Class 12 Mathematics Dynamics Notes and Important Questions

Class 12 Dynamics Notes (Mathematics)

Newton’s Law of Motion Exercise 19.1 . 1. Find the impulse of a force acting on a mass of 50 kg producing a change in velocity from 5 ms–1 to 15 ms–1. Soln: mass (m) = 50 kg initial velocity (u) = 5 ms-1 final velocity (v) = 15 ms-1  Impulse = mv - mu = 50 × 15 - 50 × 5 = 500 kg ms-1. 2. A girl on a bicycle, total mass 50 k, has a velocity of 1 m/s and paddling faster for 5 seconds, the velocity increases to 3 ms-1. Find the average force exerted. Soln: mass (m) = 50 kg initial velocity (u) = 1 ms-1 final velocity (v) = 3 ms-1 time taken (t) = 5 seconds. force exerted (F) = ?  F = mv - mu t = m(v - u) t = 50(3 - 1) 5 = 100 5 = 20 N 3. A body of mass 50 kg falling from a certain height is brought to rest after striking the ground with a speed of 5 ms-1. If the resistance force of the ground is 500 N, find the duration of contact. Soln: mass of the body (m) = 50 kg initial velocity (u) = 0 final velocity (v) = 5 ms-1 force on the ground (F) = 500 N duration of contact (t) = ? Now, F = mv - mu t  t = m(v - u) F = 50(5 - 0) 5 = 250 500 = 0.5 sec 4. A cart is pushed on a frictionless smooth plane with an average force of 20 N for 5 seconds. If the cart with mass 50 kg is at rest in the beginning, find the velocity acquired by the cart. Soln: average force (F) = 20 N mass of the cart (m) = 50 kg time taken (t) = 5 secs. initial velocity (u) = 0 final velocity (v) = ? Now, F = mv - mu t  v = Ft m + u = 20 × 5 50 + 0 = 2 ms-1.

*382* Solution Manual to Basic Mathematics Mechanics 5. a) A rocket expels gas at the rate of 0.5 kgs–1. If the velocity of the gas expelled is 200 ms–1, what is the force produced by the rocket? Soln: rate of expelling     m t = 0.5 kgs-1 initial velocity (u) = 200 ms-1 final velocity (v) = 0 force produced by the rocket (F) = ?  F = mv - mu t = m t (v - u) = 0.5(200 - 0) = 100 N b) Sand allowed to fall vertically at a steady rate hits a horizontal floor with a speed 0.05 ms–1. If the force exerted on the floor is 0.005 N, find the mass of sand falling per second. Soln: initial velocity (u) = 0 final velocity (v) = 0.05 ms-1 force exerted on the floor (F) = 0.005 N mass of sand falling per second     m t = ? Now, F = m t (v - u)  0.005 = m t (0.05 - 0)  m t = 0.005 0.05 = 0.1 kg s-1 c) Suppose a rocket moving upwards in the air loses its mass as its fuel burns. If the velocity of the rocket is reduced from 210 ms–1 to 110 ms–1 by a force of 100 N due to earth, find the mass of fuel burnt per second. Soln: initial velocity (u) = 210 ms-1 final velocity (v) = 110 ms-1 force exerted (F) = 100 N mass of fuel brunt per second     m t = ? Now, F = mv - mu t  100 = m t (110 - 210)  m t = - 100 100 = -1 kgs-1 d) Suppose a steady mass of rain falls vertically on a flat roof at the rate of 0.5 kgs–1 and then comes to rest. If the force on the roof is 2.5 N, find the velocity of raindrops just before hitting the roof. Soln: quantity of rain falling per second     m t = 0.5 kgs-1 force exerted on the roof (F) = 2.5 N velocity before hitting the roof (u) = ? velocity after hitting the roof (v) = 0

Dynamics * 383* Newton’s Law of Motion Now, F = mv - mu t  F = m t (v - u)  2.5 = 0.5(0 - u)  u = -5 ms-1  u = 5 ms-1. Exercise 19.2 1. State Newton's Laws of motion and explain how from the first law we obtain a definition of the force and from the second a measure of force. Soln: Newton's Law's of Motion First law : Every body continues in its state of rest or of uniform motion in a straight line unless compelled by some external force to change that state. Second law : The rate of change of momentum is directly proportional to the impressed force and takes place in the direction of applied force. Third law : To every action, there is an equal and opposite reaction. The first law gives the definition of the force and second law gives the measurement of the force and third law specifies the property of force. First law gives definition of force Newton's first law serves to define a force. It states that the state of rest or of uniform motion of a body will change only when some external force is applied on it. An object on the table does not move unless it is pushed or pulled. Hence force is defined as that which change or tends to change the state of rest or uniform motion of a body in a straight line. Thus, the external agency is needed which change the state of rest of uniform motion of a body. This agency is called force. Second law gives measurement of force Suppose a force F acts on a body of mass m for a time t and changes its velocity from u to v. Impressed force  Rate of change of its momentum i.e. F mv - mu t or, F  m (v - u) t But v - u t = change of velocity time = acceleration = a(say)  F  ma  F = kma, where k is proportionality constant. This equation enable us to measure of force. 2. How large a force is required to bring a 1000 kg car moving with a velocity of 100 ms-1 to rest at (a) a distance of 1000 m; (b) in 20 second. Soln: mass of the car (m) = 1000 kg initial velocity (u) = 100 ms-1 final velocity (v) = 0 force (F) = ? a) Now, v2 = u2 + 2as  02 = (100)2 + 2 . a . 1000  2000 a = - 10000  a = -5 ms-2  F = ma = 1000×5 =5000 N.

