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: First Method
72 km/h =72000/3600 m/s = 20 m/s
90 km/h =90000/3600 m/s = 25 m/s
Suppose the jeep catches up with 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.
Alternative Method
Since the jeep reaches the turning 10 seconds later, in this 10 s the bike moves a distance of =20 m/s*10 s =200 m. So when the jeep reaches the turning, the distance between the jeep and bike =200 m. This distance will be covered by the jeep with the relative speed.
The relative speed of the jeep with respect to the bike is
=25 m/s -20 m/s =5 m/s.
Time to cover this distance with the relative speed
=(200 m)/(5 m/s)
=40 s.
So the jeep will catch up with the bike in 40 s. The distance traveled by the jeep in 40 s from the turning
=(25 m/s) *(40 s)
= 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
The relative speed of the faster car with respect to the 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.
The 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 m =100/3+5 m= 33.3+5 m
=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 would the ball take to reach the ground?
Answer: Let the time the ball takes 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 -6 0=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 the 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 the 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: The height of fall of balls is given by the equation,
h = ut +½gt² =½gt² (Since u=0)
When the 6th ball is being dropped, the 5th ball has been dropped 1 s ago, the 4th ball 2 s ago, and the 3rd ball 3 s ago. Hence,
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 arm height (at 1.8 m)?
Answer: Height of fall = h = 11.8 - 1.8 = 10 m, Let the time of fall be t.
From the equation h =ut +½gt²,
Here u =0 for the kid, so
h =½gt²,
→t=√(2h/g)
=√(20/9.8) s
Now the young man has this time to run 7 m to catch the kid. So his speed should be
=Distance/Time
= 7/√(20/9.8) m/s
=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 (given 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
=speed * time =(5/3 m/s) *(1.57 s) = 2.62 m.
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 the sand and final velocity v =0, here h =10 cm = 0.10 m, Let r be the retardation of the ball in the 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.
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CHAPTER- 2 - "Vector related Problems"
CHAPTER- 2 - "Vector related Problems"
sir, in Qno 21, i didn't understand the time concept can u help?
ReplyDeleteSuppose 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.
ReplyDeleteSince 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.
sir you are really a greeat person to solve HC Verma in such an wonderful manner.
DeleteThank You student!
Deletesir 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?
ReplyDeletesir 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?
ReplyDeleteSince 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.
DeleteSir,
ReplyDeleteIn Q30, after ball starts penetrating sand, should we not consider g also.
I mean effective retardation should be r-g.
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.
DeleteSir question no 29 ..how do we get time of fflight
ReplyDeleteDear 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