Question 1

A particle moving with an initial velocity

\hat{i} - 4 \hat{j} + 10 \hat{k}

has acceleration

\hat{i} + \hat{j} - 2 \hat{k}

. Its velocity at the end of 2 seconds, points along the unit vector:

Accepted Answer

As we know  v=u+at v=(\hat i-4\hat j+10\hat k)+(\hat i+\hat j-2\hat k)	imes 2 v=(\hat i-4\hat j+10\hat k)+(2\hat i+2\hat j-4\hat k) v=3\hat i-2\hat j+6\hat k Unit vector is given as, |v|=\sqrt{3^2+(-2)^2+6^2} |v|=\sqrt{49} |v|=7 So, Velocity at the end of 2 sec, points along unit vector =\dfrac{v}{|v|}                                                                                                =\dfrac{3\hat i-2\hat j+6\hat k}{7}                                                                                                =\dfrac{1}{7}(3\hat i-2\hat j+6\hat k)

Question 2

A particle moves in XY plane according to the law  x  =  {a}{sin(t)}  and  y  =  {a}{(1-cos(t))}  where a is constant. The particle traces :

Accepted Answer

x=a\sin t y=a(1-\cos t) \cfrac{dx}{dt}=a\cos t \cfrac{dy}{dt}=a\sin t The particle is under the action and two components  V^2=(a\cos t)^2+(a\sin t )^2=a^2(\sin^2 t+\cos^2t) V=a  constant.Hence the particle trace is (A) parabola.

Question 3

Which of the following displacement  (x)  vs. time  (t)  graphs is not possible?

Accepted Answer

A is not possible, because at a particular time  t , displacement cannot have two values.

Question 4

A man is going in a topless car with a velocity of

10.8 k m / h

. It is raining vertically downwards. He has to hold the umbrella at an angle of

5 3^{o}

to the vertical to protect himself from rain. The actual speed of the rain is

(c o s 5 3^{o} = \frac{3}{5})

Accepted Answer

Let the actual velocity of the rain =  x Then , relative to man, it is coming with  10.8  km/h  in horizontal and  x  in vertical tan53^o = \dfrac{10.8}{x} = \dfrac{4}{3}     x = 3 	imes \dfrac{10.8}{4} km/h   3 	imes \dfrac{10.8}{4} 	imes  \dfrac{1}{3.6}  = 2.25  m/s

Question 5

Consider the following statements A and B and identify the correct answer.

A) The speed acquired by a body when falling in a vacuum for a given time is dependent on the mass of the falling body.

B) A stone falls freely from rest and the total distance covered by it in the last second of its motion equals the distance covered by it in the first three seconds of its motion. The stone remains in the air for

5 s

. [

g = 10 m s^{- 2}

]

Accepted Answer

A) The Kinematic equation:  v=u+at  shows that the speed of the freely falling body depends on initial velocity u, acceleration a and time t. It does not depend on the mass. Therefore, statement A is false. B) In general the distance traveled by the stone in some  n^{th}  second is given by:       S_n = u + \frac{1}{2} a(2n-1) The stone falls freely from rest  	herefore u=0 Let  n  be the last second (time stone remains in air). Now distance traveled is iven by:       S_n=\dfrac{1}{2}a(2n-1) This is also equal to the distance covered in first three seconds. This implies:  \dfrac{1}{2}a(2n-1)= ut+\frac{1}{2}at^2=\dfrac{1}{2}3^2 a = \frac{9}{2}a \Rightarrow 2n-1=9 \Rightarrow n=5 The stone remains in air for 5 sec and hence statement B is true.

Question 6

A body starts from rest and moves with an uniform acceleration. The ratio of distance covered in the  n^{th}  second to the distance covered in  n  seconds is :

Accepted Answer

Distance covered in  n  seconds,o  \rightarrow  start from rest s = ut + \dfrac {1}{2} at^2 s = \dfrac {1}{2} an^2 -----(1) Distance covered in  n^{th}  second  = s_n - s_{n-1} \Rightarrow s_n - s_{n-1} = \dfrac {1}{2} a n^2 - \dfrac {1}{2} a (n - 1)^2 ----- (2) So,  \dfrac {eq^n (2)}{eq^n (1)} = \dfrac {\dfrac {1}{2}an^2 - \dfrac {1}{2} a(n-1)^2}{\dfrac {1}{2}an^2} = \dfrac {2n - 1}{n^2} = \dfrac {2}{n} - \dfrac {1}{n^2}

Question 7

Raindrops are falling vertically with a velocity  10 m/s . To a cyclist moving on a straight road the raindrops appear to be coming with a velocity of  20 m/s . The velocity of  cyclist is

Accepted Answer

Let the velocity of the cyclist be  v_{x} Velocity of the rain is  v_{y} = 10m/s Relative velocity  = -v_{x}  along  x  and  v_{y}  along  y Hence relative velocity   = \sqrt{v_{x}^{2} + v_{y}^{2}} = 20ms^{-1} v_{x}^{2} + 100 = 400 v_{x}^{2} = 400 - 100 = 300 Therefore, the velocity of cyclist,  v_{x} = 10\sqrt{3} ms^{-1} Hence option  B  is correct.

Question 8

A body is thrown vertically upwards. Which one of the following graphs correctly represents the velocity vs time?

Accepted Answer

Velocity of a body thrown vertically upwards can be given as: v_f = v_i+at where  a=-g  (always acts in downward direction) 	herefore v_f=v_i-gt for   t\dfrac{v_i}{g} ,  v_f  is negative.also the slope of  v  vs  t  graph is  -g  [Negative] From the above  explanations, we can conclude that the velocity starts from a positive value, starts deceasing, reaches zero and keeps on decreasing for be a negative value.Hence, the correct option is  (D)

Question 9

A ball is dropped vertically from height  h  and is bouncing elastically on the floor (see figure). Which of the following plots best depicts the acceleration of the ball as a function of time.

Accepted Answer

During free fall of the ball, gravitational force acts on the body in downward direction always and so the acceleration of the ball remains in vertically downward direction which is constant with time. Its acceleration becomes positive only at single instant of time during the collision of ball with the ground. The ground imparts an upward impulse and hence the upward acceleration on the ball which makes the ball to move upward, but as soon as the ball looses contact with the ground, its accelerates in the downward direction again. Hence, option B is correct.

Question 10

Position (Km) - Time (min.) graph is as shown for two cars

A

and

B

. Both collide at time

t = 150

minute. Then the distance of position

R

of accident from the starting point

Q

of car

A

will be: (Initial distance between the two cars is

500

km) (Position in the graph shows the distance of the two cars from the point

Q

)

Accepted Answer

We need to calculate the distance of position 'R' of accident from the starting point 'Q' of car A i.e. the y coordinate of point R in the graph.It is given that the angle of PR with the y-axis is  37^o Thus angle of PR with x-axis will be  127^o Thus slope of line PR is  tan(127^o)=-\dfrac { 4 }{ 3 }  Equation of line will be y=\dfrac { -4 }{ 3 } x+500\ \Rightarrow \quad y=\dfrac { -4 }{ 3 } 	imes 150 + 500\ \Rightarrow \quad y=300  km (x=150 min)

Motion In A Straight Line