Newton Law of Motion Solved Example

NEWTON LAW OF MOTION Page # 14 SOLVED EXAMPLE Ex.1 Ans. Ex.2 Ans. An astronaut accidentally gets separated out of his small spaceship baccelerating in inter stellar space at a constant rate of 100 m s-2. What is the acceleration of the astronaut the instant after he is outside the spaceship ? (Assume that there are no nearby stars to exert gravitational force on him.) Since there are no nearby stars to exert gravitational force on him and the small spaceship exerts negligible gravitational attraction on him, the net force acting on the astronaut, once he is out of the spaceship, is zero. By the first law of motion the acceleration of the astronaut is zero. A bullet of mass 0.04 kg moving with a speed of 90 m s-1 enters a heavy wooden block and is stopped after a distance of 60 cm. What is the average resistive force exerted by the block on the bullet? The retardation ‘a’ of the bullet (assumed constant) is given by a = 90  90 m/s2 2  0.6 = – 6750 m s–2 Ex.3 Ans. Ex.4 The motion of a particle of mass m is 1 described by y = ut  gt2 Find the force 2 acting on the particle. 1 We know – y = ut + gt2 2 dy Now, v = =4+g dt Then the force is given by F = ma = mg Thus the given equation describes the motion of a particle under acceleration due to gravity and y is the position coordinate in the direction of g. A batsman hits back a ball straight in the direction of the bowler without changing its initial speed of 12 m s–1. If the mass of the ball is 0.15 kg, determine the impulse imparted to the ball. (Assume linear motion of the ball) Ans. Change in momentum = 0.15 × 12–(–0.15×12) = 3.6 N s, Impulse = 3.6 N s, in the direction from the batsman to the bowler. Ex.5 Two identical billiard balls strike a rigid wall with the same speed but at different angles, and get reflected without any change in speed, as shown in Fig. What is (i) the direction of the force on the wall due to each ball? (ii) the ratio of the magnitudes of impulses imparted to the balls by the wall ? Ans. An instinctive answer to (i) might be that the force on the wall in case (a) is normal to the wall, while that in case (b) is inclined at 30° to the normal. This answer is wrong. The force on the wall is normal to the wall in both cases. How to find the force on the wall? The trick is to consider the force (or impulse) on the ball due to the wall using the second law, and then use the third law to answer (i). Let u be the speed of each ball before and after collision with the wall, and m the mass of each ball. Choose the x and y axes as shown in the figure, and consider the change in momentum of the ball in each case : Impulse is the change in momentum vector. Therefore, x-component of impulse = – 2 m u y-component of impulse = 0 Impulse and force are in the same direction. Clearly, from above, the force on the ball due to the wall is normal to the wall, along the negative x-direction. Using Newton’s third law of motion, the force on the wall due to the ball is normal to the wall along the positive xdirection. The magnitude of force cannot be ascertained since the small time taken for the NEWTON LAW OF MOTION collision has not been specified in the problem. Page # 15 Ex.7 Ans. Note, while px changes sign after collision, py does not. Therefore, x-component of impulse = –2 m u cos 30° y-component of impulse = 0 The direction of impulse (and force) is the same as in (a) and is normal to the wall along the negative x direction. As before, using Newton’s third law, the force on the wall due to the ball is normal to the wall along the positive x direction. The ratio of the magnitudes of the impulses imparted to the balls in (a) and (b) is 2  1.2 2 mu /(2 mu cos 30º) = 3 Ex.6 Ans. A mass of 6 kg is suspended by a rope of length 2 m from the ceiling. A force of 50 N in the horizontal direction is applied at the midpoint P of the rope, as shown. What is the angle the rope makes with the vertical in equilibrium ? (Take g = 10 m s–2 ). Neglect the mass of the rope. Consider the equilibrium of the weight W. Clearly,T2 = 6 × 10 = 60 N. Consider