Chapter – 4
Laws of Motion
1. Give the magnitude and direction of the net force acting on
(a) a drop of rain falling down with a constant speed
(b) a cork of mass 10 g floating on water
(c) a kite skillfully held stationary in the sky
(d) a car moving with a constant velocity of 30 km/h on a rough road
(e) a high-speed electron in space far from all material objects, and free of electric and magnetic fields.
Solution
(a) The raindrop is falling at a constant speed. Therefore, acceleration will become zero. When acceleration is zero, the force acting on the drop will become zero since F = ma.
(b) The cork is floating on water, which means the weight of the cork is balanced by the upthrust. Therefore, the net force on the cork will be zero
(c) Since the car moves with a constant velocity, the acceleration becomes zero. Therefore, the force will be zero.
(d) The net force acting on the high-speed electron will be zero since the electron is far from the material objects and free of electric and magnetic fields.
2. A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble,
(a) during its upward motion
(b) during its downward motion
(c) at the highest point where it is momentarily at rest. Do your Solutions change if the pebble was thrown at an angle of 45° with the horizontal direction? Ignore air resistance.
Solution
(a) During the upward motion of the pebble, the acceleration due to gravity acts downwards, so the magnitude of the force on the pebble isF = mg = 0.05 kg x 10 ms-2 = 0.5 NThe direction of the force is downwards
(b) During the downward motion also the magnitude of the force will be equal to 0.5 N and the force acts downwards
(c) If the pebble is thrown at an angle of 45° with the horizontal direction, it will have both horizontal and vertical components of the velocity. At the highest point, the vertical component of velocity will be zero but the horizontal component of velocity will remain throughout the motion of the pebble. This component will not have any effect on the force acting on the pebble. The direction of the force acting on the pebble will be downwards and the magnitude will be 0.5 N because no other force other than acceleration acts on the pebble.
3. Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg,
(a) just after it is dropped from the window of a stationary train
(b) just after it is dropped from the window of a train running at a constant velocity of 36 km/h
(c ) just after it is dropped from the window of a train accelerating with 1 m s-2
(d) lying on the floor of a train which is accelerating with 1 m s-2, the stone being at rest relative to the train. Neglect air resistance throughout.
Solution
(a) Mass of stone = 0.1 kg Acceleration = 10 ms – 2 Net force, F = mg = 0.1 x 10 = 1.0 NThe force acts vertically downwards
(b) The train moves at a constant velocity. Therefore, the acceleration will be equal to zero. So there is no force acting on the stone due to the motion of the train. Therefore, the force acting on the stone will remain the same (1.0 N)
(c) When the train accelerates at 1 m/s 2, the stone experiences an additional force of F’ = ma = 0.1 x 1 = 0.1 N. The force acts in the horizontal direction.But as the stone is dropped, the force F’ no longer acts and the net force acting on the stone F = mg = 0.1 x 10 = 1.0 N. (vertically downwards).
(d) As the stone is lying on the train floor, its acceleration will be the same as that of the train.Therefore, the magnitude of the force acting on the stone, F = ma = 0.1 x 1 = 0.1 N.It acts along the direction of motion of the train.
4. One end of a string of length l is connected to a particle of mass m and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed v the net force on the particle (directed towards the centre) is :
(i) T
(ii) T – mv 2/l
(iii) T + mv 2/l
(iv) 0 T is the tension in the string. [Choose the correct alternative].
Solution
(i) TThe net force acting on the particle is T, and it is directed towards the centre. It provides the centripetal force required by the particle to move along a circle.
5. A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 m s-1. How long does the body take to stop?
Solution
Here
,Force = – 50 N (since it is a retarding force)Mass m = 20 kgv = 0 u = 15 m s – 1 Force F = maa = F/m = -50/20 = – 2.5 ms- 2 Using the equation, v = u + at 0 = 15 + (- 2.5) tt = 6 s
6. A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 ms-1 to 3.5 ms-1 in 25 s. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?
