Conservation momentum questions help

philadelphia

1. A 1.0 kg mass moving with a velocity v m/s strikes a stationary mass (assume no frictional losses).
a. Is it possible for both masses to move in the same direction and at the same speed after the collision? If so, find the value of the stationary mass.
b. Is it possible for both masses to move in OPPOSITE directions but with the same speed after the collision? If so, find the value of the stationary msss.

2. Relevant equations
m₁v₁=m₂v₂

3. The attempt at a solution
Not sure if this is the right path for (a)
m₁ = 1kg
v₂ = 0 m/s

a. after collision
v₁ = v₂ = v = 5(random variable to plug)
(1kg)(5m/s) = m₂(5m/s)

b. i dont get help help me with a step

2. A space probe of mass 100kg is traveling due East at 5000m/s. It fires a thruster for 7 minutes to change direction. If the engine produces 500 Newtons of force in a direction 45 degrees North of West.
a)Find the impulse
b)FinD the new momentum vector of the spacecraft
c)Find the new speed of the space craft and its direction
d) If the fuel is ejected at 3000m/s find the mass of the propellant used in the burn


2. Relevant equations
F[tex]\Delta[/tex]t = m[tex]\Delta[/tex]V
m= 100 kg
v=5000 m/s
t= 7min = 420 secs

500N 45 degrees NoW
3. The attempt at a solution
a) m[tex]\Delta[/tex]V = 100kg(5000m/s) = 500000 kg m/s
b) Im stuck afterwards Im guessing its cos(45)500N * 420 seconds
c) answer from b divide by 100kg (F[tex]\Delta[/tex]t / m = V)
d) I dont know

Mthees08

1b) is simply if it is an elastic collision, meaning Vf1=-Vf2. Entirely possible,

2 is just a stupid question. There is no east in space....
a) your answer looks correct...
b) P=mv use conservation of momentum and the impulse you found to find its new momentum. I=Pfinal - Pinitial
c) you seem to have a handle on.
d) I am assuming this is more simple than I am thinking. However the simplest I can think of is that impulse is the integral of force, Impulse is also change in momentum, which gives you that by newtons 2nd and 3rd laws the force exerted by the fuel acts equally on the rocket. The force of thrust is Dm/dt times V so that may help you. If none of that rant helps say so I am going to go solve this as an exercise of my own so I will have it shortly.