Momentum, Kinetic Energy, And Vector Analysis In Physics Problems

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Momentum, Kinetic Energy, and Vector Analysis in Physics Problems For problem #5: I don’t understand why it says these are incorrect. This is most certainly a conservation of momentum problem. The momentum before the string is burned is zero, so the momentum after the string is burned must also be zero. (m + 3m)*0 = 3m(2.80) + m(v) 0 = 8.40 + v v = - 8.40 m/s perhaps you need to put it in as - 8.40 for three significant figures? This is right. The kinetic energy formula is (1/2)mv^2 The total kinetic energy is (1/2)(0.5)( - 8.40)^2 + (1/2)(3*0.5)(2.80)^2 = 23.5 J Again I think it may be that it just wants three significant figures. These numbers are right. For problem #6 part a: I used 1.20 instead of 1.25 The answer for the first part should be 2.18 kg m/s For problem #10 part b: Made a stupid mistake here. Moved the decimal the wrong way. The correct answer is 4.26*10 - 14 J
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