A vehicle traveling due east at 3m/sec collides with and sticks to a vehicle having twice the mass that is tra

Question

A vehicle traveling due east at 3m/sec collides with and sticks to a vehicle having twice the mass that is traveling due west at 2 m/sec. Immediately after the collision, the vehicular combination is moving due west at 1/3 m/sec. The mass of the smaller vehicle is most nearly (kg).
a) 100
b) 250
c) 900
d) 1100
e) 2500
f) 4000
g) 4900
h) 6000
i) No single numerical answer is true for this situation.

Answer 1

This is an interesting question. Just simply write a conservation of momentum equation to solve.
Let's say west is positive and east is negative. The first vehicle is traveling east so we say at -3m/s and the other vehicle is traveling at 2 m/s since it is heading west. The mass of the first vehicle is x and the mass of the second vehicle is 2x and the resulting mass is 3x.

The equation becomes:

2x(2m/s)-x(3m/s) = 3x(1/3m/s)
This equates to:
4x-3x = x or x=x

This means that for any weight you specify the answer will be correct. So choose i)
..

Answer 2

Taking due west as positive direction:
m = mass

using conservation of momentum:
2m x 2 + m x (-3) = 3m x 1/3

there is an m in each term, therefore m could be anything or cancelled out - the answer is i)

Answer 3

The mass of the vehicles is not important .
I mean , whatever it will be , there will be no change to the answer.
so : "i" is the correct answer.

Answer 4

-3m + 4m = (3/3)m
The m's cancel, leaving no way to determine m.