Solution:
Given,Mass, m = 3.0 Kgu = 2.0 m/sv = 3.5 m/st = 25 sF = maF = m [(v-u)/t] (since a = (v -u)/t )F = 3 [ (3.5 – 2)/25] = 0.18 NThe force acts in the direction of motion of the body
7. A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Give the magnitude and direction of the acceleration of the body.
Solution:laws of motion image 1 Given,Mass, m = 5 kg Force, F1 = 8 N and F2 = 6 N The resultant force of the body Acceleration, a = F/ma = 10/ 5 = 2 m/s in the direction of the resultant force The direction of the acceleration,tan β = 6/8 = 0.75β = tan-1 (0.75)β = 37 0 with 8 N
8. The driver of a three-wheeler moving at a speed of 36 km/h sees a child standing in the middle of the road and brings his vehicle to rest in 4.0 s just in time to save the child. What is the average retarding force on the vehicle? The mass of the three-wheeler is 400 kg, and the mass of the driver is 65 kg.
Solution:
Given,Initial velocity, u= 36 km/h Final velocity, v = 0 Mass of the three-wheeler, m1= 400 Kg Mass of the driver, m2 = 65 Kg Time taken to bring the vehicle to rest = 4.0 s Acceleration, a = v- u/t = (0 – 10)/ 4 = – 2.5 m/sNow, F = (m1 + m2)/ a = (400 + 65) x (-2.5)= – 1162.5 N = – 1.16 x 103 NThe negative sign shows that the force is retarding
9. A rocket with a lift-off mass of 20,000 kg is blasted upwards with an initial acceleration of 5.0 ms-2. Calculate the initial thrust (force) of the blast.
Solution:
Given,Mass of the rocket, m = 20000 kg = 2 x 104 kg Initial acceleration = 5 m s – 2 g = 9.8 m/s 2 The initial thrust (force) should give an upward acceleration of 5 ms-2 and should overcome the force of gravity.Thus, the thrust should produce a net acceleration of 9.8 + 5.0 = 14.8 ms-2.Using Newton’s second law of motion, the initial thrust acting on the rocket Thrust = force = mass x acceleration F = 20000 x 14.8 = 2.96 x 105 N
10. A body of mass 0.40 kg moving initially with a constant speed of 10 ms-1 to the north is subject to a constant force of 8.0 N directed towards the south for 30 s. Take the instant the force is applied to be t = 0, the position of the body at that time to be x = 0, and predict its position at t = -5 s, 25 s, 100 s.
Solution:
Given,Mass of the body = 0.40 kg Initial velocity, u = 10 m/s Force, f = -8 N (retarding force)Using the equation S = ut + (½) at 2(a) Position at the time t = – 5 sThe force starts acting on the body from t = 0 sSo the acceleration of the body when the time is – 5 s is 0 S 1= (10)(-5) + (½) (0) (-5)2 = – 50 m(b) Position at the time t = 25 sThe acceleration of the body due to the force acting in the opposite direction Acceleration, a = F/a = -8 /0.4 = -20 m s – 2 S 2= (10)(25) + (½) (-20) (25)2 = – 6000 m(c) Position at the time t = 100 sFor the first 30 sec, the body will move under the retardation of the force, and after that, the speed will remain constant.Therefore, distance covered in 30 sec S 3= (10)(30) + (½)(-20)(302)= 300 – 9000 = – 8700mThe speed after 30 sec isv = u + atv = 10 – (20 x 30) = 590 m/sThe distance covered in the next 70 sec is S4 = – 590 x 70 + (½) (0) (70)2 = – 41300 m Therefore the position after 100 sec = S3 + S4 = – 8700 – 41300 = – 50000m
11. A truck starts from rest and accelerates uniformly at 2.0 ms-2. At t = 10 s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What are the (a) velocity, and (b) acceleration of the stone at t = 11 s? (Neglect air resistance.)
