NCERT Solutions Class 11 Physics chapter 2 Motion in a Straight line with Video Explanation - Param Himalaya
Exercise Solution:
2.1 In which of the following examples of motion, can the body be considered approximately a point object:
(a) a railway carriage moving without jerks between two stations.
(b) a monkey sitting on top of a man cycling smoothly on a circular track.
(c) a spinning cricket ball that turns sharply on hitting the ground.
(d) a tumbling beaker that has slipped off the edge of a table.
Solution :
When the size of an object is very small compared to the scale of observation, the object can be considered as a point object.
(a) The size of railway carriage is very small compared to the distance between the two stations. Therefore , the carriage can be considered a point object.
(b) The size of the monkey is very small compared to the distance convered by the cyclist therefore , monkey can be considered a point object.
(c) The size of the spinning cricket ball is comparable to the distance through which the ball may turn as it hits the ground . Therefore , cricket ball cannot be considered a point objects.
(d) The size of the beaker is not negligible as compared to the height of the table . Therefore , in this case , the beaker cannot be considered a point object.
2.2 The position-time (x-t) graphs for two children A and B returning from their school O to their homes P and Q respectively are shown in Fig. 2.9. Choose the correct entries in the brackets below ;
(a) (A/B) lives closer to the school than (B/A)
(b) (A/B) starts from the school earlier than (B/A)
(c) (A/B) walks faster than (B/A)
(d) A and B reach home at the (same/different) time
(e) (A/B) overtakes (B/A) on the road (once/twice).
Solution :
(a) From the graph , it is clear that P is closer to O than Q . Therefore, child A lives closer to the school than the child B.
(b) It is clear that child A starts at t= 0 , while the child B starts at later Time.
(c) The x-t graph for child B is steeper than that of child A. In other words , slope of x-t graph for child B is greater than that of child A. Since the slope of x-t graph is equal to the speed , child B walks faster than child A.
(d) It is clear from the graph that both children reach their homes at t1 , i.e., they reach home at the same time.
(e) The x-t graphs for the children intersect only once at R. Therefore , B overtakes A on the road only once.
2.3 A woman starts from her home at 9.00 am, walks with a speed of 5 km h–1 on a straight road up to her office 2.5 km away, stays at the office up to 5.00 pm, and returns home by an auto with a speed of 25 km h–1. Choose suitable scales and plot the x-t graph of her motion.
Solution :
Time taken to Reach office = $\frac{Distance}{Speed} = \frac{2.5 km}{5 km/h} = 0.5 hr$
Time taken to return from office = $\frac{Distance}{Speed} = \frac{2.5 km}{25 km/h} = 0.1 hr = 6 minutes $
Therefore, the woman reaches the office at 9:30 A.M. and return home at 5.06 P.M
2.4 A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1 m long and requires 1 s. Plot the x-t graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start.
Solution :
Assigning the forward motion positive and backward motion negative.
x= 5m + (-3m) + 5m + (-3m) + 5m + (-3m) + 5m + (-3m) + 5m
Total number of steps = 8 + 8 + 8 + 8 +5 = 37 steps
Since time taken for one step is 1 a , the total time taken to fall in pit 13 m away from the start = 37 × 1 = 37 s.
2.5 A car moving along a straight highway with speed of 126 km h–1 is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop ?
Solution :
Given
$v_{0} = 126 km/h = 126 \times \frac{5}{18}m/s = 7 \times 5 m/s = 35 m/s$
Distance s = 200 m , v = 0 , a = ? , t = ?
$v^{2} = v_{0}^{2} + 2as$
$a = \frac{v^{2} -v_{0}^{2}}{2s}$
$a = \frac{ 0^{2}-35^{2}}{2 \times 200}$
$a = - \frac{1225}{400}$
$a = - 3.06 ms^{-2}$
on using first equation of motion
$v = v_{0} + at $
$t = \frac{v-v_{0}}{a}$
$t = \frac{0 -35}{-3.06} = \frac{35}{3.06} = 11.43 sec$
2.6 A player throws a ball upwards with an initial speed of 29.4 m s–1.
(a) What is the direction of acceleration during the upward motion of the ball ?
(b) What are the velocity and acceleration of the ball at the highest point of its motion ?