*384* Solution Manual to Basic Mathematics Mechanics b) Now, v = u + at  0 = 100 + a . 20  a = -5 ms-2  F = ma =1000 × 5 = 5000 N. 3. A constant force of 10 N acting on an object reduces its velocity from 15 ms-1 to 5 ms-1 in 2s. Find the mass of the object. Soln: force (F) = 10 N initial velocity (v) = 15 ms-1 final velocity (v) = 5 ms-1 time taken (t) = 2 sec mass of the object (m) = ? Now, acceleration = v - u t =5 - 15 2 = -5 ms-2 The negative sign shows the retardation  F = m × a  10 = m × 5  m = 2 kg. 4. A 12.0 g bullet is accelerated from rest to a speed of 700 ms-1 as it travels 20 cm in the barrel of a gun. Assuming the acceleration to be uniform, find how large the accelerating force was ? Soln: mass of the bullet (m) = 12.0 g = 0.012 kg initial velocity (u) = 0 final velocity (v) = 700 ms-1 distance covered (s) = 20 cm = 0.2 m accelerating force (F) = ? Now, v2 = u2 + 2as  (700)2 = 02 + 2 × a × 0.2  490000 = 0.4a  a = 490000 0.4 = 1225000 ms-2 Hence, the required accelerating force, F = m × a = 0.012 × 1225000 N = 14700 N. 5. A bullet moving at 250 ms-1 penetrates 5 cm into a tree trunk before coming to rest. Assuming that the force exerted by the tree trunk is uniform, find its magnitude. Mass of the bullet is 10 g. Soln: velocity of the bullet (u) = 250 ms-1 mass of the bullet (m) = 10 g = 0.01 kg initial velocity (u) = 250 ms-1 final velocity (v) = 0 ms- 1 Let a is the retardation, then v2 = u2 – 2as  02 = (250)2 - 2 × a × 0.05  0.1a = (250)2  a = 625000 ms-2  F = ma = 0.01 × 625000 N = 6250 N. 6. A force of 25 newtons acts on a mass of 0.50 kg starting from rest. Find a) the acceleration in ms-2 Soln: force acting (F) = 25 N mass of the body (m) = 0.50 kg initial velocity (u) = 0 Now, F = m × a  a = F m = 25 0.50 = 50 ms-2.

Dynamics * 385* Newton’s Law of Motion b) the final velocity after 20 s. Soln: Let, v be the velocity after 20 seconds.  v = u + at = 0 + 50 × 20 = 1000 ms-1. c) the distance moved in 20 s. Soln: Let s is the distance moved in 20 secs.  s = ut + 1 2 at2 = 0 + 1 2 × 50 × (20)2 = 10000 m. 7. A train of mass 327 tones moves at the rate of 108 kmh-1, after the steam is shut off, it is brought to rest by the brakes in 50 m. Find the force exerted, assuming it to be uniform and assuming g = 9.81 ms-2. Soln: mass of the train (m) = 327 tones initial velocity (u) = 108 kmh-1 = 108 × 1000 60 × 60 = 30 ms-1 final velocity (v) = 0 distance described (s) = 50 m force exerted (F) = ? Let, a be the retardation due to brakes, then v2 = u2 - 2as  02 = (30)2 - 2 . a . 50  a = 900 100 = 9 ms-1  F = mass × acceleration = 327 × 9 tones ms-2 = 327 × 9 9.81 tones wt. = 300 tones wt. 8. On turning a corner, a motorist rushing at 36 km/hr finds a child 51 m ahead, he stops the car within one metre of the child by the application of the brakes. Calculate the retarding force and the time required to stop the car. The total mass of the car and the passenger = 2000 kg. Soln: distance covered (s) = 50 m velocity of the car (v) = 36 kmh-1 = 36 × 1000 60 × 60 = 10 ms-2 mass of the car (with passenger) (m) = 2000 kg Now, v2 = u2 - 2as  0 = 102 - 2 × a × 50  a = 1 ms-2 F = m × a = 2000 × 1 = 2000 N. Again, if t is the required time, then v = u + at  0 = 10 - 1 × t  t = 10 secs.. 9. A force equal to a weight of 1 kg acts on a body continuously for 10 secs and causes it to distance 10 metres in that time, find the mass of the body. Soln: force (F) = wt. of 1 kg = 1 × 9.8 = 9.8 N Initial velocity (u) = 0 m/s  acceleration of object (a) = ? Now, s = ut + 1 2 at2  10 = 1 2 × a × (10)2  a = 0.2 m/s2 Again, F = ma 9.8 = m × 0.2 m = 49 kg.

*386* Solution Manual to Basic Mathematics Mechanics 10. Due to the application of a force of 12 kg.wt. a body of mass 5 kg changes its velocity from 12 m/s to 20 m/s. Find the distance through which the body described. Soln: mass (m) = 5 kg Force (F) = 12 kg wt = 12 × 10 N = 120 N Now, F = ma  120 = 5 × a  a = 24 m/s2 We have, v2 = u2 + 2as  202 = 122 + 2.24.s  s = 5 1 3 m. 11. Find the velocity of a 4 kg shot that will just penetrate through a wall 25 cm thick, the resistance being 36 tones wt. (g = 9.8 ms-2) Soln: mass of shot (m) = 4 kg penetrated space (s) = 25 cm = 1 4 m resistance (F) = 36 tones wt = 36 × 1000 × 9.8 N final velocity (v) = 0 m/s initial velocity (u) = ? If a is the retardation produced by the wall Then, F = -ma  a = - F m = 36 × 1000 × 9.8 4 = - 88200 m/sec2 Again, v2 = u2 + 2as  02 = u2 - 2 × 88200 × 1 4  u2 = 44100 u = 44100 ms-1 = 210 ms-1. 12. A body of mass 1 kg is falling under gravity at the rate of 28 ms-1. What is the uniform force that will stop it; i) 0.1 second, ii) 20 cm (g = 10 ms-2) Instead of falling under gravity if the body is moving at the rate of 20 ms-1 along a horizontal line, what will be the force required in above two cases. Soln: mass of the body (m) = 1 kg initial velocity (u) = 28 ms-1 final velocity (v) = 0 The forces acting on the body are (a) weight of the body, mg = 1 g in downward direction (b) the force F acting vertically upward direction Case (i) let a be the retardation after striking the ground before coming to the rest for 0.1 s. Then, v = u – at 0 = 28 - a × 0.1  a = 28 0.1 = 280 m/sec2 Then by second law of motion we have F - mg = ma  F = m(g + a) = 1 × (10 + 280) = (10 + 280) N = 290 N = 290 10 kg wt = 29 kg wt. Case (ii) let a' be the retardation after striking the ground before coming to the rest at the distance of 20 cm = 0.2 m Then, v2 = u2 - 2a's  02 = 282 - 2 × a' × 0.2  0.4a' = 784  a' = 784 0.4 = 1960 ms-2

