Detailed Solutions To H.c Verma With Short Answer And Objective Type Chapter 3 5l61c

JEE (MAIN & ADV.), MEDICAL + BOARD, NDA, X & IX

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[KINEMATICS] Chapter 3 1.

2.

3. 4. 5.

6. 7. 8.

9. 10.

11.

QUESTIONS FOR SHORT ANSWER According to Galileo earth moves around the sun and according to the church sun moves around the earth . If the absolute motion has no meaning the two view points are not equally wrong Because , if the earth or sun are not absolutely stationary , they have no effect in the circulation of one around the other . When a particle moves with constant velocity i.e. it’s velocity at any instant is same The magnitude and direction are same at any moment .The average vectocity gives the instantaneous velocity equal to the speed . The car has velocity 60 km per hour towards east . Another has velocity 60 km/hr . towards north . The velocity are not equal Because their velocity are perpendicular to each other and they cann’t be related V = 20m / sec . When a particle moves towards west in following order it has no acceleration in any direction . 1st sec . 2nd sec. 3rd sec. 4th sec. 5th sec . 5ms −1 5ms −1 6ms −1 8ms −1 9ms −1 When the projectile is at it’s highest point , in the projectile motion it’s speed is the smallest . They covered same distance . If a particle is accelerating it is either speeding up or speeding down . Yes , we agree with this statement , because with the speeding of it has positive acceleration and with the speeding down it has negative acceleration The path of the rocket as seen from the plane it is in a straight line where as it is in a parabolic path when it is seen from the ground , If someone asks then the actual path is the parabolic path . (a) When a body moves in a circular path , it’s acceleration is not zero ; but the velocity of the particle is zero . (b) When a body exhibits SHM it’s velocity is opposite to it’s acceleration . (c) When the body moves in a circular motion it’s radial velocity is perpendicular to acceleration . (a) At time t = t1 , Vx = + ve, ax = + ve

(b) At t = t2 , Vx = + ve, ax = 0

(c) At time t = t3 , Vx = −ve, C y = −ve

12. 13. 14.

1

They both have the same displacement . The car can be accelerated by changing the direction or motion Sun rays are electro magnetic waves . They can travel in a straight line but with the filling of rain they may be tilted or vertical . OBJECTIVE – I (b) 70km / hr. towards south west

2.

(d) The Particle moves at a constant velocity upto a time t0 and then stops

3.

(d) The information is insufficient . V +V (a) V = 1 2 2 (c) 2 / v = 1/ v1 + 1/ v2

4.

5.

As v =

2x 2v v ⇒ 1 2 ⇒ v ( v2 + v1 ) = 2v1v2 ⇒ 2 / v = 1 / v1 + 1/ v2 x / v1 + x / v2 v2 + v1 th

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2

BY R. K. MALIK’S NEWTON CLASSES

6.

(d) g down ward

7.

(c) VA = VB

8. 9.

(c) In a projectile motion the velocity is perpendicular to the acceleration for one instant only . (c) Both will reach simultaneously

10.

(d) x = v 2

sin 2θ v 2 sin 30 50 × g ⇒ 50 = ⇒ v2 = sin 30 g g

When θ = 45°

x= 11. 12. 13. 1. 2.

3.

v 2 sin 2θ 50 × g sin 90 50 × 1 = × = = 100m. g sin 30 g 1/ 2

(d) Because , the projectile velocity is unknown (a) Due north . (b) The mass M moves upward with a speed u / cos θ OBJECTIVE – II (a) In one hour the displacement is = 0 (d) The average velocity is = 0 X = u ( t − 2s ) + a ( t − 2s )

2

(c) The acceleration of the particle is 2a . (d) at t = 2 s. particle is at the origin . (a) Average speed of a particle in a given time is never less than the magnitude of the average velocity . (b) It is possible to have a situation in which dv / dt ≠ 0, but v = 0

9.

(c) The average velocity of a particle is zero in a time interval . It is possible that the instantaneous velocity is never zero in the interval . An object may have (b) Varying velocity without having varying speed . (d) Non – zero acceleration withoput having varying speed . (a), (b) , (d) (b), (c) , (d) (a) The magnitude of the velocity of a particle is equal to it’s speed (a) The particle has a constant acceleration . (d) The average speed in the interval 0 to 10 sec , is same as the average speed in the time interval 10 sec , to 20 sec . (a) The particle has come to rest 6 times .

10.

