Explanation Given: The mass of the each disks A and B is m A = m B = 1 kg . The initial velocity of the disk A is v A = 4 m/s . The initial velocity of the disk B is v B = 3 m/s . The coefficient of the restitution is e = 0.5 . 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). Here, the impact occurs along n − a x i s . Write the formula for coefficient of restitution ( e ) . e = ( v A ' ) n − ( v B ' ) n ( v B ) n − ( v A ) n (II) Here, v A ' , v B ' is the velocity of the disks A and B after impact, v A , v B is velocity of the disks A and B and the suffixes n , t indicates the n and t axes. Conclusion: From Figure (1). The line of impact is along n − a x i s . The velocity of the disk A along the n , t − a x e s is v A = 4 m/s ( v A ) n = − 4 ( 4 5 ) = − 3.2 m/s ( v A ) t = − 4 ( 3 5 ) = − 2.4 m/s Here, the negative direction indicates, the disk A moves in opposite direction to the disk B (facing each other). The velocity of the disk B along the line of impact is v B = 3 m/s ( v B ) n = 3 ( 3 5 ) m/s = 1.8 m/s ( v B ) t = − 3 ( 4 5 ) m/s = − 2.4 m/s Refer Figure (1). After impact the disks A and B moves with the velocity of v A ' , v B ' in the direction of x and y axes respectively. From Figure (1). After Impact: The velocity of the disk A along n − a x i s is ( v A ' ) n The velocity of the disk A along t − a x i s is ( v A ' ) t The velocity of the disk B along n − a x i s is ( v B ' ) n The velocity of the disk B along t − a x i s is ( v B ' ) t According law conservation of linear momentum: Consider when the disk A and disk B collides. At Initial: Calculate the initial momentum. ∑ m v 1 = m B ( v B ) n + m A ( v A ) n Substitute 1 kg for m A , m B , − 3.2 m/s for ( v A ) n and 1.8 m/s for ( v B ) n . ∑ m v 1 = ( 1 kg ) ( 1.8 m/s ) + ( 1 kg ) × ( − 3.2 ) m/s = 1.8 − 3.2 = − 1.4 N ⋅ s Calculate the final momentum. ∑ m v 2 = m A ( v A ' ) n + m B ( v B ' ) n Substitute 1 kg for m A , m B . ∑ m v 2 = 1 kg ( v B ' ) n + 1 kg ( v A ' ) n = ( v B ' ) n + ( v A ' ) n Substitute − 1.4 N ⋅ s for ∑ m v 1 and ( v B ' ) n + ( v A ' ) n for ∑ m v 2 in Equation (I)

15th Edition

ISBN: 9780137514724

Author: HIBBELER

Publisher: PEARSON

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Chapter 15, Problem 18FP

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The speeds of each disks just after the impact.

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A 100 kg body moves to the right at 5 m/s and another body of mass of W moves to the left at 3 m/s. they meet each other and after impact, the 100 kg body rebounds to the left at 2 m/s. Determine the mass of the other body if the coefficient of restitution is 0.50.

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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 speeds of each disks just after the 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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The speeds of each disks just after the impact.

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Chapter 15 Solutions