Dynamics * 387* Newton’s Law of Motion Then by second law of motion we have F - mg = ma'  F = m(g + a') = 1(10 + 1960) N = 1970 N = 1970 10 kg wt. = 197 kg wt.. Instead of falling under gravity if the body is moving along the horizontal at the rate of 28 ms-1, the required force are i) F = ma = 1 × 280 N = 280 N = 280 10 kg wt = 28 kg wt. and ii) F = ma' = 1 × 1960 N = 1960 N = 1960 10 kg wt = 196 kg wt. 13. A mass of 5 kg falls 300 cm from rest and is then brought to rest by penetrating 30 cm into some sand; find the average thrust of the sand on it. Soln: mass (m) = 5 kg initial velocity (u) = 0 m/s height (h) = 300 cm = 0.3 m Suppose v is the velocity of the body when it falls 300 cm from rest under gravity. Then, v2 = u2 + 2gh  v2 = 02 + 2 . g . 3  v2 = 6g ..................... (i) The velocity given by (i) is reduced to zero when the body goes to 30 cm = 0.3 m into sand. If a is the retardation of the system them 02 = v2 - 2 × a × 0.3  a = v2 0.6 = 6g 0.6 = 10 g m/sec2 Let, F be the average thrust of the sand on the body. Then the forces acting on the body are (a) The weight 5 g N of the body acting downward. (b) A force F N of the sand acting upward Then applying Newton's second law of motion we have F - mg = ma  F - 5g = 5 × 10 g  F = 55g  F = 55 kg wt.. 14. A particle of mass 15 kg falls from a height of 18 metres and penetrates into some sand. If the average resistance offered by the sand is equal to a force of 150 kg.wt., find how far it penetrates into the sand. Soln: mass(m) = 15 kg height (h) = 18 m let v be the velocity just before penetrating, then v2 - u2 = 2gh [ u = 0]  v2 = 2 × 10 × 18 = 360  v2 = 360 After striking the sand the forces acing on the body are i) The weight mg of the stone acting vertically downwards. ii) The resistance F offered by the sand acting vertically upward. Then applying Newton's second law; F - mg = ma'  F = m(g + a')

*388* Solution Manual to Basic Mathematics Mechanics  150 × 10 = 15 (10 + a')  a' = 90 Now, retardation of particle (a') = 90 while penetrating the sand s m far from the surface of sand. Thus we have, v2 = u2 - 2a's  0 = 360 - 2 × 90 × s [v2 = final velocity = 0]  s = 2m. 15. A mass 'm' kg is acted on by a constant force 'P' kg wt and in 't' secs; it moves a distance of x metres from rest and acquires a velocity of v m/s. Show that x = gt2P 2m = v2m 2gP . Soln: mass of the body = m kg force acting = P kg wt. = Pg N time taken = t secs. distance described = x metres initial velocity = 0 final velocity = v ms-1 If a is the acceleration produced, then by second law of motion we have Force = mass × acceleration produced  Pg = ma  a = Pg m Again, s = ut + 1 2 at2  x = 0 × t + 1

Pg m t2  x = gt2P 2m ................... (i) Also, v2 = u2 + 2ax  v2 = 02 + 2 . Pg m . x  v2 = 2Pgx m  x = v2m 2gP .................... (ii) Combining (i) and (ii), we get x = gt2P 2m = v2m 2gP 16. A balloon is rising with an acceleration f. Prove that the fraction of the weight of the balloon which must be emptied out of the balloon in order to double the acceleration is f g + 2f . Soln: If the mass of the balloon is m, then its weight = mg Let, f be the acceleration, F be the upward lifting force of the balloon. Then, by Newton's second law of motion we have, F - mg = mf or, F = mg + mf .............. (i)

Dynamics * 389* Newton’s Law of Motion Suppose, x be the mass to be taken out from the balloon so that it may now move up with acceleration 2f Then F - (m - x)g = (m - x)2f or, mg + mf - mg + xg = 2mf - 2xf or, xg + 2xf = 2mf - mf or, (g + 2f) x = mf  x m = f g + 2f Exercise 19.3 1. A force of 12 N acts for 5 s on a mass of 2 kg. What is the change in momentum of the mass ? What would be the change in momentum of a mass of 10 kg under the same condition ? Soln: force act (F) = 12 N time taken (t) = 5 secs. change in momentum = ? Case I : When mass (m) = 2 kg Then F = m × a  12 = 2 × a  a = 6 ms-2 Again, v = u + at  v = 0 + 6 × 5  v = 30 ms-1  Change in momentum = mv - mu = 2 × 30 - 2 × 0 = 60 kg ms-1 Case II : When mass (m) = 10 kg  Change in momentum = 10(30 - 0) kg ms-1 = 300 kg ms-1. 2. A body of mass 0.5 kg and initially at rest, is subjected to a force of 2 newtons for 1 sec. Calculate the change in momentum of the body. Soln: mass of the body (m) = 0.5 kg initial velocity (u) = 0 force exerted (F) = 2N time taken (t) = 1 sec. Then, F = m × a  2 = 0.5 × a  a = 4 ms-2 Again, v = u + at  0 + 4 × 1 v = 4 ms-1  Change in momentum = mv - mu = m(v - u) = 0.5(4 - 0) = 2 Ns. 3. A bullet of mass 0.006 kg travelling at 120 ms-1 penetrates deeply into a fixed target, and is then brought to rest in 0.01 s. Find a) Change in momentum of the bullet Soln: mass of the bullet (m) = 0.006 kg velocity of the bullet (u) = 120 ms-1 time taken (t) = 0.01 sec. final velocity (v) = 0 m/s  Change in momentum = mv - mu = m(v - u) = 0.006 × (120 - 0) = 0.72 Ns. b) The average retarding force exerted on the bullet Soln: If F is the average retarding force exerted on the bullet then, F = change in momentum time taken = 0.72 0.01 N = 72 N.

