Momentum, Kinetic Energy, And Vector Analysis In Physics Problems
Find step-by-step Assignment Answers on momentum and vector physics.
Abigail Bennett
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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
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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Physics