KKMishra's Tutorials: August 2015

EXERCISES (Question number 41 to 52)

REST AND MOTION

41. A boy standing on a long railroad car throws a ball straight upwards. The car is moving on the horizontal road with an acceleration of 1 m/s² and the projection velocity in the vertical direction is 9.8 m/s. How far behind the boy will the ball fall on the car?

Answer: First of all we should know the time duration for which the ball is in the air. Let this time be t, vertical displacement in this time is zero. and given that u=9.8 m/s. From the equation h=ut-½gt², We get,

0=9.8t-½9.8*t²,

→ ½t²-t=0,

→ t(t-2)=0, → t=0 or 2 s. t=0 corresponds to the time when it is being projected. So the time duration for which it remains in air = 2 s.

Since the ball is projected straight upwards its horizontal velocity with respect to the frame of a railroad car is zero at the instant of projection. But the frame is itself moving with an acceleration of 1 m/s² in the horizontal direction so with respect to this accelerating frame the ball will have equal but backward acceleration during the flight. So to calculate the horizontal displacement with respect to railroad car we have the following data,

u=0, a=-1 m/s², t=2 s, So displacement x=ut+½at²

=0-½*2² =-2 m. Negative sign shows backward displacement.

So the ball will fall 2 m behind the boy on the car.

42. A staircase contains three steps each 10 cm high and 20 cm wide (figure 3-E9). What should be the minimum horizontal velocity of a ball rolling off the uppermost plane so as to hit directly the lowest plane?

Answer: To hit directly the lowest plane the ball will have to clear point B. From A to B the ball will move a horizontal distance equal to 40 cm = 0.4 m and the vertical distance equal to 20 cm =0.2 m. Let t be the time taken to reach the point be. In this time vertical movement under the gravity has following data,

u=0, h=0.2 m, from the equation h=ut+½gt², we get,

0.2 = 0+½*9.8 t² → t²=0.4/9.8 = 0.04 → t=0.2

Let v be the horizontal velocity of the ball, then the horizontal distance traveled by the ball = 0.2 v. In order to clear point B, this must be at least =0.4,

0.2 v= 0.4 → v=2 m/s. So minimum velocity required to hit the lowest floor is 2 m/s.

43. A person is standing on a truck moving with a constant velocity of 14.7 m/s on a horizontal road. The man throws a ball in such a way that it returns to the truck after the truck has moved 58.8 m. Find the speed and the angle of projection (a) as seen from the truck, (b) as seen from the road.

Answer: Time of flight of the ball =58.8/14.7 s = 4 s

(a) Since the ball returns back to the truck its angle of projection is vertically upwards with respect to the truck. Let u be the speed of projection, we have t=4 s, displacement h=0 m, using equation

h=ut-½gt² → 0 = 4u-½9.816 → u = 9.8*8/4 = 19.6 m/s

(b) As seen from the road the speed of the ball will be resultant of two speeds, vertical speed u = 19.6 m/s and horizontal speed given by the truck v = 14.7 m/s, we get it = √(u²+v²) = √(19.6²+14.7²)

= 24.5 m/s and the angle seen from the road will be

= tan^-1 19.6/14.7 = tan^-1 1.33 = tan^-1 (4/3) = 53° with horizontal.

44. The benches of a gallery in a cricket stadium are 1 m wide and 1 m high. A batsman strikes the ball at a level one m above the ground and hits a mammoth sixer. The ball starts at 35 m/s at an angle of 53° with the horizontal. The benches are perpendicular to the plane of motion and the first bench is 110 m from the batsman. On which bench will the ball hit?

Answer: Distance of first bench = 110 m, Let us assume that the ball strikes the nth bench. The height above the strike level = (n-1) m and horizontal distance traveled = (110+n) m

Now from the equation h=ut-½gt²

(n-1) = 35 sin 53° t-½*9.8t² →4.9t²-28 t+(n-1) =0

→ t = [28±√{28²-44.9(n-1)}]/9.8 (we shall take the +ve sign because the ball will strike the bench when falling.