(d) The acceleration of s2 with respect to s1 may be anything between 0 and 8m / sec 2 .

4.

5. 6. 7. 8.

th

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1.

BY R. K. MALIK’S NEWTON CLASSES EXERCISE (a) The total distance he has to walk to reach the field = 50m + 40m + 20m = 110m (b) His displacement from his house to field = pv =

pu 2 + uv 2 ⇒ PV = 302 + 40 2 = 2500 = 50m

In ∆Pvs, tan θ =

θ

sv 30 3 = = ⇒ θ = tan −1 {3 / 4} north to east sp 40 4

40mr

R

20m

y

50m u

30 mr p

2.

s Total distance = 20 + 20 + 20 = 60m Total displacement = 20 − 20 + 20 = 20m Along the −ve directon .

( −20, 0 ) 3.

( 20, 0 )

From patna to Ranchi by air = 260km., From patna to Ranchi by road = 320km Time taken by aeroplane = 30 minutes = 0.5 hour , Time taken by a bus = 8 hours Total dist.covered 260km (a) Average speed of the plane = = = 520km / hr. Time taken 0.5hrs. (b) Average speed of the Bus =

Total dist.covered 320km = = 40km / hr . Time taken 8hrs.

260km = 520km / hr. 0.5m. From patna to Ranchi plane is going with a velocity 520 km/hr. (d) The straight path distance from patna to ranchi = 260km . as the bus takes time to reach = 8 hrs. 260km = 32.5km / hr. So ,the average velocity = V = 8hrs. (a) Dist travelled = 12416 − 12352 = 64km / hr Dist.travelled 64 = = 32km / hr. Average speed = Time 2 (b) Here the total displacement is zero , as he comes back to the home . Displacement 0 = = 0km / hr. Average Velocity = Time 2 (c) Average velocity of the plane =

4.

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3

4

5.

BY R. K. MALIK’S NEWTON CLASSES Time taken = t = 2.0 second , max speed = V = 18.0km / hr = 5m / s, a = ? Initral velocity = u = 0, Average acceleration = a = v − u / 1 = 5 − 0 / 2 = 2.5m. / sec 2

6.

Distance = Area under the time , 1 1 Speed 20 = bh = × 8 × 20 = 80m , Acceleration = = = 2.5m / s 2 2 2 Time 8 20 10

2

7.

4

6

8 10

t = 0, Intially u = 0, Distance travelled in first 10 second = a = 5.0 ft / s 2 , t = 10sec.

1 1 S = ut + at 2 = 0 + × 5 ×10 = 250 ft 2 2 The velocity during 10 second = V = u + at = 0 + 5 ×10 = 50 ft / sec. Distance covered during 20 sec. = s2 = ut + 1 / 2at 2 = 50 × 10 + 1/ 2 × 0 × 102 = 500 ft. Velocity during 20 second = v = u + at = 50 ft / sec. During the next 10 second acceleration const . a = 50 ft / s 2 1 1 S3 = ut + at 2 = 50 × 10 + × ( −5 ) ×10 × 10 = 500 − 250 = 250 ft 2 2 Total distance covered by the cart = S = S1 + S 2 + S3 ⇒ s = 250 + 500 + 250 = 1000 ft 8.

(a) Acceleration ==

V f − Vini t f − tini

=

8−2 6 = = 0.6m / s 2 . 10 − 0 10

(b) Distance = Area under st line = 1 / 2 DE.CE + AB.BE = 1 / 2 ×10 × 6 + 10 × 2 = 30 + 20 = 50m (c) Displacement = Distance = 50 m C

D

2

E

A0

9.

5

10

B (a) The average velocity during 0 to 10 second = Distance/time = 100/10 =10 m/second (b) Instantaneous velocity at 2 second = 50 / 2.5 = 20m / sec. At second ‘1’ the instantaneous velocity is zero . So at 5 second = 0 As it is moving with uniform velocity , So , instantaneous velocity at 8 second = 20m / s Instantaneous Velocity at 12 second = −20m / s. As the particle is coming down with a uniform velocity the answer is 20m / s having a ‘ −ve ' sign .

th

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BY R. K. MALIK’S NEWTON CLASSES X (n)

100 50 Time ( y )

10.

2.5 5 7.5 10 12.5 15 Distance travelled = sum of the area of the two ∆s

= 1/ 2 × 20 × 5 + 1/ 2 ( 40 − 20 ) 5 = 50 + 50 = 100m 11.

