# Conservation of energy using orientation

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.....

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 %

Doc Al
Mentor
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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