Friday, July 6, 2012

Class 11 - Physics - CH3 - Motion In a Straight Line (Part-2)

Motion in straight line!
(source:clker.com)

NCERT Chapter Solutions (Q14-onwards)


For Questions (1-13) see page.
Q 3.14: A man walks on a straight road from his home to a market 2.5 km away with a speed of
5 km/hr. Finding the market closed, he instantly turns and walks back home with a speed of
7.5 km /hr. What is the
(a) magnitude of average velocity, and
(b) average speed of the man
over the interval of time
(i) 0 to 30 min,
(ii) 0 to 50 min,
(iii) 0 to 40 min ?
[Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time, and not as magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero !]

Answer:





We know  Average Velocity = Total Displacement ÷ Total Time
         and  Average Speed    =  Total Distance ÷ Total Time
Given,
  Distance between home and market = 2.5 km
  Velocity from Home to Market (v1) = 5 km/hr
  Velocity from Market to home (v2) = 7.5 km/hr
   Time taken to reach market (t1) = 2.5 / 5 = 1/2 hrs (= 30min)
   Time taken to reach home (t2) = 2.5 /7.5 = 1/3 hrs (= 20 min)



(i) Case: 0 to 30 min
     In 30 mins, Distance = Displacement = 2.5 km
     Duration = 30 min = 0.5 hr
     Average Velocity = 2.5 /0.5 = 5 km/hr
     Average Speed = 2.5 / 0.5 = 5 km/hr

(ii) Case 0 to 50 min. i.e. (t1+ t2 = 30 min + 20 min = 5/6 hrs)
 In 50 mins, Distance = Path travelled from Home to Market + Path travelled from Market to Home
     Distance = 2.5 + 2.5 = 5.0 km
Average Speed = 5.0 ÷ (5/6hrs) = 5 ✕ 6 / 5 = 6 km/hr
Net Displacement = 2.5 + (-2.5km) = 0 km
Average velocity = 0 ÷ (5/6hrs) = 0 km / hr

(iii) 0 to 40 min = (30 min + 10min)
In 40 mins, total distance travelled is =
    = Distance from home to market + distance covered in 10 min from market to home.
    = 2.5km + (7.5 ✕ (1/6hrs)) = 2.5 + 1.25 = 3.75
Average Speed = 3.75 ÷ 40 min = 3.75 ÷ 40/60 hrs = 5.625 km/hr

In 40 mins, net displacement = 2.5km - 1.25km = 1.25km
Average Velocity = 1.25 ÷ 40/60 hrs = 1.875 km/hr

Q15: In Exercises 3.13 and 3.14, we have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider instantaneous speed and magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why?

Answer: Instantaneous Velocity is defined as the value approached by the average velocity when the time interval for that measurement is approaching to zero, i.e. Δt →0

Similarly instantaneous speed is the speed of a distance covered at that instant. 

In both cases the magnitude of time is so small ( Δt →0), For such small time interval, we assume object does not change its direction. We can consider magnitude of displacement is same as distance. Thus both instantaneous speed and instantaneous velocity are the same in this case.


Q16: Look at the graphs (a) to (d) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.


Answer: None of these graphs represent 1-d motion of a particle.
(a) Particle cannot have two positions at the same time.
(b) A particle cannot have two values of velocity at the same time.
(c) Speed cannot be negative.
(d) Total path length can never decrease with increase in time.


Q17: Figure 3.21 shows the x-t plot of one-dimensional motion of a particle. Is it correct to say from the graph that the particle moves in a straight line for t < 0 and on a parabolic path for t >0 ? If not, suggest a suitable physical context for this graph.


Answer: Let us try to answer this question first i.e. 'Does Displacement/Distance(x)-time (t) graph always show the trajectory i.e. actual path covered in space w.r.t. time ?' Answer is NO.

Similarly, the given graph does not tell if object is moving in a straight line for t < 0. It simply tells that its position (x) is zero for t ≤ 0. The parabola curve indicates the object moves with constant acceleration. In general, a stone is at rest for some time before it is dropped off, can be depicted by the given graph.

Interesting Note: On the other hand, the above graph seems to hold correct. Consider a scenario of projectile motion (2d motion), an aeroplane is moving in a straight line and drops off a bomb. 

If we plot the vertical distance(position) of the bomb w.r.t. time, the graph comes out to be same as asked. Note the graph represents 1d-motion. What's wrong with this example? Revisit the question, motion of the article is asked for  the same dimensional-reference. In this example, before dropping off, the bomb's position is zero for vertical dimension but non-zero for horizontal dimension.