∴ The displacement is zero , so the aera velocity is zero . At , t = 0, Distance covered = S = 20m

Let us consider a point P at time , t = 12sec. at S = 20m . So , change in displacement = Distance between Final position – Distance between initial position = 20 − 20 = 0 So, Average Velocity = Displacement / time = 0 /12 = 0 So , at t = 12sec. the average velocity of the particles zero . x in m 0 p 20 10

0

12.

12 20

10 −t in sec

At position B instaneous velocity has direction along BC . For average velocity between A and B Vavg = Displacement / time

(

)

= AB / t = t in C . We can see that AB is along BC

i.e. they are in same direction . The point is B ( 60m,3m ) y A B 4 C 2 2

13.

4

6

Given : V = 4.0m / s,

x

a = 1.2m / s 2 ,

t = 5.0 sec.,

S =?

S = ut + 1 / 2at 2 ⇒ S = 4 × 5 + 1/ 2 × 5 × 5 = 20 + 15 = 35m. 14.

Given : V = 0,

U = 43.2km / h = 12m / s.

V 2 −U 2  a = −6.0m / s 2  ⇒S= 2a 15.

Given : u = 0, a = 2.0m / s 2 ,

V 2 − U 2 = 2as

∴S = 0 − t = 30sec,

0 − 144 144 = = 12m . 2 × ( −6 ) 12 v = 0, t = minute = 60sec

(a) Total distance moved by the train in the first case . Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1, Ph.: 0651-2562523, 9835508812, 7488587412

5

6

BY R. K. MALIK’S NEWTON CLASSES S = ut + 1 / 2at , ⇒ S1 = 0 + 1/ 2 × 2 × 30 × 30 = 900m 2

Velocity at that time V = u + at = 0 + 2 × 30 = 60m / s. In the second case , when the train starts decelerating , U = 60m / s, V = 0, t = 60 sec V = u + at S2 =

⇒ v −u / t = a

⇒ a = −60 / 60

= −1m / s 2

20 V 2 − u2 = 0 − 60 × = 1800m 2a 2 × ( −1)

Total distance covered = s = s1 + s2 = 900 + 1800 = 2700m = 2.7 km. (b) The maximum speed attained by the train = 60m / s (c) At half the maximum speed = 60 / 2 = 30m / s, the position of the train = v = 30m / s, u = 27 s, u = 0

16.

v − u 30 − 0 = = 15sec, a 2 ⇒ s = 0 + 1/ 2 × 2 × 15 × 15 = 225m Given : v = 16m / s, u = 0,

17.

V 2 − u2 v − u 0 − 16 = −320m, v = u + at ⇒ t = v − u = 2as ⇒ a = = = 0.05sec 2s a −320 u = 350m / s, s = 5.0cm = 0.05m, v = 0, a = ?

v = u + at ⇒ t =

2

s = ut + 1/ 2at 2 s = 0.4m

2

v2 − u 2 0 − 350 × 350 ⇒a= = 12, 25, 000 = 12.25 × 105 m / s 2 2as 2 × 0.05 v = 18km / hr = 18 × 1000 / 3600 = 20 / 4 = 5m / s

We know , v 2 − u 2 = 2 as ⇒ a = 18.

Given that : u = 0, t = 5.0 s,

v−u 5−0 1 = = 1m / s 2 ∴ s = ut + at 2 t 5 2 = 0 + 1/ 2 × 1× 5 × 5 = 25 / 2 = 12.5m / s, Avg velocity = 12.5 / 5 = 2.5m / s a=

19.

(b) The dist . travelled by the particle during this period is 2.5m / s a = −6.0m / s, V = 0, Given : t = 0.2sec, u = 54km / h = 20m / s,

S =?

Distance travelled in the reaction time = u = s / t ⇒ s = u × t ⇒ s = 15 × 0.2 = 3m . Distance travelled after applying the brakes , s =

v 2 − u 2 0 − 15 × 15 225 = = = 18.75m 2a 2 × ( −6 ) 12

∴Total distance travelled = 18.75 + 3m = 21.75m. 20. Car model Driver x [Reaction time 0.20s] (a) (Deceleration on hard Speed = 54km / hr. Braking 2 distance , a = 19 m braking = 6.0m / s ) Total stopping dist , t = 22m, Speed = 54km / hr (b) (Deceleration on hard Braking dist = e = 15m. braking = 7.5m / s 2 ) Total stopping dist f = 18m

th

Driver y [Reaction 0.30s ] Speed = 72km / hr Braking distance , c = 33m . Total stopping dist = d = 39m speed = 72km / hr Braking dist , g = 27 m Total stopping dist . h = 33m

