Conservation of energy using orientation

  • #26
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Given the data in the problem statement, your velocity is correct. (Unless you made an error somewhere.) But that would imply that the kinetic energy increased as a result of the collision, which I do not think was your instructor's intention in setting up the problem.

Oh so ... i understand what you are trying to say.. so the kinetic energy needs to be lost rather then being increased.....
 
  • #27
217
0
Given the data in the problem statement, your velocity is correct. (Unless you made an error somewhere.) But that would imply that the kinetic energy increased as a result of the collision, which I do not think was your instructor's intention in setting up the problem.

iven:
m1= 80 kg
m2 = 90 kg
v1 = 7.5 m/s [ E 20 S]
v1` = ?
v2= 0m/s
v2` = 3m/s [ E 30 N]

X component
m1v1 = m1v1` + m2m2`
(80kg)(7.047694656m/s[E]) = (80kg)(v1`) + (90kg)(2.598076211[E])
563.8155725 kgm/s [E] - 233.826859 kgm/s[E] = (80kg)(v1`)
329.9887135 [E] / 80kg = v1`
4.124858919 m/s [E] = v1`

y components
m1v1 = m1v1` + m2v2`
(80kg)(2.565151075m/s) = (80kg)(v1`) + (90kg)(1.5m/s[N])
205.212086kgm/s-135m/s[N] = (80kg)(v1`)
70.212086kgm/s/80kg = v1`
4.252651075m/s = v1`

4.124858919 m/s [E] + 4.252651075m/s = v1`
5.924 m/s [S 44° E] = v1`

b) Ektot = 1/2m1v1^2
Ektot= 1/2(80kg)(7.5m/s)^2
Ektot=2250 J

Ektot` = 1/2m1v1`^2 + 1/2m2v2`
Ektot` = 1/2(80kg)(5.924m/s)^2 + 1/2(90kg)(3m/s)^2
Ektot` =1403.75104 J + 405 kg J
Ektot` = 1808.75104 J

Ektot - Wdef = Ektot`
2250 J - Wdef = 1808.75104 J
Wdef = 441.24896 J

The Wdef is the energy lost during the collision

(441.248896 J / 2250 J) x 100%
19.6 %


I guess i should just re-post my answer on this thread...
 
  • #28
Doc Al
Mentor
45,140
1,439
iven:
m1= 80 kg
m2 = 90 kg
v1 = 7.5 m/s [ E 20 S]
v1` = ?
v2= 0m/s
v2` = 3m/s [ E 30 N]
Looks like you did have a mistake in the problem statement, which you've corrected. Now it should make sense.
 

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