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Page 1 of 29, FUNDAMENTAL OF PHYSICS, for IIT-JEE, NEET, 11th & 12th, 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, = ma., balanced by the upthrust., (b) The cork is floating on water, which means the weight of the cork, 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, was thrown at an angle of, is momentarily at rest. Do your solutions change if the pebble, 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 is, F = mg = 0.0, kg x 10 ms2 = 0.5 N, The direct, 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., for IIT-JEE, NEET, 11th & 12th, R.K.Sharma (8507152734)

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Page 2 of 29, FUNDAMENTAL OF PHYSICS, for IIT-JEE, NEET, 11th & 12th, 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 with1 m s?, (d) lying on the floor of a train which is accelerating with 1 m s2, 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 N, The 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 1m/s2, the stone experiences an additional force of F' = ma, = 0.1 x 1 = 0.1 N. The force acts in the horizontal direçtion., 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, motion of the train., 4. One end of, 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:, ing of length I is connected to a particle of mass m and the other to a small, (iii) T+ mv?//, (iv) 0, T is the tension in the string. [Choose the correct alternative]., Solution:, (i) T, for IIT-JEE, NEET, 11th & 12th, R.K.Sharma (8507152734), UNDAMBNTAOR PHYSICČS

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Page 3 of 29, FUNDAMENTAL OF PHYSICS, for IIT-JEE, NEET, 11th & 12th, The 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 s1. How long does the body take to stop?, Solution:, Here,, Force = - 50 N (since it is a retarding force), Mass m = 20 kg, V = 0, u = 15 m s-1, Force F = ma, a = F/m = -50/20 = - 2.5 ms2, Using the equation, v = u + at, 0 = 15 + (- 2.5) t, t = 6 s, 6. A constant force acting on a body, 1 in 25 s. The direction of the motion, mass 3.0 kg changes its speed from 2.0 ms1 to 3.5 ms, the body remains unchanged. What is the magnitude and, direction of the force?, Solution:, Given,, Mass, m 3.0 Kg, u = 2.0 m/s, V = 3.5 m/s, SUNDAMENAL OF PHYSICS, F = m [(v-u)/t] (since a = (v -u)/t), F = 3[ (3.5 - 2)/25] = 0.18 N, The 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., for IIT-JEE, NEET, 11th & 12th, R.K.Sharma (8507152734)

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Page 4 of 29, FUNDAMENTAL OF PHYSICS, for IIT-JEE, NEET, 11th & 12th, Solution:, 6N A, TA, Given,, Mass, m 5 kg, Force, F, = 8N and F2 = 6N, The resultant force of the body, VF + F = V&, 64+36, F =, 10N, %3D, Acceleration, a F/m, a = 10/ 5 = 2m/s in the direction of the resultant force, The direction of the acceleration,, tan B = 6/8 = 0.75, B = tan-1, B= 37° with 8N, 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,, for IIT-JEE, NEET, 11th & 12th, R.K.Sharma (8507152734), İL, DAMENTAL O PHYSICS

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Page 5 of 29, FUNDAMENTAL OF PHYSICS, for IIT-JEE, NEET, 11th & 12th, Initial velocity, u= 36 km/h, Final velocity, v = 0, Mass of the three-wheeler, m;= 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/s, Now, F = (m1 + m2)/ a (400 + 65) x (-2.5), = - 1162.5 N = - 1.16 x 103 N, The negative sign shows that the force is retarding, 9. A rocket with a lift-off mass of 20,000 kg is blasted upwards, ms2. Calculate the initial thrust (force) of the blast., an initial acceleration of 5.0, Solution:, Given,, Mass of the rocket, m 20000 kg 2 x 104 kg, Initial acceleration = 5 ms2, g = 9.8 m/s?, The initial thrust (force) should give an, gravity., vard acceleration of 5 ms2 and should overcome the force of, Thus the thrust should produce, acceleration of 9.8 + 5.0 = 14.8 ms-2., Using Newton's second law, motion, the initial thrust acting on the rocket, Thrust = force = mass x acceleration, F = 20000 x 14,8 = 2, mass 0.40 kg moving initially with a constant speed of 10 ms1 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), for IIT-JEE, NEET, 11th & 12th, R.K.Sharma (8507152734), NDAENTAL OF RHYSICS