Explanation Given: The mass of the bullet, m b = 20 g . The velocity of bullet, v b = 400 m/s The mass of a stationary block, m B = 2 kg The coefficient of kinetic friction between the block and the plane, μ k = 0.2 The acceleration due to gravity, g = 9.81 m/s 2 Show the free body diagram of the block as in Figure (1). Apply the conservation of linear momentum. m b v b + m B v B = ( m b + m B ) v (I) Here, the final velocity of a stationary block is v , initial velocity of a stationary block is v B . Apply the principle of impulse and momentum equation for y component. m ( v 1 ) y + ∑ ∫ t 1 t 2 F y d t = m ( v 2 ) y m ( v 1 ) y + N t − ( m b + m B ) g t = m ( v 2 ) y (II) Here, the initial and final velocity of the stationary block in y component is ( v 1 ) y and ( v 2 ) y respectively, initial and final linear momentum of the stationary block in y component is m ( v 1 ) y and m ( v 2 ) y , impulses acting on the stationary block from time limits t 1 to t 2 is ∑ ∫ t 1 t 2 F d t , and variable in time is dt . Apply the principle of impulse and momentum equation for x component. m ( v 1 ) x + ∑ ∫ t 1 t 2 F x d t = m ( v 2 ) x ( m b + m B ) ( v 1 ) x + [ − μ k N t ] = ( m b + m B ) v (III) Here, the coefficient of kinetic friction between the block and the surface is μ k , the initial and final velocity of the wooden block in x component is ( v 1 ) x and ( v 2 ) x respectively, initial and final linear momentum of the wooden block in x component is m ( v 1 ) x and m ( v 2 ) x , and impulses acting on the wooden block from time limits t 1 to t 2 is ∑ ∫ t 1 t 2 F d t . Determine the displacement of the block by integrating d s = v d t with the initial conditions s = 0 at t = 0. ∫ 0 s d s = ∫ 0 t v d t (IV) Conclusion: Substitute 20 g for m b , 0 for v B , 2 kg for m B , and 400 m/s for v b in Equation (I). ( 20 g ) ( 400 m/s ) + ( 2 kg ) ( 0 ) = ( 20 g + 2 kg ) v ( 20 g × 10 − 3 kg g ) ( 400 m/s ) = ( 20 g × 10 − 3 kg g + 2 kg ) v 0

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Author: HIBBELER

Publisher: PEARSON

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Chapter 15, Problem 43P

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The sliding distance of the stationary block before the bullet stops.

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Chapter 15, Problem 42P

Chapter 15, Problem 44P

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