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BY R. K. MALIK’S NEWTON CLASSES

02 − 152 = 19m. So , b = 0.2 × 15 + 19 = 33m this is a very good animals was that is a very good 2 ( −6 ) wander full re dig was letter was that is a very good was that is a very good was wander full re dig was letter was that is a very goodwas this aia Similarly others can be calculated Braking distance : Distance travelled when brakes are applied Total stopping distance = Braking distance + distance travelled in Feaction time Velocity of motor bike = vm = 72km / hr 72000 = = 20m / s velocity of joop = Vi = 90km / hr = 25m / s t = 10sec 3600 Let after ‘t’ second the jeep will catchup the bike ⇒ Dist covered by the jeep in t second = Dist covered by the bike in 10 sec+ dist . covered by jeep in t sec . 10v1 10 × 20 200 = v2t = v1 × 10 + v1 × t = t ( v2 − v1 ) = 10v1 = t = = = = 40 sec 5 v2 − v1 25 − 20 a=

21.

Distance travelled by the jeep = v2t ⇒ s = 25 × 40 = 1000m = 1km 22.

Accl . is zero due to its const speed . Total dist travelled by the car = 5 + 5 = 10m Relative velocity = 60 − 42 = 18km / hr = 5m / sec, But , s = ut ,

∴ a = 0,

s = ut + 1/ 2at 2 ,

t = s / v = 10 / 5 = 2 s

60 × 1000 × 2 200 = = 33m 3600 6 Total dist for overtake is = S A + its own length = 33 + 5 = 38m In 1st 2 sec.S A = U A × t =

V2 V1 10m After crossing

V2 V1

5m 5m Before crossing

23.

u = 50m / s,

.

g = a = −10m / s 2 ,

v = 0,

v 2 − u 2 0 − 50 × 50 = = 12 5m −2 × 10 2a (b) Time to reach the maximum height , v = u + at v − u 0 − 50 ⇒ t = v = u + at ⇒ = = 5sec. a −10 (c) The speed at half the maximum height , s = 125 / 2 = 62.5m .

(a) s =

v 2 − u 2 = 2as ⇒ v = 2as + u 2 = 2 × ( −10 ) × 62.5 + 50 × 50 = 1250 + 2500 ⇒ v = 1250 = 35.35 = 35m / s 24.

7

Given that : u = 7 m / s,

s = 60m, a = −10m / s 2

s = ut + 1/ 2at 2 ⇒ −60 = 7t − 1/ 2.10.t 2 7 ± 35.34 ∴t = = 4.2 sec. ∴ t = 4.2 s 10

= 5t 2 − 7t − 60 = 0

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8

25.

BY R. K. MALIK’S NEWTON CLASSES Given : u = 28m / s, v = 0, a = −g = −10m / s 2 (a) The maximum height reached by the stone , 0 − ( 25 ) v2 − u 2 v − u = 2as ⇒s= ⇒S= = 40m. −2 × 10 2a (b) velocity 1 sec before reaching the maximum height , The maximum height = s = 40m. v = ? u = 28m / s 2

2

2

Time taken to reach the maximum height = v = u + at = t =

v − u 0 − 28 = = 2.8sec −10 a

Time 1sec before = t1 = 2.8 − 1 = 1.8sec1 Velocity 1sec before = v = u + at = 28 + ( −10 ×1.8) = 10m / s 26.

27.

When the 6th ball is being dropped 5th ball has moved for 1sec , 4th ball for 2 sec & 3rd ball for 3sec . u = 0 for all the balls . 1 2 1 Distance tavelled by 5th ball is , gt = 9.8 × 1 = 4.9m. 2 2 th Distance tavelled by 4 ball is , 1 / 2.9.8.2.2 = 19.6m Distance tavelled by 3rd ball is , 1 / 2.9.32 = 44m s = 7 m, Let ‘t’ be the time taken for the child to fall 1 2 1 1 gt ⇒ H − h = gt 2 ⇒ (11.8 − 1.8 ) = × 10 × t 2 2 2 2 10 × 2 2 ⇒ = t ⇒ t 2 = 2 ⇒ t = 2 sec. is the time taken by the child to fall 10 Distance travelled by the man to reach the child is 7m s ⇒ u = ⇒ u = 7 / 2 = 4.9m / s. t

s=

11.8m 1.8m 7m

28.

