Wednesday, August 12, 2015

HC Verma solutions, Concepts of Physics, Part 1,Chapter 3, REST AND MOTION : KINEMATICS",'EXERCISES - Q21 to Q30

EXERCISES (Question number 21 to 30)

REST AND MOTION

21. A police jeep is chasing a culprit going on a motorbike. The motorbike crosses a turning at a speed of 72 km/h. The jeep follows it at a speed of 90 km/h, crossing the turning ten seconds later than the bike. Assuming that they travel at constant speeds, how far from the turning will the jeep catch up with the bike? 

Answer: 72 km/h =72000/3600 m/s = 20 m/s

               90 km/h =90000/3600 m/s = 25 m/s 

Suppose the jeep catches the bike at a distance of x m from the turning. Time taken by the bike in covering x m = x/20 s. 

Since the jeep starts 10 s later, so it takes 10 s less time to cover the equal distance. So time taken by the jeep in covering x m = (x/20)-10 s. But from the jeep's speed, it comes to be x/25 s. These two times must be equal. Equating both times we get,

(x/20)-10 = x/25, → (x-200)/20=x/25 → (x-200)/4=x/5 

→ 5x-1000=4x → x=1000 m = 1.0 km. 




22. A car traveling at 60 km/h overtakes another car traveling at 42 km/h. Assuming each car to be 5.0 m long, find the time taken during the overtaking and the total road distance used for the overtake. 

Answer: 60 km/h = 60000/3600 m/s =50/3 m/s 

42 km/h = 42000/3600 m/s = 35/3 m/s 

Relative speed of faster car with respect to slower car 

= 50/3 - 35/3 m/s =15/3 m/s = 5 m/s. Total distance to be covered in overtaking = Total length of both cars = 10.0 m. So the time taken in overtaking = 10/5 s = 2 s. 

Total road distance used for the overtaking will be the distance covered by overtaking the car in this time plus the length of the car = (50/3)x 2+5 = 100/3+5 = 33.3+5 

=38.3 m ≈ 38 m  (It can be illustrated by the following diagram)
Diagram for Question no. 22
 

23. A ball is projected vertically upward with a speed of 50 m/s. Find (a) the maximum height, (b) the time to reach the maximum height, (c) the speed at half the maximum height. Take g= 10 m/s².

Answer: u=50 m/s, v=0 (at maximum height) , Let Maximum height be h . The relationship is given by the equation,

v²=u²-2gh, → 0=u²-2gh, → h=u²/2g =2500/20 m =125 m. 

If t is the time to reach the maximum height, we can use the following equation, 

h=ut-½gt², →125=50t-½.10t², →125=50t-5t², →25=10t-t² 

→t²-10t+25=0, (t-5)²=0 →t=5 s. 

Half height=s=125/2 m,    Speed at half height v is given by the equation, v²=u²-2gs =50²-2.10.125/2 =2500-1250 =1250

→ v=√1250 = 35.3 m/s ≈ 35 m/s.




24. A ball is dropped from a balloon going up at a speed of 7 m/s. If the balloon was at a height of 60 m at the time of dropping the ball, how long will the ball take in reaching the ground? 

Answer: Let the time taken by the ball to reach the ground be t. We have u=-7 m/s (Taking downward direction as +ve), h=60 m and g=10 m/s². Using equation h=ut+½gt²,

→60=-7t+½.10.t²    → 5t²-7t-60=0  We solve this quadratic equation for t and take only the positive value 

→ t={7±√(49+4.5.60)}/2*5 

→ t= 4.23 s




25. A stone is thrown vertically upward with a speed of 28 m/s. (a) Find the maximum height reached by the stone, (b) Find its velocity one second before it reaches the maximum height, (c) Does the answer of part (b) change if the initial speed is more than 28 m/s such as 40 m/s or 80 m/s. 

Answer: (a) As in question number 23, maximum height h=u²/2g. Here u=28 m/s, take g=9.8 m/s²

we get h=28²/19.6 = 40 m 

(b) Let t be the time to reach the maximum height when the final velocity is v=0 m/s. From the equation v=u-gt, we have  0=u-gt, → t=u/g. Now the time taken one second before maximum height = u/g-1. To get velocity at this time we again use the equation v=u-gt,

Now v=u-g(u/g-1) =u-u+g =g =9.8 m/s 

(c) No, because the velocity in (b) is independent of u.



