HC Verma solutions, Concepts of Physics, Part 1,Chapter 3, REST AND MOTION : KINEMATICS",'EXERCISES - Q31 to Q40

EXERCISES (Question number 31 to 40)

REST AND MOTION

31. An elevator is descending with uniform acceleration. To measure the acceleration, a person in the elevator drops a coin at the moment the elevator starts. The coin is 6 ft above the floor of the elevator at the time it is dropped. The person observes that the coin strikes the floor in 1 second. Calculate from these data the acceleration of the elevator.

 

Answer: Let the acceleration of the elevator be a'. Since the coin falls freely, its acceleration is due to gravity, g. As viewed from the elevator frame, the observer sees the coin's relative acceleration due to the frame change. This relative acceleration is = g -a'. Now using the equation h =ut +½at², we have u =0,   h = 6 ft, taking g =32 ft/s², t = 1 s,

→ 6 =½(g -a').1² 

→ g -a'= 12, → a' = g -12 =32 -12 = 20 ft/s².

32. A ball is thrown horizontally from a point 100 m above the ground with a speed of 20 m/s. Find (a) the time it takes to reach the ground, (b) the horizontal distance it travels before reaching the ground. (c) the velocity (direction and magnitude) with which it strikes the ground.

Answer: (a) Let the time to reach the ground be t. Since the ball is thrown horizontally, its downward velocity component will be zero, which means u=0. h = 100 m. From the equation of accelerated motion h = ut +½gt², we get, 

100 = ½*9.8*t² → t²=100/4.9 → t =4.5 s 

(b) The horizontal movement will be a uniform speed. Distance traveled before reaching the ground = speed x time

= 20 m/s x 4.5 s = 90 m. 

(c) When it strikes the ground, the horizontal component of velocity will be  Vx = 20 m/s.

The vertical component of the velocity 

Vy = u +gt = 0+ 9.8*4.5  

→Vy =44.1 m/s .

 So the magnitude of the velocity before it strikes the ground 

= √{(Vx)²+(Vy)²}= √(20²+44.1²) = 48.5 m/s ≈ 49 m/s 

If θ is the direction of velocity with horizontal.

tanθ =Vy/Vx =44.1/20 = 2.205, θ = 65.61 ≈66°. 

  

 

33. A ball is thrown at a speed of 40 m/s at an angle of 60° with the horizontal. Find (a) the maximum height reached and(b) the range of the ball. Take g=10 m/s². 

 

Answer: Given V=40 m/s, θ = 60° So horizontal component of the velocity 

Vx= V cosθ = 40. cos 60° =20 m/s . This component of the velocity will remain unchanged during the flight because there is no acceleration in the horizontal direction. 

(a) Vertical component of the velocity,

Vy =V sin θ =40 sin 60° =40*√3/2 =20√3 m/s.

    As we have seen in previous problems, the maximum height is given by h=u²/2g. Here, u =20√3 m/s, g =10 m/s². So we get  

h =(20√3)²/20 = 60 m.  

(b) Range of the ball = (u² sin 2θ)/g, Here u =V = 40 m/s, 

So range x =40² sin 120° /10 = 1600/10* √3/2 m =80 √3 m.

          or you can solve it as below:-

Time of flight, T = 2t =2√(2h/g) =2√(2*60/10) =4√3. 

Range =Horizontal speed*time =20*4√3 m =80√3 m. 

 

34. In a soccer practice session, the football is kept at the center of the field 40 yards from the 10 ft high goalposts. A goal is attempted by kicking the football at a speed of 64 ft/s at an angle of 45° to the horizontal. Will the ball reach the goalpost? 

 

Answer: u =64 ft/s, θ = 45°, Horizontal distance of goalpost =40yards =120 ft,

 Range = (u² sin 2θ)/ g

 = 64² sin 90°/32 = 128 ft, So the football will touch the ground beyond the goalpost. We need to know the height of the ball when it is vertically in line with the goalpost. 

Horizontal velocity = u cos 45° =64/√2 ft/s 

Time taken by the ball to cover the distance of 40 yards 

= 120 ft up to the goalpost =120÷(64/√2) = 15√2/8 s  

Now calculate the height of the ball after this time. The vertical component of the velocity 

= u sin 45° =64/√2 ft/s 

from h=ut-½gt² 

h = 64/√2 *15√2/8 - ½*32*(15√2/8)²   

   = 120 -16*450/64 =120 - 450/4 =120 - 112.5 ft = 7.5 ft,

Since the height of the goalpost is 10 ft, So, Yes, the football will enter the goalpost.

 

35.  A popular game in Indian villages is Goli which is played with small glass balls called Golis. The Goli of one player is situated at a distance of 2.0 m from the Goli of the second player. This second player has to project his Goli by keeping the thumb of the left hand at the place of his Goli, holding the Goli between his two middle fingers and making the throw. If the projected Goli hits the Goli of the first player, the second player wins. If the height from which Goli is projected is 19.6 cm from the ground and the Goli is to be projected horizontally, with what speed should it be projected so that it directly hits the stationary Goli without falling on the ground earlier?  