29.

6 × 1000 10 5 = t= t 3600 6 3 In t sec . the derry falls from a height of 12.1m to hit the cadet’t head . 1 1 12.1 h = gt 2 ⇒ 12.1 = × 9.8 × t 2 ⇒ t2 = ⇒ t = 2.46 = 157 m. 2 2 4.9 ∴ S = 5 / 3 ×1.57 = 2.62m. Now distance to be covered by cadet is 1.57 m. Let after ‘t’ time distance covered by the cadet is S = ut =

t = 0.2sec,

s = 6.0m,

g = a = 10m / s 2

Let the height from which the ball was dropped is ( h + 6 ) m and the time taken by the ball to cover the dist of h metre be ‘t’ sec so , time taken to cover th total distance

( h + 6) m

is ( t + 0.2 ) sec. we know , h = 1/ 2 gt 2 .......(1)

as u = 0 and ( h + 6 ) = 1/ 2 g

( t + 0.2 )

2

......(2)

Substracting (1) – from (2) we get , h + 6 − h = 1 / 2 g ( t + 0.2 ) − 1/ 3 gt 2 2

th

⇒ 0.2 gt + 02 g = 6

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BY R. K. MALIK’S NEWTON CLASSES ⇒ 2t + 0.2 = 6 ⇒ 2t = 5.8 ⇒ t = 5.8 / 2 = 2.9sec

9

Putting the value of ‘t’ in eqn (1) we get , h = 1/ 2 × 10 × 2.9 × 2.9 ⇒ h = 42. 05m . The height from which the ball was dropped is h + 6 So Total height = 42.05 + 6 = 48.05m

h

t sec

(h + 6)

(1 + 0.2 ) 6m 0.2sec.

30.

Given : u = 0,

s = 5m.,

a = 10,

⇒ v = u 2 + 2as = 2 × 9.8 × 5 = 98 = 9.899 10 s= m. 100

So , v 2 − u 2 = 2as

When it gets buried , u = 10m / s, v = 0,

v2 − u 2 ∴ v − u = 2as ⇒ a = 2as 2

2

0 − ( 9.899 ) 97.99 ⇒a= = = 97.99 × 5 = 489.95 = 490m / s 10 1/ 5 2× 100 For elevator accl . of coin u = 0, Let the elevator descends down ward with accln . ‘a’’ The coin has to move more dist that is 1.8m . to strike the floor , Time taken = t = 1sec 2

31.

12 1 1 1 2 1 2  S = ut + = 0 + a 12 = a = + = + = SC  ut at 0 g (1)  1/ 2 g , e 2 2 2 2 2   1 a Total distance covered by coin is , 1.8 + 1/ 2a = g ⇒ 1.8 + = 9.8 / 2 = 4.9 2 2

( )

a 6f

6 f = 1.8m

1

a ⇒ = 4.9 − 18 = 3.1m ⇒ a = 6.2m / s 2 = 6.2 × 3.28 = 20.34 ft / s 2 2 32.

2

a

1 1 2s 2.100 (a) s = ut + at 2 ⇒ s = 0 + at 2 ⇒ t = = = 20 = 4.5 J 2 2 a 10 1 1 (b) R = Ut + at 2 ⇒ R = ( u cos θ ) t + 1/ 2at 2 ⇒ R = 20 20 + .0.t 2 2 2 = 20 20 = 90m. (c) VX = 20m / s, Vy = Uy + at = 0 + 9.8 × 4.5 = 44.1m / s.

V = VX 2 + Vy 2 = 202 + ( 44.1) = 48.42m / s. tan θ = 2

Vy Vx

=

44.1 ⇒ θ = tan −1 ( 2205 ) = 66 20

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10

BY R. K. MALIK’S NEWTON CLASSES

Vx Vy

33.

θ = 60° ,

u = 40m / s,

a = g = 10m / s 2

(a) maximum height reached = u 2 sin 2 θ / 2 g =

θ = 45° ,

Given : g = 32 ft / s 2 ,

Horizontal range = ( u cos θ ) t =

1 y = ut + at 2 2 35.

36.

