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

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

m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2} \Rightarrow m_{1} (u_{1} - v_{1}) = m_{2} (v_{2} - u_{2})

............(1) And according to the law of conservation of kinetic energy

\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} (u_{1}^{2} - v_{1}^{2}) = m_{2} (v_{2}^{2} - v_{1}^{2})

.............(2) Now dividing the equation

(u_{1} + v_{1}) = (v_{2} + u_{2}) \Rightarrow u_{1} - u_{2} = v_{2} - v_{1}

Therefore, this is, relative velocity of approach is equal to relative velocity of separation.

Question 3

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	imes 3.5=2	imes 3.14	imes 100	imes 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 4

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

Question 5

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 6

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	imes 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 7

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 8

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^{−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

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 	imes 10+\cfrac{1}{2} 	imes 4 	imes 10^{2}     =0+200=200\ m

Question 10

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 	imes 3.14 	imes 42250 = 265464.58\ km Therefore speed of satellite, v=\dfrac{	ext{distance travelled}}{	ext{time taken}} = \dfrac{265464.58}{24 	imes 60 	imes 60} = 3.07\ km\ s^{-1}