of the ball will be g. Initial velocity will be 0. in T sec. body travels h mts.  by applying equations of motion we get$s= ut +(1/2)gT^2$$h = (1/2)gT^2$       ------[1]in $T/3$ sec       $h_1 = (1/2)gT^2=(1/2)g(\dfrac{T}{3})^2=(1/2)g(\dfrac{T^2}{9})$     -------[2]from $[1]$ and $[2]$ we get $h_1 =\dfrac{h}{9}$ distance from point of release.therefore distance from ground is $h-\dfrac{h}{9} =\dfrac{8h}{9}$

Question 5

The position $x$ of a particle varies with time as $x = at^{2} - bt^{3}$. The acceleration of particle is zero at time $T$ equal to

Accepted Answer

$x=at^2-bt^3$$\Rightarrow$ Velocity $=\dfrac{dx}{dt}=2at-3bt^2$Acceleration$=\dfrac{d^2x}{dt^2}=\dfrac{d}{dt}(2at-3b^2)=2a-6bt$or $a=2a-6bt$Making $a=0$$\Rightarrow 2a=6bt$$\Rightarrow t=dfrac{2a}{6b}=\dfrac{a}{3b}$.

Question 6

A stone is thrown in vertically upward direction with a velocity of $5\ ms^{-1}$. If the acceleration of the stone during its motion is $10\ ms^{-2}$ in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?

Accepted Answer

Given, initial velocity, $u=5ms^{&#8722;1}$Final velocity, $v=0$Since, u is upward & a is downward, it is a retarded motion. &#8756;$a=&#8722;10ms^{&#8722;2}$Height attained by stone, $s=?$Time take to attain height, $ t=?$(i) Using the relation, $v^2&#8722;u^2=2as$, we have$s=v^2&#8722;u^2/2a$$=(0)^2&#8722;(5)^2/2&#215;(&#8722;10)=1.25m$(ii) Using the relation, $v=u+at$$0=5+(&#8722;10)t$ or$t=5/10=0.5s$

Question 7

An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.

Accepted Answer

Distance covered by the satellite in $24$ hours$s = 2\pi r$   $=2 \times 3.14 \times 42250 = 265464.58\ km$Therefore speed of satellite,$v=\dfrac{\text{distance travelled}}{\text{time taken}} = \dfrac{265464.58}{24 \times 60 \times 60} = 3.07\ km\ s^{-1}$

Question 8

A car moves with a speed of 40 km/h for 15 minutes and then with a speed of 60 km/h for the next 15 minutes. The total distance covered by the car is :

Accepted Answer

Speed of car in first 15 minutes = 40 km/hDistance covered = speed $\times $ Time taken60 minutes = 1 hour15 minutes = $\dfrac{15}{60}$ hourDistance covered = $240\times \dfrac{15}{60}=10 km$Speed of car in next 15 minutes = 60 km/hDistance covered = $60\times \dfrac{15}{60}=15 km$$\therefore $ Total distance covered = (10+15)km = 25 km$\therefore $ 25 km is correct

Question 9

A racing car has uniform acceleration of $4\ ms^{-2}$. What distance will it cover in $10\ s$ after start?

Accepted Answer

Given: Initial velocity is: $u=0$Acceleration is: $ a=4\ ms^{2} $Time is: $ t=10\ s$Using second equation of motion:$s=ut+\cfrac{1}{2}at^{2}$$s=0 \times 10+\cfrac{1}{2} \times 4 \times 10^{2}$   $=0+200=200\ m$

Question 10

Derive an expression for the velocity of the two masses ${m}_{1}$ and ${m}_{2}$ with speeds ${u}_{1}$ and ${u}_{2}$ undergoing elastic collision in one direction.

Accepted Answer

Let two bodies of masses ${m_1}$ and ${m_2}$ moving with velocities ${u_1}$ and ${u_2}$ along the same straight line. And consider the two bodies collide and after collision ${v_1}$ and ${v_2}$ be the velocities of two masses.Before collisionMomentum of mass  ${ m_{ 1 } }={ m_{ 1 } }{ u_{ 1 } }$Momentum of mass  ${ m_{ 2 } }={ m_{ 2 } }{ u_{ 2 } }$Total momentum before collision ${p_1} = {m_1} u_1 + {m_2}{u_2}$Kinetic energy of mass  ${m_1} = \frac{1}{2}{m_1}{u_1}^2$Kinetic energy of mass  ${m_2} = \frac{1}{2}{m_2}{u_2}^2$Thus,Total kinetic energy before collision is$K.E=\frac { 1 }{ 2 } { m_{ 1 } }{ u_{ 1 } }^{ 2 }+\frac { 1 }{ 2 } { m_{ 2 } }{ u_{ 2 } }^{ 2 }$After collision Momentum of mass  ${ m_{ 1 } }={ m_{ 1 } }{ v_{ 1 } }$Momentum of mass  ${ m_{ 2 } }={ m_{ 2 } }{ v_{ 2 } }$Total momentum before collision ${ P_{ f } }={ m_{ 1 } }{ v_{ 1 } }+{ m_{ 2 } }{ v_{ 2 } }$Kinetic energy of mass  ${m_1} = \frac{1}{2}{m_1}{v_1}^2$Kinetic energy of mass  ${m_2} = \frac{1}{2}{m_2}{v_2}^2$Total kinetic energy after collision ${K_f} = \frac{1}{2}{m_1}{v_1}^2 + \frac{1}{2}{m_2}{v_2}^2$So, according to the law of conservation of momentum $\begin{array}{l} { m_{ 1 } }{ u_{ 1 } }+{ m_{ 2 } }{ u_{ 2 } }={ m_{ 1 } }{ v_{ 1 } }+{ m_{ 2 } }{ v_{ 2 } } \ \Rightarrow { m_{ 1 } }\left( { { u_{ 1 } }-{ v_{ 1 } } } \right) ={ m_{ 2 } }\left( { { v_{ 2 } }-{ u_{ 2 } } } \right)  \end{array}$  $............(1)$And according to the law of conservation of kinetic energy$\begin{array}{l} \frac { 1 }{ 2 } { m_{ 1 } }{ u_{ 1 } }^{ 2 }+\frac { 1 }{ 2 } { m_{ 2 } }{ u_{ 2 } }^{ 2 }=\frac { 1 }{ 2 } { m_{ 1 } }{ v_{ 1 } }^{ 2 }+\frac { 1 }{ 2 } { m_{ 2 } }{ v_{ 2 } }^{ 2 } \ \Rightarrow { m_{ 1 } }\left( { { u_{ 1 } }^{ 2 }-{ v_{ 1 } }^{ 2 } } \right) ={ m_{ 2 } }\left( { { v_{ 2 } }^{ 2 }-{ v_{ 1 } }^{ 2 } } \right)  \end{array}$ $.............(2)$Now dividing the equation $\begin{array}{l} \left( { { u_{ 1 } }+{ v_{ 1 } } } \right) =\left( { { v_{ 2 } }+{ u_{ 2 } } } \right)  \ \Rightarrow { u_{ 1 } }-{ u_{ 2 } }={ v_{ 2 } }-{ v_{ 1 } } \end{array}$Therefore, this is, relative velocity of approach is equal to relative velocity of separation.