(c) Choose the x = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of x-axis, and give the signs of position, velocity and acceleration of the ball during its upward, and downward motion.
(d) To what height does the ball rise and after how long does the ball return to the player’s hands ? (Take g = 9.8 m s–2 and neglect air resistance).
Solution :
(a) the ball moves under the action of gravity. Therefore , the direction of acceleration due to gravity is always vertically downwards .
(b) At the highest point , the velocity of the ball is zero and acceleration due to gravity is 9.8 ms-2 vertically downwards.
(c) When the highest point is chosen as the location for x = 0 and t=0 and vertically downward direction to be the positive direction of X-axis , we have , during upward motion. The signs of position and velocity are negative and the sign of acceleration is positive.
During downward motion , The signs of position , velocity and acceleration are positive.
(d) For the upward motion , $v_{0} = -29.4 m /s and g = +9.8 m/s^{2}$
$\therefore v^{2} - v_{0}^{2} = 2gh$
$or h = \frac{-v_{0}^{2}}{2g} = \frac{-(-29.4)^{2}}{2 \times 9.8} = -44.1 m$
the negative sign shows that the distance is covered in the upward direction.
if t is the time taken to reach the highest point, then,
$v=v_{0}+gt or 0 = -29.4+9.8t$
$\therefore t = 3s$
Now for vertical motion under gravity , time of ascent = time of descent.
$\therefore$ time taken by the ball to return to player's hands = 3 + 3 = 6s
2.7 Read each statement below carefully and state with reasons and examples, if it is true or false ; A particle in one-dimensional motion
(a) with zero speed at an instant may have non-zero acceleration at that instant
(b) with zero speed may have non-zero velocity,
(c) with constant speed must have zero acceleration,
(d) with positive value of acceleration must be speeding up.
Solution :
(a) The Statement is true. When a body is thrown vertically upward from the ground, then at the highest point , the body has zero speed but acceleration of 9.8 $m/s^{2}$ is acting vertically downwards.
(b) It is false. it is because the speed of a body is equal to magnitude of its velocity. Therefore, downwards.
(c) It is true. When a body moves in a straight line , its direction is unchanged . If the speed of the body is constant, change in velocity is zero and hence acceleration is zero.
(d) The velocity of body moving in a straight line is given by : $v = v_{0} + at$. The Statement given in the problem is not correct if $v_{0}$ is negative at t=0. However , the given statement is true if both $v_{0}$ and a are positive.
2.8 A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one tenth of its speed. Plot the speed-time graph of its motion between t = 0 to 12 s.
Solution :
Since the ball is dropped from a height of 90m, we have $v_{0} = 0 , g = 10 $m/s^{2}$ and h = 90m. Time taken by the ball to reach floor is
$h=v_{0}t+\frac{1}{2}at^{2}$
$h= \frac{1}{2}at^{2}$
$t^{2} = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2\times90}{10}} = 4.24s$
velocity of the ball at floor is $v= \sqrt{2gh} = \sqrt{2\times10 \times 90}v= 30\sqrt{2}$ m/s
Rebound velocity of ball is
${v}' = \frac{9}{10}v = \frac{9}{10} \times30\sqrt{2} = 27\sqrt{2}$ m/s
Time taken by the ball to reach the highest point is
${t}' = \frac{{v}'}{g} = \frac{27\sqrt{2}}{10}$ = 3.81 s
Total time = $t+{t}'$ = 4.24 + 3.81 = 8.05
The ball takes further 3.81 s to reach the floor again where its velocity before striking the floor will be $27\sqrt{2}$ m/s
Rebound velocity of the ball is ${v}'' = \frac{9}{10} \times 27 \sqrt{2} = 24.3\sqrt{2}$ m/s
The ball will start upward motion after time = 8.05+3.81 = 11.86 s
Fiqure shows the speed-time graph of motion of the ball between t=0 and t=12s.
2.9 Explain clearly, with examples, the distinction between :
(a) magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval;
(b) magnitude of average velocity over an interval of time, and the average speed over the same interval. [Average speed of a particle over an interval of time is defined as the total path length divided by the time interval]. Show in both (a) and (b) that the second quantity is either greater than or equal to the first. When is the equality sign true ? [For simplicity, consider one-dimensional motion only].
