Explanation Given: The mass of the collar A is m A = 1 lb . The mass of the collar B is m B = 10 lb . Initially the collar B is at rest. The displacement of the collar A is s = 4 ft . The stiffness of the spring is k = 20 lb/ft . The impact between the collar A and B occurs for a time of 0.002 s . The coefficient of restitution is e = 0.5 . Draw the free body diagram of the system as shown in Figure (1). Write the principle of conservation energy. T 1 + V 1 = T 2 + V 2 (I) Here, T 1 , T 2 is the initial and final kinetic energy and V 1 , V 2 is the initial and final potential energy. Write the formula for kinetic energy. T = 1 2 m v 2 (II) Here, the collar and collar is at the datum only. Hence the gravitation potential energy V g = 0 and the total potential energy ( V ) is equal to the elastic potential energy ( V e ) . V = V e Write the formula for elastic potential energy. V = V e = 1 2 k s 2 (III) Here, k is the stiffness of the spring and s is the change in length of the spring. Write the principle of conservation of linear momentum. ∑ m v 1 = ∑ m v 2 (IV) Here, m is the mass, v is the velocity and the suffixes 1 , 2 indicates the initial and final stages. Write the formula for coefficient of restitution ( e ) . e = ( v B ) 2 − ( v A ) 2 ( v A ) 1 − ( v B ) 1 (V) Here, v is the velocity, the suffixes A , B indicates the collar A and B and the suffixes 1 , 2 indicates the initial and final stages. Write the principle of linear impulse and momentum. m v 1 + ∑ ∫ t 1 t 2 F d t = m v 2 (VI) Here, ∑ ∫ t 1 t 2 F d t is the net impulse on the collar B . Conclusion: According to principle of conservation energy. Consider the collar B only (After the collar A strikes the it). At initial: (just after the collar A strikes it) Substitute 10 lb for m and ( v B ) 2 for v in Equation (II). T 1 = 1 2 ( 10 lb × 1 32.2 ft/s 2 ) ( v B ) 2 2 = 0.1553 ( v B ) 2 2 When the spring is unstretched the elastic potential energy become zero. V 1 = 0 At final: When the motion of the collar was momentarily stopped, the kinetic energy becomes zero. T 2 = 0 Calculate the final potential energy. From Figure (1). The extension length of the spring is s = 4 2 + 3 2 − 3 ft = 5 − 3 = 2 ft Substitute 20 lb/ft for k and 2 ft for s in Equation (III). V 2 = 1 2 ( 20 lb/ft ) ( 2 ft ) 2 = 40 lb ⋅ ft Calculate the velocity of the collar B . Substitute 0 for V 1 , T 2 , 0.1553 ( v B ) 2 2 for T 1 and 40 lb ⋅ ft for V 2 in Equation (I). 0 + 0.1553 ( v B ) 2 2 = 0 + 40 lb ⋅ ft ( v B ) 2 2 = 40 0.1553 ( v B ) 2 = 257.566 ( v B ) 2 = 16.05 ft/s According to law conservation of linear momentum: At Initial: (before collision) Calculate the initial momentum. ∑ m v 1 = m A ( v A ) 1 + m B ( v B ) 1 When the collar B is at rest, the velocity of the collar B is zero. Substitute 1 lb for m A , 10 lb for m B and 0 for ( v B ) 1 . ∑ m v 1 = ( 1 lb × 1 32.2 ft/s 2 ) ( v A ) 1 + ( 10 lb × 1 32.2 ft/s 2 ) ( 0 ) = 1 32.2 ( v A ) 1 Calculate the final momentum. ∑ m v 2 = m A ( v A ) 2 + m B ( v B ) 2 Substitute 1 lb for m A , 10 lb for m B and 16

Engineering Mechanics: Dynamics (14th Edition)

14th Edition

ISBN: 9780133915389

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 15.4, Problem 83P

To determine

The velocity of the collar A

just after the impact and the average force exerted between the collars during impact.

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Chapter 15.4, Problem 82P

Chapter 15.4, Problem 84P

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The velocity of the collar A just after the impact and the average force exerted between the collars during impact.

Solution Summary: The author determines the velocity of the collar after collision and the average force exerted between the two collars during impact.

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The velocity of the collar A just after the impact and the average force exerted between the collars during impact.

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Chapter 15 Solutions

The velocity of the collar A

just after the impact and the average force exerted between the collars during impact.