*390* Solution Manual to Basic Mathematics Mechanics c) The distance of penetration of the target. Soln: If a is the retardation of the system Then F = m × a  a = F m = 72 0.006 = 12000 ms-2 If s is the required distance of penetration of the target Then v2 = u2 – 2as 0 = 1202 – 2.1200.s  s =

2400 = 0.6 m. 4. An inflated balloon contains 2.0 g of air which is allowed to escape from a nozzle at a speed of 4.0 ms-1. Assuming that the balloon deflates at a steady rate in 2.5 s, what is the force exerted on the balloon ? Soln: mass of the air (m) = 2 gm = 0.002 kg initial speed of air (u) = 4 ms-1 final velocity of air (v) = 0 time of duration (t) = 2.5 sec Then F = mass × acceleration = mass × change in velocity time taken = 0.002 × 4 2.5 = 0.0032 N. 5. A bullet of mass 10 g is fired from a rifle of mass 1000 g with a velocity of 50 kmh-1. Find the velocity of the recoil of the rifle. Soln: mass of the bullet (m) = 10g = 10 1000 kg = 0.01 kg mass of the rifle (M) = 1000 g = 1 kg muzzle velocity of the bullet (m) = 50 kmh-1 recoil velocity of the rifle (V) = ? Using the principle of conservation of linear momentum we have, mv = MV  0.01 × 50 = 1 × V  v = 0.5 kmh-1. 6. A ball of mass 0.1 kg moving with a velocity of 6 ms-1, collides directly with a ball of mass 0.2 kg at rest. Calculate their common velocity if both balls move together in the same direction. Soln: Here, m1 = 0.1 kg, m2 = 0.2 kg, u1 = 6 m/s u2 = 0 m/s Combined mass of balls = (m1 + m2) = (0.1 + 0.2) kg Let V be the velocity of the combination. Then Using the principle of conservation of linear momentum we have, m1 u1 + m2 u2 = (m1 + m2) V  0.1× 6 + 0 = (0.1 + 0.2) V  0.3 V = 0.6 V = 2 ms-1. 7. Two bodies of masses 8 and 4 kg move along the x-axis in opposite directions with velocities 11 ms-1 and -7 ms-1 respectively. They collide and stick together. Find their velocity just after collision. Soln: Here, m1 = 8 kg, m2 = 4 kg u1 = 11 m/s u2 = - 7 m/s

Dynamics * 391* Newton’s Law of Motion Let V be the velocity of the combination. Then using principle of conservation of linear momentum, the total momentum of the system remains constant. Hence momentum before collision = momentum after collision. m1 u1 + m2 u2 = (m1 + m2) V  8 × 11 + 4 × (-7) = (8 + 4) V  12V = 88 - 28 V = 60 12 = 5 ms-1. 8. A 10 g bullet is fired from a kilogram gun suspended to move freely. This bullet now enters a block of wood of mass 990 g. If the speed of the bullet is 500 ms-1, find the speed of the gun and the common velocity of the wood. Soln: mass of the bullet (m) = 10 g = 10 1000 = 0.01 kg mass of the gun (M) = 1 kg muzzle speed of the bullet (v) = 500 ms-2 Speed of the gun (V) = ? Using principle of conservation of linear momentum, we have mv = MV  0.01 × 500 = 1 × V  V = 5 ms-1. Again, combined mass of bullet and gun = 10g + 990g = 1 kg. Let V1 be the common velocity of bullet and wood then momentum of the system before firing = momentum of the system of after firing,  0.01 × 500 + 0.9 × 0 = 1 × V1  5 + 0 = v  v = 5 ms-1. 9. A gun of mass 400 kg fires a shot of mass 3 kg, with a velocity of 200 ms-1, find the constant force which acting on the gun would stop it after a recoil of 2.5 metres. Soln: Now, momentum of the shot = mv = 3 × 200 momentum of the gun = MV = 400 × V  Momentum of the shot = momentum of the gun  3 × 200 = 400 × V  V = 3 × 200 400 = 6 4 ms-1 = 3 2 ms-1 Let, a be the retardation of recoil of gun. Then, 0 = V2 - 2as  V2 = 2as      3

= 2 × a × 2.5 9 4 = 2 × 2.5 × a  a = 9 4 × 2 × 2.5 = 9 20 ms-2 Hence, the constant force acting on gun = F = ma = 400 × 9 20 = 180 N. 10. A shot of 2 kg is discharged by a gun of mass 400 kg with a velocity of 800 m/sec. Find the constant force which would be required to stop the recoil of the gun in (i) 2 metres, (ii) 1

4 sec. Soln: mass of shot(m) = 2 kg velocity of shot(v) = 800 m/s mass of gun (M) = 400 kg recoil velocity of gun ( V) = ?

*392* Solution Manual to Basic Mathematics Mechanics Now, mv = MV  2 × 800 = 400 × V  V = 4 m/sec Let F be the constant force required to stop the recoil of gun (i) at distance 2 metres Then 02 = V2 - 2as  16 = 2a2  a = 4 m/s2  F = m × a = 400 × 4 = 1600 N. (ii) at time 1 1 4 = 5 4 seconds Then 0 = V - at  4 = a.5 4 a =

 F = m × a = 400 ×

5 = 80 × 16 = 1280 N. 11. A gun of mass 1 metric ton, fires a shot of mass 14 kg and recoils up smooth inclined plane, rising a height of 1.6 m, find the initial velocity of the projectile. Soln: Suppose u be the initial and v be the final velocity of the gun. Now, 02 = u2 - 2gh  u2 = 2 × 9.8 × 1.6 = 31.36  u = 31.36 = 5.6 m/sec Again, we have mass of the gun (M) = 1 metric ton = 1000 kg velocity of the gun (V) = 5.6 m/sec mass of the shot (m) = 14 kg velocity of the shot (v) = ? Now, MV = mv  1000 × 5.6 = 14 × v  v = 1000 × 5.6 14 = 400 m/sec. 12. A cricket ball of mass 150 g is moving with a velocity of 12 ms-1 and is hit by a bat so that the ball is turned back with a velocity of 20 ms-1. The force of the blow acts for 0.01 s. Find the impulse and the average force exerted on the ball by the bat. Soln: Here, mass of the ball (m) = 150 g = 0.15 kg initial velocity of the ball (u) = 12 ms-1 final velocity of the ball (v) = 20 ms-1 time duration (t) = 0.01 sec Now, impulse of the force = change in momentum = mass × change in velocity = 0.15 × (120 - 12) = 0.15 × 32 = 4.8 MKS units of impulse. Again Let, R be the average force exerted on the ball by the bat Then impulse of the force = R × t  4.8 = R × 0.01  R = 4.8 0.01 = 480 N.

