need help with PHYSICS problem (conservation of momentum) pls help :(?

Question

Need help with PHYSICS problem (conservation of momentum)?
Here is the problem:

The longest bicycle in the workld was built in New Zealand in 1988 it is more than 20 m in length, has a mass of 3.4 x 10² kg, and can be ridden by four people at a time. Supposed four people are riding the bike south-east when they realize that the street truns and that the bike won't make it around the corner. All four riders jump off the bike at the same time with the same velocity (9.0 km/h to the northwest, as measured relative to Earth.) The bicylce continues to travel forward with the velocity of 28 km/h to the southeast. If the combined mass of the riders is 2.5 x 10² kg, what is the velocity of the bicycle and riders immediately before the riders escape?












i think its asking for v(initial), right?

does the change of direction affect the conservation of momentum??

the formula is m1x v1 (initial) + m2 x v2 (initial) = m1x v1 (final) + m2 x v2 (final)

please sovle and explain a little

Answer 1

You can consider the four people and the bicycle as one item before they jump off the bike.
M0V0 represent the momentum of this item
M1V1 represent the momentum of the bike after the four riders jump off
M2V2 represent the momentum of the the riders after they jump off

M0V0 = M1V1 + M2V2
(3.4 x 10² + 2.5 x 10²) V0 = (3.4 x 10²) (28) + (2.5 x 10²) (-9)
V0 = 12.3 km/h or 3.42 m/s

Answer 2

The radius is not given so I hope this problem focuses on Linear momentum conservation .20 * 2 = .5 *1 + .2 *v .4 - .5= .2 * v v= -.05m/s the first ball moves back at .5 m/s according to this equation but physically visualising this thing seems not to be possible... so the fault seems to be that the .5 Kg ball can't move at 1 m/s after collision, unless the initial velocity of the .2 kg ball is greater

Answer 3

I answered already http://answers.yahoo.com/question/index?qid=20071212000628AA4RDQf&r=w#RM14CzO.ADK6wRMDzTlH7Xh_Xco9E_0JQlu3EhDoI_Ijya8CdE78