( y − ht )

40 yard = 40 × 30 = 120 ft [As 3 ft = 1 yard]

120 15 2 = = 265sec 1 8 64 × 2

1 2 ⇒ y = ( 64sin 45° ) 265 − × 9.8 × ( 2.65 ) 2

Given : y = 19.6um,

y = ut + 1 / 2at 2

R = ut = v × 2 y / a ,

⇒ u = 2 1250 / 49

⇒ y = 1/ 2at 2 ,

⇒ y = 7.08 ft

⇒ t = 2y / a

= 2 / 7 1250 = 10.10m / s

Total horizontal distance = Range + its length = 11.7 + 5 = 16.7 ft , θ = 15, y = x tan θ − gx 2 / 2u 2 cos 2 θ = 0

⇒ u2 =

∴ x tan θ = gx 2 / 2u 2 cos 2 θ

gx gx 32 × 16.7 = = = 64 ×16.7 2 cos θ .sin θ sin 2θ sin 30

⇒ u 2 = 1068.8 37.

2 × 10

3 4 = 60m

40 × 40 × 3 = 160 × 3m 10

(b) The range of the ball = x = 42 sin 2θ / g 34.

40 × 40 ×

⇒ u = 32.69 ft / s

tan θ = h / x′ = 171/ 228 = 3 / 4 y = x tan θ −

gx 2 2u 2 cos 2 θ

⇒ x = 35.78 ft

∴θ = 37°

3 ⇒ −171 = × − 4

32 x 2

4 2 × 225 ×   5

2

≃ 30 ft.

Now , x′ − x = ( 228 − 36 ) ft = 192. ft. Thus , the packet would fall 192 ft short of his friend .

θ

X 5 ft / s

171 ft

x x 38.

x 2.28 ft

⇒ 0 = u cos θ + at

V = u + at

⇒ u cos θ = gt

15.1/ 2 3 = s Range , R = V cos θ t 10 4 = 15 × 1/ 2 × 3 / 4 = 5.6m ∴ Yes .

⇒t =

th

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BY R. K. MALIK’S NEWTON CLASSES

11

5m / s

60°

39.

5m Total of flight T = 2u sin θ / g ,

Average velocity = change in displacement / Time

From the fig , AB is horizontal . So there is no effect of vertical Component of the velocity during this displacement . So , the body moves at a const spped ‘u cos θ ' in horizontal direction The average velocity is this much .

A H /2H 40.

B

H /2 During the motion of bomb the horizontal velority ‘u’ remains const . and is same as that of aeroplne at every point of it’s path suppose the bomb explode i.e. reach the ground in time t . Distance travelled in horizontal direction by bomb = u = the dist , travelled by aeroplane . so bomb explode ventically below the aeroplane u

θ fig1

41.

(ii) suppose the aeroplane move making angle θ with horizontal . For both bomb & aeroplane Horizontal dist . is ‘u cos θ t '.t → Time to reach the ground so , in this case also , the bomb will explode vertically below the aeroplane . Let ‘T’ be the time of flight of ball . In ‘T’ time the rail road car covers = sa = ut + 1 / 2at 2 Due to inertia of motion the ball covers a horizontal dist . of sb = ut 2

2

 2v   2 × 9.8  Now , SC − Sb = ∆S = 1 / 2at = 1/ 2a   = 1/ 2   = 2m  9.8   q  2

42.

Range = nb,

H = nh, R = 4

2H g

If u is the min speed , x = 40, b

h

& θ = 0. book miss

b h

43.

y = −20,

b

0 (a) As seen from truck the ball move from vertically upwards & come back . 58.8 / 14.7 = 4 sec Time taken by the truck to cover 58.8m T = s / v

u = ?,

v = 0,

g = −9.8m / s 2 ,

t = 4 / 2s

= 2 sec.

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12

BY R. K. MALIK’S NEWTON CLASSES ⇒ u = 9.8 × 2 = 19.6m / s vertically upwards .

∴ v = u + at

(b) As seeen from the road , time of fight (T) = 2v2 / g 4 × 9.8 = 19.6m / s tan θ = v2 / v1 = 19.6 /14.7 = 4 / 3 2 Let the ball fall in the end of nth bench . Then , x = 110 + n, y = n − 1, Given : u = 35m / s

⇒V =

44.

(

)

10 110 + n 2 gx 2 ⇒ n − 1 = (110 + n ) tan 53 − θ = 53°, y = x tan θ − 2 2 2 2u cos 2 θ 2 ( 35 ) cos ( 53 )

⇒ ( n − 1) = (110 − n )

4 3

(

−10 12100 + n 2 + 220n 2450 ×

9 25

)

∴ n − 1 = 5, n = 6 (∴ 6th bench )

1m 110

45.