Dynamics * 393* Newton’s Law of Motion 13. A gun of mass 40 metric tones resting on an incline of 3 in 5, fires a shot of 100 kg horizontally with a velocity of 700 m/sec. Find the velocity of the recoil of the gun and the distance it moves up the incline before coming to rest. (g = 9.8 ms-2) Soln: mass of the gun (M) = 40 metric tones = 40000 kg Mass of shot (m) = 100 kg Velocity of the shot (v) = 700 m/sec. Velocity of gun (V) = ? If  is the angle of inclination of the gun with the inclined plane. Then from figure, sin  = 3 5 and cos  = 4

velocity of the shot in the direction of the inclined plane = 700 cos  m/sec. Therefore its momentum = 100 × 700 × 4 5 = 56000 ........... (ii) Now, mv = MV  40000V = 56000  V = 7 5 ms-1 = 1.4 ms-1. Let s be the distance travelled by the gun along the horizontal, then acceleration due to gravity in the direction of horizon = -g sin Now, v2 = u2 + 2as  02 =     7

+ 2(-g sin)s  0 = 49 25 - 2 × 9.8 × 3 5 × s  s = 49 × 5 25 × 2 × 9.8 × 3 = 49 294 = 1 6 m. Hint and solution of MCQ's 1. Force is a vector quantity 2. When force acting on a body is zero, then momentum is constant. 3. The pull on a body of mass 5 kg due to the earth is Mg N = 5 × 9.8N = 49N 4. Mg = 60  m = 60 g = 60 10 = 6 kg 5. F = 20N m = 4kg a = F m = 20 4 = 5m/s2 6. m = 1gm a = 1 cm/s2 F = ma = 1dyne (the CGS unit of force is dyne) 7. The momentum of body = mass × velocity 8. m = 16kg, v = 20 m/s  momentum = mv = 320 kgm/s

*394* Solution Manual to Basic Mathematics Mechanics 9. F = 5N, m = 0.2kg, t = 15 F = mv – mu t  5 = 0.2v – 0 1  0.2v = 5  v = 25m/s 10. Impulse = Force × time = Ft 11. F = 15N, v = 18m/s, u = 6m/s, t = 4s F = mv – mu t = m    v – u t  15 = m 12 4  m = 5 kg Impulse = Ft = 15 × 4 = 60Ns = 60kg m/s 12. m = 50gm = 0.05kg v = 20m/s u = 0m/s F = 2.5 N  F = m    v – u t  2.5 = 0.05 × 20 t  t = 0.05 × 20 2.5 = 0.4 sec 13. m = 24 kg, v = 15 m/s, u = 0 m/s, t = 5s  F = m    v – u t = 24 × 15 5 = 72N 14. F = 200 kg u = 50m/s v = 0m/s s = 25m Now, v2 = u2 – 2as [ a = average retardation]  02 = 502 – 2a.25  a = 50 × 50 50 = 50m/s2  F = ma = 200 × 50 = 10,000 N 15. m = 49 kg, F = 10kg, wt = 10 × 9.8 N, u = 10m/s, v = 0m/s Now, – F = m    v – u t (Retarding force)  t = m(v – u) –F = 49 × – 10 – 9.8 = 5s 16. Force of body = mg N, acting vertically down, Force of lift = ma, acting vertically up  Pressure on lift moving up = ma + mg = m(a + g)

Dynamics * 395* Newton’s Law of Motion 17. Force A body = 60g N, acting down Force of lift = 60N, acting down  Pressure on lift moving down = mg – ma = 60g – 60 ×0.8 = 60 × 9.8 – 48 = 540N 18. m = 4.9 kg, u = 50m/s, v = 0m/s let a be the retardation for the body coming to rest then v = u – at  0 = 50 – a × 5  a = 10m/s2 Then using Newton's law Resultant force, F = mg + ma = m(g + a) = 4.9 × 20 = 98N 19. m = 4.9 kg F = 12 tonnes wt. = 12 × 1000 × 9.8N Let a be the retardation then F = ma  12 × 1000 × 9.8 = 4.9 × a  a = 24,000 m/s2 20. m = 0.4kg (mass of bullet) M = 24 kg (mass of gun) v = 360 m/s (velocity of bullet) V = ? (recoil velocity of gun) using Newton's third law MV = mv  24 × V = 0.4 × 360  V = 6 m/s 21. m = 0.5 kg, F = 2N, t = 15 Change in momentum = F × t = 2 × 1 Ns = 2Ns 22. m = 0.05 u = 200m/s v = 150m/s  Impulse = m [v – (–u)] [being opposite] = 0.05 [v + u] = 0.05 × 380 = 19

*396* Solution Manual to Basic Mathematics Projectile Exercise 19.4 . 1. A particle projected upwards from the level ground at an angle of 60° with the horizon has an initial speed of 40 3 ms-1. (g = 10 ms-1) (a) How long will it be before it hits the ground ? (b) How far from the starting point will it strike ? Soln: initial velocity (u) = 40 3 ms-1 angle of projection () = 60° (a) Let, T be the time of flight  T = 2u sin g = 2 × 40 3 × sin 60° 10 = 2 × 4 3 × 3 2 = 12 secs. (b) Let, H he the horizontal range  H = u2 sin2 g = ( ) 40 3 2 × sin (2 × 60°) 10 = 40 3 40 3 3 10 2    240 3 m. 2. A ball is projected with an initial upward velocity component of 20 ms-1 and a horizontal velocity component of 25 ms-1. (g = 9.8 ms-2) (a) Find the position and the velocity of the ball after (i) 2 s (ii) 3 s (iii) 4 s. (b) How much time is required to reach the highest point of the trajectory ? (c) How high is the point? (d) How much time (after launch) is required for the ball to return to its original level? (e) How far has it travelled horizontally during this time? Illustrate your answer with neat and sufficiently large sketch. Soln: (a) Let, V be the velocity and P(x, y) be position of projectile at time t. Let, v be the velocity and  be angle of projection v cos = 25, v sin  = 20 x = (v cos) ................ (i) y = (v sin)t - 1 2 gt2 .......... (ii) Then V is the resultant of two velocities Vx and Vy given by Vx = v cos (along horizontal) ................ (iii) and Vy = v sin - gt (along vertical) .............. (iv) (i) After 2 second Substituting value of t = 2, on (i), (ii), (iii) and (iv), we get x = 25 × 2 = 50 m. y = 20 × 2 - 1 2 × 9.8 × 22 = 40 - 19.6 = 20.4 m Vx = 25 ms-1 Ans Vy = 20 - 9.8 × 2 = 20 - 19.6 = 0.40 ms-1 (ii) After 3 seconds Substituting the value of t = 3 on (i), (ii), (iii) and (iv), we get x = 25 × 3 = 75 m y = 20 × 3 - 1 2 × 9.8 × 32 = 60 - 44.1 = 15.9 m

