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HW1_prob1_GPS01_G04

HW1_prob1_GPS01_G04 - d p dt = F(5 where p is the particle...

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Problem 1.4 (3 rd ed. 1.1) Start with Newton’s 2nd Law for a particle with constant mass m , m d v dt = F . (1) Next, take the dot product of Eq.(1) with the particle velocity v to get m v · d v dt = F · v . (2) Now expand the left dot product of Eq.(2) as m v · d v dt = m ( v x dv x dt + v y dv y dt + v z dv z dt )= m 2 d dt ( v 2 x + v 2 y + v 2 z )= m 2 dv 2 dt = dT dt , (3) where T is the kinetic energy of the particle ( mv 2 2 ), to obtain the desired result, dT dt = F · v . (4) When the mass is not constant, we use Newton’s 2nd Law in the form,
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Unformatted text preview: d p dt = F , (5) where p is the particle momentum vector. Take the dot product of this equation with p to obtain p · d p dt = F · p , (6) which can be rearranged in a fashion similar to that of Eq.(3) to give 1 2 dp 2 dt = F · p . (7) Since p 2 = 2 mT , Eq.(7) can be rewritten in the desired form d ( mT ) dt = F · p . (8) 1...
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