When the apple just touches the end B to the boat , x = 5m, v = 10m / sθ = ?, g = 10m / s 2 x = u 2 sin 2θ / g

⇒ 5 = (10 ) sin 2θ / 10 2

= sin 2θ = 1 / 2 = sin 30° or , sin150°

⇒ 5 = 10sin 2θ ⇒ θ = 15° or 75° .

Similarly for end c, x = 6m. Then 2θ = sin −1 ( gx / u 2 ) = sin −1 ( 0.6 ) = 180° or 71° ∴ So for successful short , θ may varies from 15° to 71° to 75° 10m / s

0

46.

(a) t =

ℓ 400 = = 40.s. v 10

(b) Let the boat reach pt . x. away from the just opposite from the fig .,

ℓ′ ℓ x = = v′ v u

ℓ ∴ x = .u ⇒ u = t.u = 40 × 2 = 80m v u = 2m / v

ℓ = 400

47.

ℓ′

v′

(a) Velocity = v = 3sin θ ℓ 500 /1000 0.5 10 t= = km = hr = min v cos θ 3sin θ 3sin θ sin θ (b) shortest possible time = ℓ / v = 0.5 / 3 = 1 / 6hr = 10 min th

Office.: 606 , 6 Floor, Hariom Tower, Circular Road, Ranchi-1, Ph.: 0651-2562523, 9835508812, 8507613968

48.

BY R. K. MALIK’S NEWTON CLASSES ℓ ℓ′ x Let it travels x dist ∴ = = v v′ u ℓ 5 2500 ⇒ x = ⋅ u = t.u = 10 × 60.5 × = m = 833.33m v 18 3 X ℓ

13

ℓ

u V

49.

V

(a) V p = 150m / s,

V = 20m / s from south , Vres = V p + Vw = Vp = Vres − Vw

Using sine law we get , V p / sin 30 = −Vw / sin α = Vres / sin (150 − α ) ∴V p / V2 = Vw / sin α

∴ sin ρ = 20 / 150.1/ 2 = 1/15

⇒ sin −1 (1/ 15) east of the car . B −Vw 30° (150 − α )

30°α

VP

A

(

(b) V p / V2 = Vres / sin (150 − ρ )

50.

)

Vres = 300sin 150 − sin −1 1 /15  speed = dist / time   500 500 ∴t = s / v = 1 1   300sin  150 − sin −1  3sin 150sin −1  sec 15  15    Velocity of sound = v, Velocity of air = u , Distnace between A and B is X In the first case , resultant velocity of sound = v + u. ( v + u ) t1 = x In the second case , resultant velocity of sound = v − u ,

⇒ ( v + u ) = x / t1..... (1)

∴ ( v − u ) t2 = x, ⇒ ( v − u ) = x / t2 ….. (2)

From (1) and (2) 2v = x / t1 + x / t2 = x (1/ t1 + 1/ t2 ) ⇒ V = x / 2 (1/ t1 + 1/ t2 ) From (1) u =

x x  x x  x x x1 1  −v = − + − =  +  = t1 t1  2t1 2t2  2t1 2t2 2  t1 t2 

Velocity of air V = x / 2 (1/ t1 + 1/ t2 ) , & Velocity of wind u = x / 2 (1/ t1 + 1/ t2 ) y

u u

51.

B X

V A X Velocity of sound v, Velocity of air u Velocity of sound be in direction AC . so it can reach B with resultant velocity AD

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14

BY R. K. MALIK’S NEWTON CLASSES Angle between v & u is θ > π / 2,

Resultant AD = v 2 − u 2

Here time taken by light reach ‘B’ is neglected . so time between seeing & hearing = time to here the drum sound .

t=

displacement x = ,= velocity v2 − u 2

x

( v + u )( v − u ) = x

x x = t1t2 t1 t2

θ

v

u X A B The particles meet at the centroid ‘o’ of the triangle any instant the particles will from a equilateral ABC with same centriod Consider the motion of particle A . At any instant its velocity makes 30° with AO . The component of this velocity along AO & v cos 30° This component is the rate of decreases of the dist AO . A

52.

2

a 1 Initially AO = a −   + 3 2 2 3

2

9 3 Therefore , the time taken for AO to become zero = V cos 30° A

O B

C

th

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