PHYSICS help wanted?

Question

A raliroad diesel engine weighs four times as much as a greight car. If the diesel engine coasts at 5 km/h into a greight car that is at rest how fast do the two coast after they couple?


A 5kg fish swiming at 1 m/s swallows an absent minded 1-kg fish at rest. What is the speed of the large fish immediatly after lunch? what would its speed be if the small fish were swimming towart it at 4 m/s?

Answer 1

Think about the momentum of the railroad engine/freight car system. Momentum is always conserved and momentum of its object is equal to its mass times its velocity. p=mv (remember momentum and velocity are VECTORS). Also, think of what kind of collision is occuring.

Answer 2

The first engine has momentum = mv = 4m * 5 km/hr = 20m kg-km/hr, and the second engine has zero momentum because it isn't moving. After they collide, the mass has increased to 5m, so the velocity will decrease to 4 km/hr.

Initial momentum mv = 5; after lunch, mass=6, so velocity=5/6 m/s.

Momentun of 5kg fish is 5 and of the 1kg fish is -4, so combined is 1. The mass now = 6, so the velocity is now 1/6 m/s in the same direction the larger fish was swimming.

Answer 3

(1) Momentum conservation law for the system: M*v1=(M+m)*v2, M & m are masses, M being =4*m. M*v1 momentum before impact, (M+m)*v2 momentum after impact. Thus v2=v1*4*m/(4*m+m)=v1*4/5, v1 being5km/h, v2=4km/hour.
(2)Assume M=5kg fish, m=1kg fish, v1=4m/s & M*v1=(M+m)*v2 – the same as above. Thus
v2=4*5/(5+1)=10/3=3.3333 m/s
(3) M*v1+m*v3=(M+m)*v2, v3=-v1. Thus v2=v1*(M-m)/(M+m)=4*4/6=8/3=2.6667 m/s