  

Answer: Let the required velocity be u. The vertical downward component of the velocity will be zero. Let it hit the ground in t seconds to drop vertically to a height of 19.6 cm =0.196 m. From the equation h = ut +½gt², we get, 

0.196=½*9.8* t²   →t=√(0.196/4.9) = 0.2 s 

Now it will hit the ground at a distance = ut = u*0.2= 0.2u m.

To hit the Goli this distance should be = 2 m, Equating we get, 

0.2u = 2    

→ u = 10 m/s.

 

36. Figure (3-E8) shows an 11.7 ft wide ditch with the approach roads at an angle of 15° with the horizontal. With what minimum speed should a motorbike be moving on the road so that it safely crosses the ditch? Assume that the length of the bike is 5 ft, and it leaves the road when the front part runs out of the approach road. 

The figure for Problem Number 36

Answer: To safely cross the ditch the motorbike should have a minimum range =11.7 ft + 5 ft =16.7 ft. Let the speed of the motorbike for this range be u. Given, θ =15°, We know, g =32 ft/s², So

Range x =(u² sin 2θ )/g 

→ 16.7 = (u² sin 30°)/32

→ u² = 16.7*32*2 

→ u =√1068.8 = 32.7 ft/s, is the minimum speed required to cross the ditch. 

 

37. A person standing on the top of a cliff 171 ft high has to throw a packet to his friend standing on the ground 228 ft horizontally away. If he throws the packet directly aiming at the friend with a speed of 15.0 ft/s, how short will the packet fall? 

 

Answer: Let us draw a diagram to understand the problem easily.

(Chapter-3, Rest & Motion: Kinematics)

    The person standing at A throws the packet along AC with a velocity u = 15 ft/s. Let the angle of velocity with horizontal be θ. 

Now, tan θ = 171/228 = 3/4, This gives sin θ = 3/5  and  cos θ =4/5. The vertical component of the velocity =u sin θ = 15*3/5 = 9 ft/s. This is the initial velocity of the packet for vertical accelerated motion. If the time to reach the ground is t. Using the equation h=ut+½gt² , we get, 

171 = 9t+½*32*t² 
→ 16t²+9t-171=0 
→t={-9±√(9²+4*16*171)}/32

taking only + ve sign for t, we get, t = (-9+105)/32 = 3 s. 
       In this time the packet will travel with the horizontal component of the velocity uniformly.

      Horizontal component of velocity = u cos θ = 15 *4/5 =12 ft/s. 

So the horizontal distance BD = 12 ft/s x 3 s = 36 ft. So the packet falls a distance CD ft short of C. CD = BC - BD = 228 -36 = 192 ft.

 

  

38. A ball is projected from a point on the floor with a speed of 15 m/s at an angle of 60° with the horizontal. Will it hit a vertical wall 5 m away from the point of projection and perpendicular to the plane of projection without hitting the floor? Will the answer differ if the wall is 22 m away?  

  

Answer: We have u= 15 m/s and θ = 60°

So range x = (u² sin 2θ )/g  = (225*√3/2)/9.8 = 19.9 m, 

So, Yes, the ball will hit the wall at a distance of 5 m without hitting the floor. 

Yes, the answer will differ because the ball will not hit the wall at a distance of 22 m away without hitting the floor.    

39. Find the average velocity of a projectile between the instants it crosses half the maximum height. It is projected with a speed u at an angle θ with the horizontal.  

Answer: Between the instants, the projectile crosses half the maximum height, its displacement is horizontal. So the direction of average velocity will be horizontal. In the horizontal direction, the projectile has no acceleration, so it moves horizontally with a uniform velocity equal to 'u cosθ'. So the magnitude of average velocity between the instants the projectile crosses half the maximum height remains equal to this uniform velocity "u cosθ".

Or we may solve it like this→ 

Let the time taken to cross the instants between half the maximum height = t, then the displacement =u cosθ*t. 

Average velocity =Displacement/time 

                       =(u cosθ*t)/t 

                       =u cosθ.

       

           

40. A bomb is dropped from a plane flying horizontally with uniform speed. Show that the bomb will explode vertically below the plane. Is the statement true if the plane flies with uniform speed but not horizontally? 

 

Answer: When the plane flying horizontally with a uniform speed, say u, drops a bomb, it gives a uniform horizontal speed u also to the bomb. So the bomb will move horizontally just below the plane but their vertical separation will increase with the time. If it explodes in the air, the bomb will be vertically below the plane. See the diagram below.

Diagram showing the plane and the bomb
at different instants of time

       If the plane flies with an angle θ with the horizontal with uniform speed u, both the plane and the bomb will have same speed and direction at the time of dropping the bomb. Both the plane and the bomb will have a horizontal component of speed equal to 'u cosθ' and vertical component of speed equal to 'u sinθ' at the time of separation. The vertical component of speed 'u sinθ' for the plane remains constant but for the bomb varies according to uniform accelerated motion under the gravity. So their vertical separation varies with time. But the horizontal component of the speed 'u cosθ' remains the same, so both the plane and the bomb move with same speed horizontally and even in this case the bomb will explode vertically below the plane. Hence the statement is still true. See the diagram below. 

Path of the plane and the bomb when the plane moves
with a uniform speed u but not in the horizontal direction.

   

==<<o>>==

Next and final set of solutions to problems 'Question numbers 41 to 52 in the next blog.. !!!

==<<0>>==

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