- #1
Rijad Hadzic
- 321
- 20
Homework Statement
So my book doesn't have an answer for this problem at the back, was just wondering if someone could check my work..
Two bumper cars at the county fair are sliding toward one another. Initially bumper car 1 is traveling to the east at 5.62 m/s and bumper car 2 is traveling 60 degrees south of west at 10 m/s. After they collide, bumper car 1 is observed to be traveling west with a speed of 3.14 m/s. Fiction is negligible between the cars and the ground. If the masses of bumper cars 1 and 2 are 596 kg and 625 kg respectively, what is the velocity of bumper car 2 immediately after the collision?
Homework Equations
Ki = Kf
[itex] V_f = ((m_1 - m_2)/(m_1 + m_2)) (V_{im_1}) + ((2m_2)/(m_1 + m_2))(V_{im_2}) [/itex]
The Attempt at a Solution
So the x velocity of bumper car 2 according the the eq ^ up there.
[itex] V_x = ((625 -596) / (625 + 596)) 10cos(240) + \frac {2(596)}{(625 + 596) }(5.62) [/itex]
x velocity = 5.62 m/s
[itex] V_y = ( (625-596)/(625+596) ) 10sin(240) + 0 [/itex]
y velocity = -.2057 m/s
I put 0 here because the y velocity of car 1 doesn't exist
Now I find magnitude:
[itex] ( (5.3682)^2 + (-.2057)^2 )^{1/2} [/itex]
= 5.3721 m/s at 357 degrees (using arctan)
Can anyone check my work and make sure its correct? Am I doing the right thing here?
I also got another answer around 50 m/s using Kinetic energy initial = Ke final but that doesn't seem right so I will use this one since 5.37 m/s seems about right to me.