the equilibrium of the point P under the action of three forces - the tensions T1 and T2, and the horizontal force 50 N. The horizontal and vertical components of the resultant force must vanish separately : T1 cos  = T2 = 60 N T1 sin = 50 N which gives that Determine the maximum acceleration of the train in which a box lying on its floor will remain stationary, given that the coefficient of static friction between the box and the train’s floor is 0.15. Since the acceleration of the box is due to the static friction, the static friction, ma = fs  s N = s m g i.e. a s g  amax = s g = 0.15 x 10 m s–2 = 1.5 m s–2 Ex.8 A mass of 4 kg rests on a horizontal plane. The plane is gradually inclined until at an angle = 15° with the horizontal, the mass just begins to slide. What is the coefficient of static friction between the block and the surface ? Ans. The forces acting on a block of mass m at rest on an inclined plane are (i) the weight mg acting vertically downwards (ii) the normal force N of the plane on the block, and (iii) the static frictional force f s opposing the impending motion. In equilibrium, the resultant of these forces must be zero. Resolving the weight mg along the two directions shown, we have m g sin = fs , m g cos = N As increases, the self-adjusting frictional force fs increases until at = max, fs achieves its maximum value, (fs)max f = us N. Therefore, tan max = s or max = tan–1 s When becomes just a little more than max , there is a small net force on the block and it begins to slide. Note that max depends only on us and is independent of the mass of the block. For max = 15°, us = tan 15° = 0.27 Ex.9 What is the acceleration of the block and trolley system shown in a diagram (a), if the coefficient of kinetic friction between the trolley and the surface is 0.04? What is the tension in the string? (Take g = 10 m s-2). Neglect the mass of the string. NEWTON LAW OF MOTION Page # 16 v2  s R g Now, R = 3 m, g = 9.8 m s-2, s = 0.1. That is, s R g = 2.94 m2 s-2. v = 18 km/h = 5 m s-1; i.e., v2 = 25 m2 s-2. The condition is not obeyed. The cyclist will slip while taking the circular turn. Ex.11 A circular racetrack of radius 300 m is banked at an angle of 15°. If the coefficient of friction between the wheels of a racecar and the road is 0.2, what is the (a) optimum speed of the racecar to avoid wear and tear on its tyres, and (b) maximum permissible speed to avoid Ans. As the string is inextensible, and the pully is smooth, the 3 kg block and the 20 kg trolley both have same magnitude of acceleration. Applying second law to motion of the block 30 – T = 3a Apply the second law to motion of the trolley T – fk = 20 a. Now fk = k N, Here k = 0.04, N = 20 x 10 = 200 N. Thus the equation for the motion of the trolley is T – 0.04 x 200 = 20 a Or T – 8 = 20a. These equations give a = 22 23 m s–2 = 0.96 m s-2 and T = 27.1 N. Ex.10 A cyclist speeding at 18 km/h on a level road takes a sharp circular turn of radius 3 m without reducing the speed. The coefficient of static friction between the tyres and the road is 0.1. Will the cyclist slip while taking the turn ? Ans. On an unbanked road, frictional force alone can provide the centripetal force needed to keep the cyclist moving on a circular turn without slipping. If the speed is too large, or if the turn is too sharp (i.e. of too small a radius) or both, the frictional force is not sufficient to provide the necessary centripetal force, and the cyclist slips. The condition for the cyclist not to slip is given by Eq. (5.18) : slipping ? Ans. On a banked road, the horizontal component of the normal force and the frictional force contribute to provide centripetal force to keep the car moving on a circular turn without slipping. At the optimum speed, the normal reaction’s component is enough to provide the needed centripetal force, and the frictional force is not needed. The optimum speed vo is given by v0 = (R g tan)1/2 Here R = 300 m,  = 15°, g = 9.8 m s-2; we have v0 = 28.1 m s-1. The maximum permissible speed max is given  u  tan    = 38.1 ms–1 by max   Rg s 1  us tan   