ch09-p043 - v 2 = – 0.15. Thus, the velocity of block C...

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43. (a) With SI units understood, the velocity of block L (in the frame of reference indicated in the figure that goes with the problem) is ( v 1 – 3)i ^ . Thus, momentum conservation (for the explosion at t = 0) gives m L ( v 1 – 3) + ( m C + m R ) v 1 = 0 which leads to v 1 = 3 m L m L + m C + m R = 3(2 kg) 10 kg = 0.60 m/s. Next, at t = 0.80 s, momentum conservation (for the second explosion) gives m C v 2 + m R ( v 2 + 3) = ( m C + m R ) v 1 = (8 kg)(0.60 m/s) = 4.8 kg·m/s. This yields
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Unformatted text preview: v 2 = – 0.15. Thus, the velocity of block C after the second explosion is v 2 = –(0.15 m/s)i ^ . (b) Between t = 0 and t = 0.80 s, the block moves v 1 Δ t = (0.60 m/s)(0.80 s) = 0.48 m. Between t = 0.80 s and t = 2.80 s, it moves an additional v 2 Δ t = (– 0.15 m/s)(2.00 s) = – 0.30 m. Its net displacement since t = 0 is therefore 0.48 m – 0.30 m = 0.18 m....
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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