Explanation Given: The mass of the balls A , B , C , D is m . Initially the ball A and B each moves with the velocity of v . The collision elastic and the coefficient of the restitution is e = 0.5 . Here the collision takes place in following order. Ball B collides with ball C 1) . Ball C collides with ball D 2) . Ball A collides with ball B 3) . Here, mass of the all balls are equal. Write the principle of conservation of linear momentum. ∑ m v 1 = ∑ m v 2 m [ ( v x ) 1 + ( v y ) 1 ] = m [ ( v x ) 2 + ( v y ) 2 ] (I) Here, m is the mass, v is the velocity, the suffixes x , y indicates the balls which involves in collision and the suffixes 1 , 2 indicates the initial and final stages. Write the formula for coefficient of restitution ( e ) . e = ( v y ) 2 − ( v x ) 2 ( v x ) 1 − ( v y ) 1 (II) Here, v is the velocity, the suffixes x , y indicates the balls which involves in collision and the suffixes 1 , 2 indicates the initial and final stages. Conclusion: According law conservation of linear momentum: 1) Ball B collides with ball C : At Initial: (Before collision) Calculate the initial momentum. The initial momentum of the system is ∑ m v 1 = m [ ( v B ) 1 + ( v C ) 1 ] The ball C is at rest until the ball B collides with it. Hence the initial velocity of the ball C is zero. ( v C ) 1 = 0 Substitute v for ( v B ) 1 and 0 for ( v C ) 1 . ∑ m v 1 = m ( v + 0 ) = m v At Final: (After collision) Calculate the final momentum. The final momentum of the system is ∑ m v 2 = m [ ( v B ) 2 + ( v C ) 2 ] Calculate the final velocity of the balls after collision. Substitute m v for ∑ m v 1 and m [ ( v B ) 2 + ( v C ) 2 ] for ∑ m v 2 in Equation (I). m v = m [ ( v B ) 2 + ( v C ) 2 ] v = ( v B ) 2 + ( v c ) 2 (III) Calculate the coefficient of restitution ( e ) . Refer Equation (II). The coefficient of restitution is e = ( v C ) 2 − ( v B ) 2 ( v B ) 1 − ( v C ) 1 Here, e = 0.5 . Substitute 0.5 for e , v for ( v B ) 1 and 0 for ( v C ) 1 . 0.5 = ( v C ) 2 − ( v B ) 2 v − 0 0.5 = ( v C ) 2 − ( v B ) 2 v 0.5 v = ( v C ) 2 − ( v B ) 2 (IV) Solve the Equations (III) and (IV) by elimination method. Add Equations (III) and (IV). 0.5 v + v = ( v C ) 2 − ( v B ) 2 + ( v B ) 2 + ( v c ) 2 1.5 v = 2 ( v C ) 2 ( v C ) 2 = 1.5 2 v ( v C ) 2 = 0.75 v Substitute 0.75 v for ( v C ) 2 in Equation (IV). 0.5 v = 0.75 v − ( v B ) 2 ( v B ) 2 = 0.75 v − 0.5 v ( v B ) 2 = 0.25 v 2) Ball C collides with ball D : At Initial: (Before collision) Here, the initial velocity of the ball C is ( v C ) 1 = 0.75 v . Calculate the initial momentum. The initial momentum of the system is ∑ m v 1 = m [ ( v C ) 1 + ( v D ) 1 ] The ball D is at rest until the ball C collides with it. Hence the initial velocity of the ball D is zero. ( v D ) 1 = 0 Substitute 0.75 v for ( v C ) 1 and 0 for ( v D ) 1 . ∑ m v 1 = m [ ( 0.75 v ) + 0 ] = m ( 0.75 v ) At Final: (After collision) Calculate the final momentum. The final momentum of the system is ∑ m v 2 = m [ ( v C ) 2 + ( v D ) 2 ] Calculate the final velocity of the balls after collision. Substitute m ( 0.75 v ) for ∑ m v 1 and m [ ( v C ) 2 + ( v D ) 2 ] for ∑ m v 2 in Equation (I)

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ISBN: 9780137514724

Author: HIBBELER

Publisher: PEARSON

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Chapter 15, Problem 63P

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The velocity of the each balls after first three collisions if the coefficient of restitution is e=0.5 .

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Chapter 15, Problem 62P

Chapter 15, Problem 64P

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The velocity of the each balls after first three collisions if the coefficient of restitution is e = 0.5 .

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