lastic Collisions: measurements: case ma (g) mBs (g) v (m/s) Ve (m/s) v (m/s) Vi
ID: 1552198 • Letter: L
Question
lastic Collisions: measurements: case ma (g) mBs (g) v (m/s) Ve (m/s) v (m/s) Vie (m/s) Zero Complete the table below with the total momentum initially (p) and finally (pr). Repeat for the kinetic energies. Make sure that all quantities have the correct sign calculations case pi (g m/s) p (g m/s) Ap (g m/s) KE (mJ) KEfr (mJ) AKE (mJ) 4 -1.2755 21. 2977 28. 5727 20.Y 72 203 T20. 26 Was momentum (mostly) conserved in these three cases? If not, check with instructor. Was kinetic energy (mostly) conserved in these three cases? If not, discuss why not.Explanation / Answer
Data given in the tables is incorrect or there is an experimental error
Total momentum is conserved i.e.
momentum of A and B before collison = momnetum of A and B after collision
case (3) momentum before collision
pa = ma*via = 513.11*0.4857 = 249.22
pb = mb*vib = 516.03*0 = 0
Total momentum before collission = 249.22
pa = ma*vfa = 513.11*0.2466 = 126.53
pb = mb*vfb = 516.03*0.2455 = 126.69
Total momentum before collission = 253.22
both shall be same, check with the data
case (4) momentum before collision
pa = ma*via = 513.11*0.2826 = 145.0
pb = mb*vib = 516.03*0 = 0
Total momentum before collission = 145.0
pa = ma*vfa = 513.11*0.02080 = 10.67
pb = mb*vfb = 516.03*0.02059 = 10.62
Total momentum before collission = 21.29
both shall be same, incorrect data collection
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