Q18: A police van moving on a highway with a speed of 30 km/h fires a bullet at a thief’s car speeding away in the same direction with a speed of 192 km/h. If the muzzle speed of the bullet is 150 m/s, with what speed does the bullet hit the thief’s car ? (Note: Obtain that speed which is relevant for damaging the thief’s car).

Answer:  Speed of Police van (vp)= 30km/h = 30 ✕1000/3600 = 8.33 m/s
Speed of the thief's car (vt) = 192 km/h = 192 ✕1000/3600 = 53.33 m/s
Muzzle speed of bullet is actually the relative speed of bullet wrt to police van (vbp) = 150 m/s
∵ relative speed of bullet (vbp) =  (bullet speed) vb- (police speed) vp
⇒  vb = vbp + vp = 150 + 8.33 = 158.33 m/s

Relative speed of bullet wrt to thief (vbt) = vb - vt = 158.33 - 53.33 = 105 m/s

Q19: Suggest a suitable physical situation for each of the following graphs:

Answer:
Case a: Initially the object is at rest (I), it goes in motion (II)  and then reverse its direction with uniform motion (III). It crosses the reference frame (IV) and stops after some time.
It may happen if a striker kicks a ball, ball goes and strikes to a wall and rebounds. It comes back to strikers, crosses it and then stops.

Case b: A ball is thrown upwards with certain velocity(a). As it proceeds upwards, its velocity decreases and becomes zero. Then it starts falling and is accelerated. It strikes a ground with same velocity(b) as initial one. After striking the ground, it rebounds with some velocity (but less than previous one) and moves upward. The sequence goes on, till velocity of the ball reaches zero.

Case c: As the graph shows, acceleration is zero. Then it increases and decreases for a small amount of time. After that acceleration is zero again. The non-zero acceleration period indicates for that duration and external force is in touch with the object (called impulse). For example, when a moving ball is hit by a bat or a hammer strikes on a nail.


Q20: Figure below gives the x-t plot of a particle executing one-dimensional simple harmonic motion. (You will learn about this motion in more detail in Chapter14). Give the signs of position, velocity and acceleration variables of the particle at t = 0.3 s, 1.2 s, – 1.2 s.



Answer:

Before finding the signs of x, v and t in the asked graph. Let us try to understand the Simple Harmonic Motion using a pendulum. As shown in figure, the bob moves from its mean position (O) to maximum positions (a) and (b) either side.




When the bob, moves from its mean position (O) to max. position at (a), its displacement increases and then the bob comes back to its original position. Correspondingly, the displacement decreases and reaches zero and then it increases to other side. The x-t graph depicts the behaviour.


When displacement  of the bob is maximum at (a) and (b), the velocity at that point is zero, because it is going to reverse its direction. When the bob starts from its mean position, its velocity is maximum and then it decreases to zero at (a). Velocity of bob reverses its direction and keeps on increasing till it reaches mean position (O). from there, it starts decreasing to zero till it reaches the other side max. position (b). The v-t graph shows the behaviour.


When the displacement of the bob is maximum at position (a) and (b),  the speed is zero but acceleration has maximum magnitude and is changing its direction. Similarly when bob starts moving from its mean position (O) towards (a), its speed starts decreasing. Correspondingly, magnitude of acceleration is zero at mean position and starts increasing on the opposite direction. The a(t)-t graph shows the acceleration behaviour at positions (o), (a) and (b).

We can deduce, at position (a), x > 0, v < 0 and a < 0.

Now we can easily find out signs of displacement, velocity and acceleration at various positions in the question - graph.

Time         Displacement     Velocity        Acceleration
t = 0.3sx < 0v < 0a > 0
t = 1.2sx > 0v > 0a < 0
t = -1.2sx < 0v > 0a > 0


Q21: Figure gives the x-t plot of a particle in one-dimensional motion. Three different equal intervals of time are shown. In which interval is the average speed greatest, and in which is it the least ? Give the sign of average velocity for each interval.



Answer:  The slope of x-t graph of time-interval represents the average velocity. 
In general, Avg Velocity = ( x2 - x1) / (t2 - t1)

(a) The slope is maximum is case-3 and is minimum in case-2. Therefore, average velocity is maximum in case-3 and is minimum in case-2.

(b) In case-1 and case-2, (x2 > x1), average velocity is positive.
In case-3, x2 < x1, therefore average velocity is negative.