Dynamics * 397* Projectile Vx = 25 ms-1 Vy = 20 - 9.8 × 3 - 9.4 ms-1 (iii) After 4 seconds Substituting the value of t = 4 on (i), (ii), (iii) and (iv), we get x = 25 × 4 = 100 m y = 20 × 4 - 1 2 × 9.8 × 42 = 80 - 78.4 = 1.6 m Vx = 20 ms-1 Vy = 20 - 9.8 × 4 = 20 - 39.2 = -19.2 ms-1 (b) Let, t be required time to reach the highest point, then t = v sin g = 20 9.8 = 2.04 sec. (c) Let, H is the greatest height attained by the ball, then H = v2 sin2 2g = (v sin)2 2g = 202 2 × 9.8 = 400 19.6 = 20.4 m. (d) Let T be the time of flight, then T = 2v sin g = 2 × 20 9.8 = 40 9.8 = 4.08 sec. (e) Let R be the Horizontal range, then  Horizontal range, H =

v sin2α 2.v cos α.vsin α 2.25.20 = g 9.8 g  = 102.04 m. 3. A helicopter flying horizontally with a speed of 30 ms-1 at an altitude of 500 m has to drop a food packet for a person standing on the ground. At what distance from the person should the packet be dropped ? The man stands in the vertical plane of the helicopter's motion. (g = 10 ms-2) Soln: Let, H be the position of the helicopter at an altitude of 500 m from the ground and M be the position of the man. Since the helicopter flying horizontally, its vertical velocity component is zero. So, we have h = 1 2 gt2  500 = 1 2 × 10 × t2  10t2 = 1000  t2 = 100  t = 10 sec Let, x be the required distance of the dropped packet from the man. Also, the horizontal velocity component = 30 ms-1  x = ut = 30 × 10 = 300 m. 4. Prove that a projectile will rise three times as high when its angle of elevation is 60° as when it is 30°, but will cover the same horizontal distance. Soln: Let, u be the velocity attained by the projectile. Case I : For the height attained by the projectile Let, H1 and H2 be the height attained by the projectile when angle of elevation are 60° and 30° respectively. Then H1 = u2 sin2 60° 2g = u2 2g ×     3

= 3u2 8g ..... (i)

*398* Solution Manual to Basic Mathematics Mechanics H2 = u2 sin2 30° 2g = H2 = u2 2g ×     1

= u2 8g ..... (ii) Dividing (i) by (ii), we get H1 H2 = 3 1  H1 = 3H2  H1 is the three times greater than H2. Case II : For horizontal range Let, R1 and R2 be the horizontal range of the projectile when angle of elevation are 60° and 30° respectively. Then R1 = u2 sin (2 × 60°) g = u2 g × 3 2 = 3 u2 2g ..... (iii) R2 = u2 sin (2 × 30°) g = u2 g × 3 2 = 3 u2 2g ....... (iv) From (iii) and (iv), we get R1 = R2  Two horizontal range are same. 5. A spring gun projects a golf ball at an angle of 45° above the horizontal. The horizontal range is 10 m. (g = 10 ms-2) (a) What is the maximum height to which the ball rises ? (b) For the same initial speed, what are the two angles of departure for which the range is 5 m ? Soln: Let, u be the velocity and T be the time of flight of the projectile and the horizontal range (R) = 10 m. Then T = 2u sin45° g and R = u cos .T  10 = u cos45° × 2u sin45° g  10 = u × 1 2 × 2u g × 1

 u2 = 10g = 10 × 10  u = 10 ms-1 (a) Let, H be the maximum height, Then, H = u2 sin2 2g = 102 × (sin45°)2 2 × 10 = 100 20 ×     1

= 5 2 = 2.5 m. (b) Let,  is the angle of departure when the range 5m. Then R = u2 sin2 g  sin2 = gR u2 = 10 × 5 10 × 10 = 1

 sin2 = sin30° and sin2 = sin150°  2 = 30° and 2 = 150°   = 15°. and  = 75°.

Dynamics * 399* Projectile 6. A ball is thrown upwards at an angle of 30° to the horizon and lands on the top edge of a building that is 10 3 m away. The top edge is 5 m above the throwing point. How fast was the ball thrown ? (g = 10 ms-2) Soln: Let, A be the top edge of the building AB. Suppose u be the velocity of the ball making the angle  = 30° with horizon and t be the time taken by the ball to reach the top of the building. Also from figure, OB = 10 3 , AB = 5 The vertical component of velocity of projection = u sin30°.  h = (u sin30°) t - 1 2 gt2 or, 5 = u × 1 2 t - 1 2 × 10 × t2  ut - 10t2 = 10 ......................... (i) Again horizontal component of velocity of projection = u cos30°t or, 10 3 = ut 3

 ut = 20 ................. (ii) Eliminating ut from (i) and (ii) we get 20 - 10t2 = 10  10t2 = 10  t = 1 Therefore, from (ii) u = 20 t = 20 1 = 20 ms-1. 7. Find the angle of projection where the range on a horizontal plane is (a) 4 (b) 4 3 times the greatest height. Soln: Let,  be the angle of projection, u is the velocity of the projection, R is the horizontal range and H is the greatest height attained. We have, H = u2 sin2 2g , R = u2 sin2 g (a) Here, horizontal range is 4 times the greatest height Then, R = 4H  u2 sin2 g = 4 × u2 sin2 2g  sin2 = 2sin2  2sin.cos = 2sin2  tan = 1  = 45°. (b) If horizontal range is 4 3 times the greatest height Then R = 4 3 H  u2 sin2 g = 4 3 × u2 sin2 2g  sin2 = 2 3 sin2  2sin.cos = 2 3 sin2  tan = 1

 tan = tan30°  = 30°.

