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LAWS OF MOTION 57 in pairs. This is an important fact that the force on the wall due to the ball is normal to enables the second law to be applicable to a the wall along the positive x-direction. The body or a system of particles (See Chapter 6). magnitude of force cannot be ascertained since the small time taken for the collision has not ⊳Example 4.5 Two identical billiard balls been specified in the problem. strike a rigid wall with the same speed but Case (b) atdifferentangles,andgetreflectedwithout any change in speed, as shown in Fig. 4.6. ( ) = m u cos 30▯,( y initial− m u sin 30▯ x initial 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 ( x final –m u cos 30 , ( yfinal− m u sin 30 balls by the wall ? Note, while p xhanges sign after collision, p y 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 Fig. 4.6 Answer An instinctive answer to (i) might be 2 mu /(2mu cos30 ▯)= 2 ≈1.2 ⊳ that the force on the wall in case (a) is normal to 3 the wall, while that in case (b) is inclined at 30° 4.7 CONSERVATIONOFMOMENTUM to the normal. This answer is wrong. The force on the wall is normal to the wall in both cases. The second and third laws of motion lead to an important consequence: the law of How to find the force on the wall? The trick is to consider the force (or impulse) on the ball conservation of momentum. Take a familiar due to the wall using the second law, and then example. A bullet is fired from a gun. If the force use the third law to answer (i). Let u be the speed on the bullet by the gun is F, the force on the gun by the bullet is – F, according to the third law. of each ball before and after collision with the wall, and m the mass of each ball. Choose the x The two forces act for a common interval of time and y axes as shown in the figure, and consider t. Accordingtothesecondlaw,F tisthechange the change in momentum of the ball in each in momentum of the bullet and – F t is the change in momentum of the gun. Since initially, case : both are at rest, the change in momentum equals Case (a) the final momentum for each. Thus if p isbthe momentum of the bullet after firing and p is the ( x initialu ( y initial g recoil momentum of the gun, p g – p b.e. p b p g ( x = −mu ( y final0 = 0. That is, the total momentum of the (bullet + final gun) system is conserved. Thus in an isolated system (i.e. a system with Impulse is the change in momentum vector. Therefore, no external force), mutual forces between pairs of particles in the system can cause momentum x-component of impulse = – 2 m u change in individual particles, but since the y-component of impulse = 0 mutual forces for each pair are equal and Impulse and force are in the same direction. opposite, the momentum changes cancel in pairs Clearly, from above, the force on the ball due to and the total momentum remains unchanged. the wall is normal to the wall, along the negative This fact is known as the law of conservation x-direction. Using Newton’s third law of motion, of momentum : Reprint 2026-27