Q22: Figure below gives a speed-time graph of a particle in motion along a constant direction. Three equal intervals of time are shown. In which interval is the average acceleration greatest in magnitude ? In which interval is the average speed greatest ? Choosing the positive direction as the constant direction of motion, give the signs of v and a in the three intervals. What are the accelerations at the points A, B, C and D ?

Answer
(i) Acceleration is given by the slope of speed-time graph. Here slope |s3 - s4| / Δt is maximum. For interval 2, average acceleration is the greatest.

(ii) The above figure is a speed-time graph. y-axis represents the speed. For interval 3, the speed is the greatest.

(iii)  v is positive for intervals 1, 2 and 3. a is positive for intervals 1 and 3 but it is negative for interval 2 (s3 > s4).

(iv)  At points A, B, C and D, slope is zero. Therefore a = 0.


Q23: Find the distance travelled by an object in nth second.

Answer:  We know equation S = ut + ½at2
∴ Distance travelled by an object in n seconds is,
   Sn = un + ½an2
and Distance travelled by the object in (n - 1) seconds is:
   Sn-1 = u(n-1) + ½a(n-1)2
Distance travelled by the object in nth second is,
Snth = Sn - Sn-1 = un + ½an2 - (u(n-1) + ½a(n-1)2)
Snth = un + ½an2 -un + u - ½a(n2+ 1 -2n)
Snth =u + ½a(2n -1)


Q24: A three-wheeler starts from rest, accelerates uniformly with 1 m s–2 on a straight road for 10 s, and then moves with uniform velocity. Plot the distance covered by the vehicle during the nth second (n = 1,2,3….) versus n. What do you expect this plot to be during accelerated motion : a straight line or a parabola ?

Answer: We know that distance travelled in nth second is,
    Snth =u + ½a(2n -1)
where u = initial velocity,
a = acceleration and n is the nth second.

For the given three-wheeler, u = 0  and a = 1m/s2, the equation becomes:
    Snth = ½(2n -1) = n - 0.5

Possible values of  Snth  vs time (n seconds) are:

n 0   1    2    3    4    5    6    7    8    9   10 
Snth    0   0.5    1.5   2.5   3.5   4.5    5.5   6.5    7.5   8.5   9.5 

The plot for Snth  vs time is for accelerated motion (upto 10s) is a straight line.


Q25: A boy standing on a stationary lift (open from above) throws a ball upwards with the maximum initial speed he can, equal to 49 m/s. How much time does the ball take to return to his hands? If the lift starts moving up with a uniform speed of 5 m/s and the boy again throws the ball up with the maximum speed he can, how long does the ball take to return to his hands ?

Answer:  Let us assume the observer is outside the lift.

Case I: When lift is not moving.
    Initial velocity of ball (uB) = 49 m/s
    When balls reaches at its max. height, final velocity (v) = 0 m/s
    v = uB + atA
 ⇒ 0 = 49 - gtA          (where g = -9.8 m/s)

 ⇒ tA= 49 /g = 49 / 9.8 = 5s

Time ascent (tA) = Time of descent (tD) = 5s
Total time taken by the ball to return to boy's hands = 10s

Case II: When lift is moving up with a uniform velocity (uL) = 5m/s
Initial velocity of ball (uB) = 49 m/s
Relative velocity of ball when  when it moves upward = uB+ uL
∴ v = (uB+ uL) - gtA
Since v = 0, Time of ascent will be tA = (uB+ uL)/g

When ball reaches at max height, its velocity is zero and it starts falling down. 
When it reaches boy's hand its final relative velocity will be = uB - uL
Using v = u + at, we have 
∴  -(uB- uL) = 0 - gtD
⇒ Time of descent (tD) =  (uB- uL)/g
∴ Total time = tA + tD = (uB + uL)/g + (uB- uL)/g = (uB- uL+ uB- uL)/g = 2uB/g 
    = 2 × 49/9.8 = 10s
(Note: In above equation, uL cancels out. ⇒ Uniform velocity of the lift does not affect the relative velocity of the ball w.r.t. lift).


(In progress...)

4 comments:

  1. This comment has been removed by the author.

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  2. I think the questions are numbered differently in the website since in the text it is numbered as 3.14,3.15 and so on but here in the website it is numbered as 13,14,15 and so on therefore I think there will be a difference in the questions as well which makes it harder for us to know where the questions are

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  3. You are too good person your answer are able to understand very easily thanx

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