need help with PHYSICS problem. . . please help :( {conservation of momentum}?

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

The problem is simpler than it looks since the direction is along a single NW - SE line. So you can directly add momentums. The momentum of the bike+riders before they jump off is (3.4*10^2 + 2.5*10^2)*V = 5.9*10^2*V. This must equal the combined momentum of bicycle and riders after they jump off. The rider's momentum is -9.0*2.5*10^2 (the minus sign is because of the change of direction), and the bike's momentum is 28*3.4*10^2; Add the latter (minding the signs) to get 72.7*10^2 kg-m/sec. This must equal 5.9*10^2*V, so

V = 72.7*10^2/5.9*10^2 = 12.32 m/sec.

This is the velocity of the bicycle and riders just before they jump off.