*400* Solution Manual to Basic Mathematics Mechanics 8. A projectile thrown from a point in a horizontal plane comes back to the plane in 4 secs at a distance of 60 m in front of the point of projection; find the velocity of projection. (g = 10 ms-2) Soln: time of projection (t) = 4 sec horizontal range (R) = 60 m velocity of projection (u) = ? Let,  be the angle of projection. Then we have t = 2u sin g and R = u2 sin2 g  4 = 2u sin 10 and 60 = u2 × 2sin cos

 u sin = 20 ......... (i) and u2sin cos = 300 ........ (ii) Dividing (ii) by (i) we get u cos = 15 or, 20 sin . cos = 15 or, tan= 4 3  sin = 4

Substituting the value of sin on (i), we get u sin = 20 or, u 4 5 = 20  u = 25 ms-1. 9. Find the velocity and the direction of projection of a shot which passes in a horizontal direction just over the top of a wall which is 250 m off and 125 m high. (g = 9.8 ms-2) Soln: Let u be the velocity,  be angle of projection of a shot. Since the shot just passes the top of the building of 125 m high.  Maximum height (H) = 125 m and horizontal range (R) = 2 × 250 m = 500 m  H = u2 sin2 2g  125 = u2 sin2 2g ........... (i) and R = u2 sin2 g  500 = u2 sin2 g ............ (ii) Dividing (i) by (ii), we get

4 = sin2 2sin2 1 4 = sin2 4sin.cos  1 = sin cos  tan = 1  = 45° Substituting the values of  = 45° on (i), we get 125 = u2 2g ×     1

 u2 = 500 g  u2 = 500 × 9.8 = 4900  u = 70 ms-1

Dynamics * 401* Projectile 10. The horizontal and vertical components of the initial velocity of a projectile are U and V respectively. If R be the range and H the greatest height attained, prove that a) 4H R = V U Soln: Let, u = initial velocity of the projectile  = angle of projection  U = u cos and V = u sin  We have, H = u2 sin2 2g and R = u2 sin2 g Now, 4H R = 4 × u2 sin2 2g u2 sin2 g = 2u2 sin2 g × g u2 sin2 = 2sin2 2sin . cos = sin cos = usinα ucosα = V U . b)     R U

= 8H g     R U

=     u2sin2 g × ucos

=     2usin.cos g × cos

= 4u2sin2 g2 = 4 g     u2sin2 2g × 2 = 8H g . 11. A projectile shot at an angle of 60° above the horizontal strikes a building 30 3 m away at a point 85 m above the point of projection. (g = 10 ms-2) a) Find the speed of projection. b) Find the magnitude and direction of the velocity of the projectile when it strikes the building. Soln: Let u be the velocity,  be angle of projection of a shot such that  = 60°. Here, horizontal distance of particle = 30 3 m. and height above the point of projection = 85 m time taken by the projectile to reach the height = t (a) horizontal velocity component of u = ucos60° The vertical component of velocity u = u sin60°.  h = u sin60° × t - 1 2 gt2  85 = u × 3 2 t - 1 2 × 10 × t2  10t2 = 3 (ut) - 170 .................. (i)  u cos60°t = 30 3  u × 1 2 t = 30 3  ut = 60 3 ....................... (ii) Eliminating ut from (ii) on (i), we get 10t2 = 3 × 60 3 - 170  10t2 = 180 – 170  t = 1 Substituting the value of t on (ii), we get u = 60 3 ms-1.

*402* Solution Manual to Basic Mathematics Mechanics (b) Let v be the striking velocity of the shot making an angle  with horizontal. vx = v cos  = 60 3 cos60° = 60 3 × 1 2 = 30 3 and vy = v sin  = u sin  - gt = 60 3 sin60° - 10 × 1 = 60 3 × 3 2 - 10 = 80 Now, v2 = vx2 + vy2 = ( ) 30 3 2+ (80)2 = 9100  v = 10 91 m/s. Also, tan = vy vx = 80 30 3 = 8 3 3  = tan-1     8 3 3 . 12. A stone is thrown horizontally with velocity 2gh from the top of a tower of height h. Find where it will strike the level ground through the foot of the tower. What will be its striking velocity ? Soln: Let, t be the time taken by the stone when it strikes the ground after falling a height h. Then, h = 0 + 1 2 gt2  t = 2h g Here, the stone was thrown horizontally. So the angle of projection,  = 0  Horizontal distance (R) = u cos t = 2gh × cos0° × 2h g = 4h2 = 2h. Let, v be the striking velocity of the stone at the level of foot of the tower Then, v2 = u2 + 2gh = 2gh + 2gh = 4gh = ( ) 2 gh 2  v = 2 gh . 13. From a point on the ground at a distance x from the foot of a vertical wall, a ball is thrown at an angle of 45° which just clears the top of the wall and afterwards strikes the ground at a distance y on the other side. Prove that the height of the wall is xy x + y . Soln: Let initial velocity = u m/s angle of projection () = 45° Let, t be the time taken by the ball to cover the horizontal distance x.  x = u cos 45° × t  t = x 2

Suppose, h be the height of wall  h = (u sin 45°) t - 1 2 gt2 = u 2 × x 2 u - 1 2 g × x2 × 2 u2 = x - x2g u2 ............ (i) Again, horizontal range = u2 g  x + y = u2 g  u2 = g(x + y) ............... (ii) Eliminating u2 from (i) and (ii), we get h = x - x2g g(x + y) = x - x2 x + y = xy x + y .