→ t = 28/9.8+√{784-19.6(n-1)}/9.8 =2.86+√{8.16-0.204(n-1)}

Horizontal distance traveled = 35 cos 53° t = 110+n

→21.06*[2.86+√{8.16-0.204(n-1)}] = 110 + n (Put value of t)

→60.24+21.06√{8.16-0.204(n-1)}=110+n

→21.06√{8.16-0.204(n-1)}=49.76+n, squaring both sides we get,

→444*{8.16-0.204(n-1)}=2476+99.52 n + n²

→n² +99.52 n + 90.58 n + 2476 - 3623+90.58 = 0

→n²+190.1 n - 1056.42 = 0, solving this quadratic equation for n and taking only the positive value we get n = 5.4

It means the ball clears the 5 th bench and hits the sixth bench.

45. A man is sitting on the shore of a river. He is in the line of a 1.0 m long boat and is 5.5 m away from the center of the boat. He wishes to throw an apple into the boat. If he can throw the apple only with a speed of 10 m/s find the minimum and maximum angles of projection for a successful shot. Assume that the point of projection and the edge of the boat are in the same horizontal level.

Answer: Since the length of the boat is 1.0 m and the man is sitting 5.5 m away from the center of the boat, for the apple to fall in the boat the range of projection should be minimum 5.0 m and maximum 6.0 m. We have u = 10 m/s.

Range x= (u² sin 2θ)/g Where θ is the angle of projection.

→sin 2θ = gx/u² for x = 5.0 m

sin 2θ = 9.8 *5/100 = 0.49

→ 2θ = 30° or 150° → θ = 15° or 75°

For x =6.0 m

sin 2θ = 9.8*6/100 = 0.59 → 2θ =36° or 144° → θ =18° or 72°

So for the successful shot minimum angle of projection is 15° and maximum 75°. In fact, the successful shots will be between angles 15° to 18° and 72° to 75°. between 18° to 72° the shot will not be successful.

46. A river 400 m wide is flowing at a rate of 2.0 m/s. A boat is sailing at a velocity of 10 m/s with respect to the water, in a direction perpendicular to the river. (a) find the time taken by the boat to reach the opposite bank. (b) how far from the point directly opposite to the starting point does the boat reach the opposite bank?

Answer: (a) Velocity of the boat perpendicular to river u = 10 m/s. Width of the river = d = 400 m. Time taken by the boat to reach opposite bank = d/u = 400/10 s = 40 s.

(b) In this 40 s time, the boat will go downstream with a velocity of 2.0 m/s. So the boat will reach opposite bank a distance = 2x40 = 80m from the point directly opposite the starting point.

47. A swimmer wishes to cross a 500 m wide river flowing at 5 km/h. His speed with respect to the water is 3 km/h. (a) if he heads in a direction making an angle θ with the flow, find the time he takes to cross the river. (b) find the shortest possible time to cross the river.

Answer: (a) The component of velocity perpendicular to direction of flow of river = 3sinθ km/h =3000sinθ/60 m/min = 50 sin θ m/min.

Time taken to cross the river = 500 m/ 50sin θ min = 10/sin θ min.

(b) For the shortest possible time the denominator "sin θ" should be maximum. And the maximum possible value of "sin θ" is 1 for θ=90°. So shortest possible time = 10/1 min =10 minutes.

48. Consider the situation of the previous problem. The man has to reach the other shore at the point directly opposite to his starting point. If he reaches the other shore somewhere else, he has to walk down to this point. Find the minimum distance that he has to walk.

Answer: Time taken to cross the river = 10/sin θ min.

Net speed of the man in the direction of flow = speed of water + component of man's speed along flow,

= 5 km/h + 3cos θ km/h

= (5+3cos θ) km/h = 1000(5+3cos θ)/60 m/min

So the distance from the point directly opposite the starting point to the point he reaches opposite bank = speed x time
y= 1000(5+3cos θ)/60 m/min X 10/sin θ min

= (1000/6).(5+3cos θ )/sin θ  m = (1000/6).(5cosecθ+3cot θ) m

For this to be minimum dy/dθ  = 0
(1000/6).(-5 cosecθ cotθ -3cosec²θ) = 0
→ cosec θ (5cotθ+3cosecθ) = 0
→ either cosecθ = 0 or 5cotθ+3cosecθ =0
since cosecθ ≠ 0 we have 5cotθ+3cosecθ =0
→ 3+ 5cos θ = 0 → cos θ = -3/5 → sin θ = 4/5

Putting these value in y=(1000/6).(5cosecθ+3cot θ) m
y= (1000/6)(25/4-9/4) = 4000/6 m =2000/3 km =2/3 km

49. An aeroplane has to go from a point A to another point B, 500 km away due 30° east of north. A wind is blowing due north at a speed of 20 m/s. The airspeed of the plane is 150 m/s. (a) Find the direction in which the pilot should head the plane to reach the point B. (b) Find the time taken by the plane to go from A to B.