Solution:
laws of motion image 2 Initial velocity, u = 0 Acceleration, a = 2 ms-2,t=10 s Using equation, v = u + at, we get v = 0 + 2 x 10 = 20 m/sThe final velocity, v = 20 m/sAt time, t = 11 sec, the horizontal component of the velocity in the absence of the air resistance remains unchanged V x =20 m/sThe vertical component of the velocity is given by the equation Vy = u + ayt Here t = 11 – 10 = 1 s and ay= a = 10 m/s Therefore, the resultant velocity v of the stone isv = ( v x 2 + v y 2)½v = (20 2 + 10 2)1/2 v = 22.36 m/stan θ = vy/v x = 10/20 = ½ = 0.5θ = tan -1 (0.5 ) = 26. 560 from the horizontal(b) When the stone is dropped from the truck, the horizontal force on the stone is zero. The only acceleration of the stone is that due to gravity which is equal to 10 m/s2 and it acts vertically downwards.
12. A bob of mass 0.1 kg hung from the ceiling of a room by a string 2 m long is set into oscillation. The speed of the bob at its mean position is 1 ms-1. What is the trajectory of the bob if the string is cut when the bob is
(a) at one of its extreme positions,
(b) at its mean position?
Solution:
(a) When the bob is at one of its extreme positions, the velocity is zero. So, if the string is cut the bob will fall vertically downward under the force of its weight F = mg.
(b) At its mean position the bob has a horizontal velocity. If the string is cut, the bob will behave like a projectile and will fall on the ground after taking a parabolic path.
13. A man of mass 70 kg, stands on a weighing machine in a lift, which is moving
(a) upwards with a uniform speed of 10 ms-1.
(b) downwards with a uniform acceleration of 5 ms-2.
(c) upwards with a uniform acceleration of 5 ms-2.What would be the readings on the scale in each case?
(d) What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?
Solution:
Mass of the man, m = 70 kg,g = 10 m/s 2 The weighing machine in each case measures the reaction R, i.e., the apparent weight.
(a) When the lift moves upwards with a uniform speed of 10 m/s, it’s acceleration= 0.R = mg = 70 x 10 = 700 N
(b) Lift moving downwards with a = 5 m s – 2 Using Newton’s second law of motion, the equation of motion can be written asR+mg = maR = m (g – a) = 70 (10 – 5) = 350 N
(c) Lift moving upwards with a = 5 m s – 2 R = m (g + a) = 70 (10 + 5) = 1050 N
(d) If the lift were to come down freely under gravity, downward . a = g:. R = m(g -a) = m(g – g) = 0 The man will be in a state of weightless ness
14. Figure shows the position-time graph of a particle of mass 4 kg. What is the (a) force on the particle for t < 0, t > 4 s, 0 < t < 4 s? (b) impulse at t = 0 and t = 4 s? (Consider one-dimensional motion only).laws of motion image 3
Solution:
When t＜0, the distance covered by the particle is zero. Therefore, the force on the particle is zero.When 0＜ t ＜4s, the particle is moving with a constant velocity. Therefore, the force will be zero.When t＞4s, the particle remains at a constant distance. Therefore, the force of the particle will be zero.Impulse at t = 0.Here, u = 0 v = ¾ = 0.75 m/sM = 4 kg Impulse= total change in momentum = mv – mu = m (v – u)= 4 (0 – 0.75) = -3 kg m/s Impulse at t= 4 su = 0.75 m/s, v = 0 Impulse = m (v – u) = 4 (0 – 0.75) = -3 kg m/s
15. Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal surface are tied to the ends of a tight string. A horizontal force F = 600 N is applied to
(i) A,
(ii) B along the direction of string. What is the tension in the string in each case?
