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Spring-Loaded collsion (momentum/center of mass) question?
A cart of mass m = 4 kg carrying a spring and moving at speed v = 3.1 m/s hits a stationary cart of mass M = 12 kg. Assume all motion is along a line and the collision is totally elastic.
i found the speed of the center of mass of the system being .775 m/s, and i need to find the speed of the big and small mass car after the collision
since the collision is elastic, we can use both momentum and energy conservation
call v1 and v2 the speeds of the smaller and larger carts respectively, after collision
momentum conservation gives us:
4kg x 3.1m/s = 4v1+ 12 v2
12.4=4v1+12v2 or
v2=(12.4-4v1)/12 (eq. 1)
energy conservation gives us
1/2(4)(3.1)^2 = 1/2(4)v1^2 + 1/2(12)v^2
divide through by 1/2(4):
3.1^2=v1^2+3v2^2
substitute eq. 1 into the equation above:
3.1^2=v1^2 +3(12.4-4v1)^2/12^2
expand the right side and get a quadratic in v1; solve this quadratic to get that
v1=-1.55m/s and v2=+1.55m/s
the vehicles have the same speed, but move in different directions