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LAWS OF MOTION 55 component of velocity normal to the force Answer The retardation ‘a’ of the bullet remains unchanged. For example, in the (assumed constant) is given by motion of a projectile under the vertical 2 gravitational force, the horizontal component a =–u = – 90 × 90 m s−2 = –6750 m s −2 of velocity remains unchanged (Fig. 4.5). 2s 2 × 0.6 3. The second law of motion given by Eq. (4.5) is The retarding force, by the second law of applicable to a single point particle. The force motion, is F in the law stands for the net external force -2 on the particle and a stands for acceleration = 0.04 kg × 6750 m s = 270 N The actual resistive force, and therefore, of the particle. It turns out, however, that the law in the same form applies to a rigid body or, retardation of the bullet may not be uniform. The even more generally, to a system of particles. answer therefore, only indicates the average In that case, F refers to the total external force resistive force. ⊳ ⊳ on the system and a refers to the acceleration Example 4.3 The motion of a particle of of the system as a whole. More precisely, a is the acceleration of the centre of mass of the mass m is described by y = ut +1 gt2. Find 2 system about which we shall study in detail in the force acting on the particle. Chapter6. Any internal forces in the system are not to be included in F. Answer We know 1 2 y =ut + gt 2 Now, v =dy =u + gt dt dv acceleration, a = dt = g Fig. 4.5 Acceleration at an instant is determined by theforceatthatinstant.Themomentaftera Then the force is given by Eq. (4.5) F = ma = mg stone is dropped out of an accelerated train, Thus the given equation describes the motion it has no horizontal acceleration or force, of a particle under acceleration due to gravity airresistanceisneglected.Thestonecarries no memory of its acceleration with the trainand y is the position coordinate in the direction a moment ago. of g. ⊳ 4. The second law of motion is a local relation Impulse which means that force F at a point in space We sometimes encounter examples where a large (location of the particle) at a certain instant force acts for a very short duration producing a of time is related to a at that point at that finite change in momentum of the body. For instant. Acceleration here and now is example, when a ball hits a wall and bounces back, the force on the ball by the wall acts for a determined by the force here and now, not by any history of the motion of the particle very short time when the two are in contact, yet (See Fig. 4.5). theforceislargeenoughtoreversethemomentum of the ball. Often, in these situations, the force and the time duration are difficult to ascertain ⊳Example 4.2 A bullet of mass 0.04 kg moving with a speed of 90 m s –1enters a separately.However, theproductofforceandtime, heavy wooden block and is stopped after a which is the change in momentum of the body remains a measurable quantity. This product is distance of 60 cm. What is the average called impulse: resistive force exerted by the block on the bullet? Impulse = Force × time duration = Change in momentum (4.7) Reprint 2026-27