Solution:
laws of motion image 4 laws of motion image 5 Given,Mass of the body A, m1 = 10 kg Mass of the body B, m2 = 20 kg Horizontal force = 600 N Total mass of the system, m = m1 + m2 = 30 kg Applying Newton’s second law of motion, we have F = maa = F/m = 600/30 = 20 m/s 2
(i) When the force is applied on A (10 kg)F – T = m 1 aT = F – m 1. a T = 600 – (10 x 20)= 600 – 200 = 400 N
(ii) When the force is applied on B (20 kg)F – T = m 2 aT = F – m 2 a= 600 – (20 x 20) = 200 N
16. Two masses 8 kg and 12 kg are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses and the tension in the string when the masses are released.
Solution:
laws of motion image 6Given,Smaller mass, m1 = 8 kgLarger mass, m2 = 12 kgTension in the string = TThe heavier mass m2 will move downwards and the smaller mass m1 will move upwards.Applying Newton’s second law,For mass m1:T – m1g = m1a —– (1)For mass m2:m2g – T = m2a ——(2)Add (1) and (2)(m2 – m1) g = (m1 + m2) aa = (m2 – m1) g/ (m1 + m2)=[ (12 – 8)/(12 + 8)] x 10 = (4 /20) x 10= 2m/sTherefore, acceleration of the mass is 2 m/s2Substituting this value of acceleration in equation (ii) we getm2g – T = m2am2g – T = m2 [ (m2 – m1) g/ (m1 + m2) ]= 2m1m2g/(m1 + m2)T = 2 x 12 x 8 x 10/ (12 + 8)T = 96 NTherefore, the tension on the string is 96 N
17. A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei, the products must move in opposite directions.
Solution:
Let m1, m2 be the masses of the two daughter nuclei and v1,v2 be their respective velocities of the daughter nuclei. Let m be the mass of the parent nucleus.Total linear momentum after disintegration = m 1 v 1 +m2 v2.Before disintegration, the nucleus is at rest. Therefore, its linear momentum before disintegration is zero.Applying the law of conservation of momentum,Total linear momentum before disintegration = Total linear momentum after dis integration 0 = m 1 v 1 + m 2 v 2 v 1=-m. 2 v 2/m 1 The negative sign indicates v1 and v2 are in opposite directions.
18. Two billiard balls, each of mass 0.05 kg, moving in opposite directions with speed 6 ms-1 collide and rebound with the same speed. What is the impulse imparted to each ball due to the other?
Solution:
Mass of each ball = 0.05 kg Initial velocity of each ball = 6 m/sThe initial momentum of each ball before the collision= 0.05 x 6 = 0.3 kg m/s After the collision, the balls change in the direction of motion without a change in the magnitude of the velocity Final momentum after collision of the first ball = – 0.05 x 6 = – 0.3 kg m/s Final momentum after collision of the second ball = 0.3 kg m/s Impulse imparted to the first ball = (-0.3) – (0.3) = – 0.6 kg m/s Impulse imparted to the second ball = (0.3) – (-0.3) = 0.6 kg m/sThe two impulses are opposite in direction.
19. A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 ms-1.What is the recoil speed of the gun?
Solution:
Mass of the shell, m = 0.020 kg Mass of the gun, M = 100 Kg Speed of the shell = 80 m/sThe initial velocity of the shell and the gun is zero. So, the initial momentum of the system is zero.Applying the law of conservation of momentum, the initial momentum is equal to the final momentum.So,0 = mv – MV Recoil speed of the gun, v = mv/Mv = (0.020 x 80)/100v = 0.016 m/s
20. A batsman deflects a ball by an angle of 45° without changing its initial speed which is equal to 54 km/h. What is the impulse imparted to the ball? (Mass of the ball is 0.15 kg.)
Solution:
laws of motion image 7 Velocity of the ball = 54 km/hThe ball is deflected back such that the total angle = 450The initial momentum of the ball is mu cos Ө = (0.15 x 54 x 1000 x cos 22. 5)/3600= 0.15 x 15 x 0.9239 along NO Final momentum of the ball = mu cos Ө along O NIm pulse = change in the momentum = mu cos Ө – (-mu cos Ө) = 2 mu cos Ө = 2 x 0.15 x 15 x 0.9239 = 4.16 kg.m/s