Dynamics * 403* Projectile 14. If R be the horizontal range of a projectile and h its greatest height, prove that its initial velocity is 2g    h + R2 16h . Soln: Let, u be the velocity of the projection and , the angle of projection. Then, greatest height (h) = u2 sin2 2g and horizontal range (R) = u2 sin2 g Now, R2 16h = u4 . (sin2)2 g2 × 1 16 × 2g u2sin2 = u2 . 4sin2 . cos2 g2 × 1 8 × g u2sin2 = u2 × 1 2g cos2 ...... (i)  2g    h + R2 16h = 2g     u2sin2 2g + u2cos2 2g = u2 (sin2 + cos2) = u2 × 1 = u Hence, initial velocity u is 2g    h + R2 16h . 15. A cannon ball has the same range R on a horizontal plane for two different angles of projection. If H and H' are the greatest heights and t1 and t2 are the time of flights in two paths for which this is possible, prove that a) R2 = 16HH' b) R = 1 2 gtt' Soln: Let, α and β are two angles of projection for same horizontal range R Then,

u sin2α 2g =

u sin2θ 2g  sin 2α = sin 2θ  sin 2α = sin2 and sin(180 - 2)  =  and 90 -  Let, H be max. height t be time of flight for  =  and H' be the max. height and t' for  = 90 -  a) Now, 16 HH' = 16     u2sin2 2g ×     u2sin2(90 - ) 2g = 4 × u2sin2 . u2cos2 g2 = 4 (2sin.cos)2u2 4g2 =     u2sin2 g

= R2  R2 = 16 HH' b) Now, 1 2 gtt' = 1 2 g ×     2usin g ×     2usin(90 - ) g = 2 u2sin . cos g = u2sin2 g = R.  R = 1 2 gtt'

*404* Solution Manual to Basic Mathematics Mechanics 45º 180m 65m 16m vy vx   16. A ball is projected with a velocity of 49.0 m/s, find the two directions along which the ball must be projected so as to have a range of 122.5 m. Soln: Initial velocity (u) = 49.0 m/s Horizontal range (R) = 122.5 Angle of projection =  We have, R = 2 2 u sin α g  122.5 = 492 9.8 sin2  sin2 = 122.5 × 9.8 49 × 49 = 1

 sin2 = sin30º or sin150º  2 = 30º or 150º   = 15º or 75º. 17. With what velocity must a body be projected at an angle of 45º from the top of a tower 65 m high, if it is to reach a point on the ground 180 m from the base of the tower. Soln: Height of tower (h) = 65m Horizontal range (x) = 180m Time of flight = t sec Angle of projection () = 45º x = u cos. t  180 = u. 1 2 . t  ut = 180 2 .......... (i) Also, taking upward direction positive –h = usin. t – 1 2 gt2  –65 = 180 2 × 1 2 × 9.8t2  4.9t2 = 180 + 65  t2 = 245 4.9 = 50  t = 5 2 see.  Putting t = 5 2 , u × 5 2 = 180 2  u = 36m/s . 18. A body is projected with a velocity of 16 2 m/sec from the top of a tower 16 m high at an elevation of 45º. Find where and with what velocity will it strike the ground? (g = 10 m/s2) Soln: Initial velocity (u) = 16 2 m/s Height of travel (h) = 16m Angle of projection () = 45º Taking upward direction positive we have, –h = u sin.t – 1 2 gt2

Dynamics * 405* Projectile  –16 = 16 2 . 1 2 . t – 1 2 × 10t2  –16 = 16t – 5t2  5t2 – 16t – 16 = 0  5t2 – 20t + 4t – 16 = 0  5t (t – 4) + 4 (t – 4) = 0  (t – 4) (5t + 4) = 0  t = 4, –4

Taking positive, t = 4 sec. Horizontal distance (x) = ucos.t = 16 2 × 1 2 × 4 = 6400 m. Let v be the velocity of striking the ground and  the angle made with horizontal. vx = v cos = u cos = 16 2 × 1 2 = 16m/s and vy = vsin = – usin + gt = – 16 2 × 1 2 + 10 × 4 = –16 + 40 = 24 m/s Now, v = vx2 + vy2 = 162 + 242 = 256 + 576 = 832 = 8 13 m/s Now, tan = vy yx = 24 16 = 3 2  = tan-1     3 2 . Hint and Solution of MCQ's 1. The way of a projectile in the air may in any direction. 2. The path of projectile is parabolic and it is called trajectory. 3. The angle made by the direction of the projected body with the horizontal line is angle of projection. 4. At the point, where a projectile attains its maximum height has horizontal velocity is zero. 5. Path of projectile, y = bx – ax2 Comparing with y = Ax2 + Bx + c we get A = – a, B = b, C = 0 The greatest height attains at x = – B 2A = b 2a . When x = b 2a , y = b. b 2a – a. b2 4a2 = b2 2a – b2 4a = 2b2 – b2 4a = b2 4a 6. Time of flight = 2u sin g  time to reach greatest height = 1 2 . 2u sin g = u sin g 7. Rmax = u2 sin2 g Which will be maximum when sin2 is max.  max. sin2 = 1  2 = 90o  = 45o 8. Let u be the velocity of projection, R the given range and  the angle of projection. Then R = u2 sin2 g = u2 sin( – 2) g = u2 sin2      2 –  g = g 2 sin u2   Two direction of projection for same horizontal range is  and  =  2 – .

*406* Solution Manual to Basic Mathematics Mechanics 9. If one direction of projectile,  =  6 then next direction =  2 –  =  2 –  6 =  3 . 10. Rmax = u2 sin2 g will be greatest for sin2 = 1  Rmax = u2 g . 11.  = 30o, u = 40m/s time taken to reach maximum height = u sin g = 40. sin30o 10 = 2 s 12. h = –uyt + 1 2 gt2 (taking upward direction positive)  60 = – usin × 4 + 1 2 × 10 × 42  60 = – 40 sin + 80  40 sin = 20  sin = 1 2  = 30o 13. Rmax = u2 g = 202 10 = 40m. 14. Rmax = u2 g = 100m for  = 45o  Hmax = u2 sin2 2g = u2 g . sin2 45o 2 = 100. 1 4 = 25m 15. R = u2 sin2 g  20 = 202. sin2 10  sin2 = 1

 2 = 30o  = 15o  two direction of projection are  and 90o –  = 15o and 75o 16. R = 4H  u2 sin2 g = 4.u2 sin2 2g  2sincos = 2 sin2  tan = 1  = 45o 17.  = 45o, t = 4 2 s , u = ? Since the projectile passes a wall of height h horizontally, h is greatest height And time taken to reach greatest height (t) = usin g  4 2 = u. sin45o

 4 2 = u 10 2  u = 80m/s. 