Answer: Speed of wind u = 20 m/s

Speed of aeroplane v = 150 m/s

Figure for problem no. 49

(a) To reach point B the pilot should head the plane in such a direction that the resultant velocity is along AB. Let us suppose that the plane heads θ° east of line AB as shown in figure. So angle between two velocities = θ°+30° ,

So tan θ° = u sin(θ°+30°)/{v+u cos(θ°+30°)}

→ sinθ° {v+u cos(θ°+30°)}=u cosθ° sin(θ°+30°)

→sinθ°{v+ucosθ°cos30°-usinθ°sin30°)} =ucosθ°sinθ°cos30°+ucosθ° cosθ°sin30°

→vsinθ+√3/2usinθ°cosθ°u/2sin²θ° =√3/2usinθ°cosθ°+u/2cos²θ

→ vsinθ°- u/2* sin²θ°= u/2*cos²θ

→ v sinθ° = u/2*(sin²θ°+ cos²θ )

→ v sinθ° = u/2

→ sinθ° = u/2v = 20/300 =1/15

→ θ° = sin^-1 (1/15)

So in order to reach point B the plane should head with an angle sin^-1 (1/15) east of line AB.

(b) Angle between u and v = α =30°+θ°

Now the resultant velocity V²= u²+v²+2uv cos(30+θ)

=20²+150²+2.20.150{cos 30° cosθ-sin30°sinθ }

=400+22500+6000{√3/2√224/15-1/21/15}

{cos θ =√(1-sin²θ)} =√224/15

=22900+6000(√672-1)/30 = 22900+200*24.9 = 27884.6

→V= 167 m/s

Distance AB = 500 km = 500,000 m

Time taken by the plane to reach B = 500,000/167 s

=500,000/(60*167) min = 50 min

50. Two friends A and B are standing a distance x apart in an open field and wind is blowing from A to B. A beats a drum and B hears the sound t₁ time after he sees the event. A and B interchange their position and the experiment is repeated. This time B hears the drum t₂ time after he sees the event. Calculate the velocity of sound in still air v and the velocity of wind u. Neglect the time light takes in traveling between the friends.

Answer: In the first case resultant velocity = u+v

So the time taken t₁ = x/(u+v)

→ u+v = x/t₁ ..................(A)

In the second case resultant velocity = v-u

Now the time taken = t₂ = x/(v-u)

→ v-u = x/t₂ ................(B)

Adding the equations (A) and (B)

2 v = x/t₁ + x/t₂

→ v = x/2 * (1/t₁+1/t₂) It is the velocity of sound in still air.

Subtracting (B) and (A) we get

2u = x/t₁ - x/t₂

→ u = x/2 * (1/t₁- 1/t₂) It is the velocity of wind.

51. Suppose A and B in the previous problem change their positions in such a way that the line joining them becomes perpendicular to the direction of the wind while maintaining the separation x. What will be the time lag B finds between seeing and hearing the drum beating by A?

Diagram for problem number 51

Answer: As seen in the above diagram the resultant velocity of sound will be along AB and its magnitude AE can be calculated from the right-angled triangle ACE.

AE²=AC²-CE² = v²-u²

AE = √(v²-u²)

So the time lag B finds between seeing and hearing the drum beating by A

= Distance AB /velocity AE

= x/√(v²-u²)

52. Six particles situated at the corners of a regular hexagon of side a move at a constant speed v. Each particle maintains a direction towards the particle at the next corner. Calculate the time the particle will take to meet each other.

Answer: Let us draw a diagram.

Diagram for problem number 52

Consider any particle say A which moves towards B with a speed of v. But the particle B is itself moving with a speed v along BC. So movement of particle B along the AB direction will be v.cos∠CBG = v.cos60° =v/2.

Now the relative speed of Particle A with respect to particle B

= v-v/2 = v/2, This is the effective speed with which they come closer. To travel a distance of 'a' the time taken

= Distance/speed

= a/(v/2) =2a/v.

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