      

26. A person sitting on the top of a tall building is dropping balls at regular intervals of one second. Find the positions of the 3rd, 4th and 5th ball when the 6th ball is being dropped.  

Answer: Height of fall of balls are given by the equation,

h=ut+½gt² =½gt²   (Since u=0) 

For the 3rd ball t=3 s, So h=½*9.8*9 =44.1 m below the top of the building. 
For the 4th ball t=2 s, So h=½*9.8*4 =19.6 m below the top of the building.  
For the 5th ball t=1 s, So h=½*9.8*1 =4.9 m below the top of the building.  

27. A healthy young man standing at a distance of 7 m from an 11.8 m high building sees a kid slipping from the top floor. With what speed (assumed uniform) should he run to catch the kid at the arms height at (1.8 m)? 

Answer: Height of fall = h = 11.8 - 1.8 = 10 m , Let time of fall be t. From the equation h=ut+½gt² =½gt², →t=√(2h/g) =√(20/9.8)  

Now the young man has this time to run 7 m to catch the kid. So his speed should be 

= 7/√(20/9.8) =7/1.428 =4.9 m/s




28. An NCC parade is going at a uniform speed of 6 km/h through a place under a berry tree on which a bird is sitting at a height of 12.1 m. At a particular instant, the bird drops a berry. Which cadet (give the distance from the tree at the instant) will receive the berry on his uniform. 

Answer: h=12.1 m, As in question 27 time of fall, 

t= √(2h/g) =√(2*12.1/9.8) =1.57 s 

Speed of NCC parade = 6 km/h =6000/3600 =5/3 m/s  

So the distance of the cadet who will receive the berry on his uniform from the tree =5/3 *1.57 = 2.62 s




29. A ball is dropped from a height. If it takes 0.200 s to cross the last 6.00 m before hitting the ground, find the height from which it was dropped. Take g=10 m/s².

Answer: Time of fall t=√(2h/g). 

Time of fall for h-6 m =√{2(h-6)/g} 

Difference of time = √(2h/g)-√{2(h-6)/g}=0.2, we solve it for h. 

→√(2h)-√{2(h-6)} =0.2√g  squaring both sides we get, 

→ 2h+ 2(h-6)-2√(4h²-24h) =0.04g

→ 4h-12-0.04g=2√(4h²-24h)  

→ 2h-6-0.2= √(4h²-24h) 

→ 2h-6.2=√(4h²-24h), Again squaring both sides we get, 

4h²-2*2h*6.2+6.2²= 4h²-24h 

→ 24.8 h- 24h =6.2² 

→ h =6.2*6.2/0.8 =48.0 m



30. A ball is dropped from a height of 5.0 m onto a sandy floor and penetrates the sand up to 10 cm before coming to rest. Find the retardation of the ball in sand assuming it to be uniform. 

Answer: Velocity just near the floor v is given by,    

v²=u²+2gh,  here u=0, h=5 m, 

v²=2g*5 =10g  

This v is the initial velocity u for the penetration in sand and final velocity v=0, here h=10 cm = 0.10 m, Let r be the retardation of the ball in sand. From the equation v²=u²-2ah, we have,

0=10g-2r*0.10  → r=10g/0.2=50g=50*9.8 =490 m/s² 

==<<O>>==

For answers-EXERCISES (Question number 31 to 40) 

visit the link below :-

Solutions to Question number 31 to 40

==<<0>>==

My Channel on YouTube  →  SimplePhysics with KK

For links to 

other chapters - See bottom of the page

Or click here → kktutor.blogspot.com 

My Channel on YouTube  →  SimplePhysics with KK

Links to the Chapters



CHAPTER- 14 - Fluid Mechanics



CHAPTER- 13 - Fluid Mechanics


CHAPTER- 12 - Simple Harmonic Motion








CHAPTER- 11 - Gravitation




CHAPTER- 10 - Rotational Mechanics




CHAPTER- 9 - Center of Mass, Linear Momentum, Collision


CHAPTER- 8 - Work and Energy

Click here for → Question for Short Answers

Click here for → OBJECTIVE-I

Click here for → OBJECTIVE-II

Click here for → Exercises (1-10)

Click here for → Exercises (11-20)

Click here for → Exercises (21-30)

Click here for → Exercises (31-42)

