Explanation Given: The mass of the disk A is m A = 3 kg . The mass of the disk B is m B = 5 kg . The initial velocity of the disk A is ( v A ) 1 = 6 m/s . The initial velocity of the disk B is ( v B ) 1 = 7 m/s . The coefficient of restitution between the disks is e = 0.65 . Draw the free body diagram of the disks as shown in Figure (1). Write the principle of conservation of linear momentum. ∑ m v 1 = ∑ m v 2 (I) Here, m is the mass, v is the velocity and the suffixes 1 , 2 indicates the initial and final stages. Refer Figure (1). Write the formula for coefficient of restitution ( e ) . e = ( v B ) x 2 − ( v A ) x 2 ( v A ) x 1 − ( v B ) x 1 (II) Here, v is the velocity, the suffixes A , B indicates the disks A and B and the suffixes x indicates along x − a x i s . Conclusion: According law conservation of linear momentum: (Along x − a x i s ) Consider when the disk A and disk B collides. At Initial: From Figure (1). ( v A ) = ( v A ) x 1 = 6 m/s ( v B ) x 1 = − 7 m/s ( cos 60 ° ) = − 3.5 m/s Here, the negative sign indicates that the velocity acting in opposite direction (leftward). Calculate the initial momentum. ∑ m v 1 = m A ( v A ) x 1 + m B ( v B ) x 1 Substitute 3 kg for m A , 5 kg for m B , 6 m/s for ( v A ) x 1 and − 3.5 m/s for ( v B ) x 1 . ∑ m v 1 = ( 3 kg ) × 6 m/s + ( 5 kg ) ( − 3.5 m/s ) = 18 − 17.5 = 0.5 N ⋅ s At Final: Calculate the final momentum. ∑ m v 2 = m A ( v A ) x 2 + m B ( v B ) x 2 Substitute 3 kg for m A and 5 kg for m B . ∑ m v 2 = 3 kg ( v A ) x 2 + 5 kg ( v B ) x 2 = 3 ( v A ) x 2 + 5 ( v B ) x 2 Calculate the velocity of the disks after collision. Substitute 0.5 N ⋅ s for ∑ m v 1 and 3 ( v A ) x 2 + 5 ( v B ) x 2 for ∑ m v 2 in Equation (I). 0.5 N ⋅ s = 3 ( v A ) x 2 + 5 ( v B ) x 2 3 ( v A ) x 2 + 5 ( v B ) x 2 = 0.5 (III) According to equation of the coefficient of restitution ( e ) between the disks. Here, e = 0.65 . Substitute 0.65 for e , 6 m/s for ( v A ) x 1 and − 3.5 m/s for ( v B ) x 1 in Equation (II). 0.65 = ( v B ) x 2 − ( v A ) x 2 6 m/s − ( − 3.5 m/s ) 0.65 = ( v B ) x 2 − ( v A ) x 2 9.5 ( v B ) x 2 − ( v A ) x 2 = 0.65 ( 9.5 ) ( v B ) x 2 = 6.175 + ( v A ) x 2 (IV) Calculate the final velocity of the disks A and B . Substitute 6.175 + ( v A ) x 2 for ( v B ) x 2 in Equation (III)

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ISBN: 9780137514724

Author: HIBBELER

Publisher: PEARSON

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Chapter 15, Problem 79P

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The velocities of the disks A and B just after impact:

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Chapter 15, Problem 78P

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The 4-kg masses A and B slide on the smooth horizontal bar with the velocities shown. Determine their velocities after they collide if their coefficient of restitution is e = 0.8

Disk A and B have a mass of 4 kg and 6 kg, respectively. If they are sliding on a smooth horizontal plane with the velocities shown, determine their speeds just after impact. coefficient of restitution between them is e =0.8. =5 m/s e 0,8 The 45° m, = 4 kg Your solution must include the following: (a) tangential component of the velocities after impact for both disks, (VAt)2 and (Vet)2, (b) normal component of the velocities after impact =3 m/s for both disks, (VAn)2 and (VBn)2, (c) the velocities after impact for both disks, (vA)2 and (vB)2, 30 ma = 6 kg [4] (a) (VAt)2 = (VBt)2 [17] (b) (Van)2 = (VBn) 2 [4] (c) (VA)2 (VB)2 = %3D %3D %3D

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The velocities of the disks A and B just after impact:

Solution Summary: The author illustrates the principle of conservation of linear momentum by drawing the free body diagram of the disks as shown in Figure (1).

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Chapter 15 Solutions