Solution :
(a) Suppose a particle moves from A to B and B to C in time t. Then the magnitude of displacement of the particle = AC whereas the total length of the path covered by the particle = AC + BC . Note that the total path length (=AB + BC ) is greater than the magnitude of displacement (=AC ). If the particle moves in a straight line , then magnitude of displacement is equal to the total path length covered by the particle in a given time.
(b) Average velocity = $\frac{Displacement}{Time} = \frac{AC}{t}$
Average Speed = $\frac{Total path length }{Time} = \frac{AB+BC}{t}$
Since ( AB + BC ) > AC , the average Speed is greater than the average velocity . If the particle moves in a straight line , displacement= Total path length in a given time so that average Speed will be equal to the average velocity . Thus in either case (a) or (b) the second quantity is either greater than or equal to first quantity. Equality sign is true when displacement is equal to distance .
2.10 A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km/h. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h–1. 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 ?
Solution :
(i) Time interval 0-30 minutes. While going to the market , the man walks at a speed of 5 km/h
\therefore Displacement in 30 min =\frac{5}{2} = 2.5 km
Average velocity =$\frac{Net Displacement}{time} = \frac{2.5 km}{0.5h} = 5 km/h$
Average Speed =$\frac{Total Distance travelled}{ Total time taken} = \frac{2.5 km}{0.5h} = 5 km/h$
(ii) Time interval 0-50 minutes. While going to the market the man walks at a speed of 5 km/h . since the market is 2.5 km from his home , he takes 30 minutes to reach the market. While coming back from the market , he walks at a speed of 7.5 km/h so that time taken to reach home = 2.5/7.5 = 1/3 hour = (1/3) × 60 = 20 minutes. This means that in 50 minutes the man walks back home . Therefore , net displacement= 0 .
average velocity =$ \frac{Net displacement}{Time} = \frac{0}{50min}$=0
Now in 50 min , the total distance travelled = 2.5 +2.5 = 5 km and total time taken = 50/60 hr = 5/6 hr.
Average speed =$ \frac{Total distance travelled }{total time taken } = \frac{5}{\frac{5}{6}}$ = 6 km/h
(iii) Time interval 0-40 minutes . in 30 minutes, the man covers a distance of 2.5 km in going from home to the market. While returning home , he covers a distance of 1.25 km in 10 minutes. ( Return speed is 7.5 km/h)
$\therefore$ Net displacement during time interval 0-40 min
= 2.5 - 1.25 = 1.25 km
Average velocity =$ \frac{Net displacement}{Time} = \frac{1.25}{40/60}$=1.875 km/h
Total distance travelled in 40 min = 2.5 + 1.25 = 3.75 km
Average speed =$ \frac{Total distance travelled }{total time taken } = \frac{3.75}{40/60}$=5.625 km/h
2.11 In Exercises 2.9 and 2.10, 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?
Solution : instantaneous velocity v = $\frac{dx}{dt}$
In such a small interval of time (dt) , the direction of motion of the particle is not supposed to change so that displacement is effectively equal to the distance . Therefore , instantaneous speed is equal to the magnitude of instantaneous velocity.
2.12 Look at the graphs (a) to (d) (Fig. 2.10) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.
Solution :
2.13 Figure 2.11 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.
Solution :
No , because the x-t graph does not represent the path taken by the particle. From the graph , at t=0 , x=0 . This graph represents the motion of a freely falling body.
2.14 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).
Solution :
Speed of police van , $v_{PE} = 30 \frac{km}{h} = 30 \times\frac{5}{18}\frac{m}{s} = \frac{25}{3} ms^{-1}$
Speed of bullet , $v_{BE} = 150 ms^{-1}$
Speed of thief's car , $v_{TE} = 192 km/h = \frac{160}{3} ms^{-1}$
When the bullet is fired from police van , the net speed of bullet is sum of $v_{PE}$ and $v_{BE}$
$\therefore$ Net speed of bullet = $v_{PE} + v_{BE} = \frac{25}{3} + 150 = \frac{475}{3} ms^{-1}$
The bullet will hit the thief's car with relative speed of the bullet w.r.t thief's car. Therefore , the speed with which the bullet thief's car is $v= \frac{475}{3} - \frac{160}{3} = \frac{315}{3} = 105 ms^{-1}$