Click here for → Exercise(43-54)

CHAPTER- 7 - Circular Motion

Click here for → Questions for Short Answer 

Click here for → OBJECTIVE-I

Click here for → OBJECTIVE-II

Click here for → EXERCISES (1-10)

Click here for → EXERCISES (11-20)

Click here for → EXERCISES (21-30)

CHAPTER- 6 - Friction

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Click here for → OBJECTIVE-I

Click here for → Friction - OBJECTIVE-II

Click here for → EXERCISES (1-10)

Click here for → Exercises (11-20)

Click here for → EXERCISES (21-31)

For more practice on problems on friction solve these- "New Questions on Friction".

---------------------------------------------------------------------------------

CHAPTER- 5 - Newton's Laws of Motion


Click here for → QUESTIONS FOR SHORT ANSWER

Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)

-------------------------------------------------------------------------------

CHAPTER- 4 - The Forces

The Forces-

"Questions for short Answers"    


Click here for "The Forces" - OBJECTIVE-I


Click here for "The Forces" - OBJECTIVE-II


Click here for "The Forces" - Exercises


--------------------------------------------------------------------------------------------------------------

CHAPTER- 3 - Kinematics - Rest and Motion

Click here for "Questions for short Answers"


Click here for "OBJECTIVE-I"


Click here for EXERCISES (Question number 1 to 10)


Click here for EXERCISES (Question number 11 to 20)


Click here for EXERCISES (Question number 21 to 30)


Click here for EXERCISES (Question number 31 to 40)


Click here for EXERCISES (Question number 41 to 52)


CHAPTER- 2 - "Vector related Problems"

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11 comments:

  1. AnonymousJuly 4, 2017 at 5:35 AM

    sir, in Qno 21, i didn't understand the time concept can u help?

    ReplyDelete
  2. K K MISHRAJuly 4, 2017 at 7:12 AM

    Suppose A is a slower vehicle that covers certain distance in 20 s from the start point. B is a faster vehicle that covers same distance in 15 s. So B can move 5 s later from the start point after A to reach the finishing point simultaneously. In other words if B comes moving to start point 5 s after A's departure then B will catch A at the finishing point. Take B as police vehicle.
    Since we have taken the required distance as unknown x, we need to form an equation to solve it for x. That is why we have calculated the time in two ways and equated it to form an equation. Then solved it for x.

    ReplyDelete
    Replies
    1. UnknownSeptember 18, 2018 at 9:14 AM

      sir you are really a greeat person to solve HC Verma in such an wonderful manner.

      Delete
    2. K K MISHRASeptember 19, 2018 at 9:27 AM

      Thank You student!

      Delete
  3. AnonymousAugust 27, 2017 at 11:04 PM

    sir in Qno 26 of Kinematics less, how is the time taken, can u explain and for the first ball what will be time taken t=0s or t=1s and why?

    ReplyDelete
  4. AnonymousAugust 27, 2017 at 11:44 PM

    sir in Qno 26 of Kinematics less, how is the time taken, can u explain and for the first ball what will be time taken t=0s or t=1s and why?

    ReplyDelete
    Replies
    1. K K MISHRAAugust 28, 2017 at 6:58 AM

      Since the balls are being dropped at regular interval of 1 s, So just when the sixth ball is being dropped means that 5th ball has been dropped 1 s earlier, 4th - 2s earlier, 3rd -3s earlier, 2nd - 4s earlier and 1st -5s earlier. So for the 1st ball t= 5s at the moment when 6th ball is being dropped.

      Delete
  5. Lets Make It.... NowApril 1, 2019 at 3:44 AM

    Sir,

    In Q30, after ball starts penetrating sand, should we not consider g also.
    I mean effective retardation should be r-g.

    ReplyDelete
    Replies
    1. K K MISHRAApril 3, 2019 at 9:23 AM

      Dear student, here we are required to find out the net/resultant retardation on the basis of velocities which includes the effect of g. We need not to analyse it.

      Delete
  6. Knowledge BuzzApril 4, 2019 at 1:06 AM

    Sir question no 29 ..how do we get time of fflight

    ReplyDelete
    Replies
    1. K K MISHRAApril 4, 2019 at 9:06 AM

      Dear student, since we found out the height from which the object was dropped, we can find out the flight-time putting the value of h in t=√(2h/g).

      Delete

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