ch24-p051

# ch24-p051 - 51. (a) The potential energy is 8.99 109 N m2 C...

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Um v m v AA BB =+ 1 2 1 2 22 . Momentum is also conserved, so 0 = + mv mv AA BB . These equations may be solved simultaneously for v A and v B . Substituting (/) BA B A vm m v =− , from the momentum equation into the energy equation, and collecting terms, we obtain Um m m m v AB A B A =+ 1 2 2 (/) ( ) . Thus, 3 333 2 2(0.225 J)(10 10 kg) 7.75 m/s. ( ) (5.0 10 kg)(5.0 10 kg 10 10 kg) B A AA B Um v mm m −−− × == = × + × 51. (a) The potential energy is U q d == ×⋅ × = 2 96 4 899 10 50 10 100 0 225 π 0 ε .. . . Nm C C m J 22 2 ch c h relative to the potential energy at infinite separation. (b) Each sphere repels the other with a force that has magnitude F q d == ×⋅ × = 2 2 96 4 899 10 50 10 0 225 π 0 ε .. . Nm C C 1.00 m N. 22 2 2 ch c h bg According to Newton’s second law the acceleration of each sphere is the force divided by the mass of the sphere. Let m A and m B be the masses of the spheres. The acceleration of sphere A is a F m A A == × = 0 225 450 3 . . N 5.0 10 kg ms 2 and the acceleration of sphere B is a F m B B == × = 0 225 22 5 3 . ..

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## This note was uploaded on 12/13/2010 for the course PHY 211 taught by Professor Ld during the Spring '10 term at NE Texas CC.

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ch24-p051 - 51. (a) The potential energy is 8.99 109 N m2 C...

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