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.

Homework Equations

m₁v₁=m₂v₂

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 don't 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 spacecraft and its direction
d) If the fuel is ejected at 3000m/s find the mass of the propellant used in the burn

Homework Equations

F$$\Delta$$t = m$$\Delta$$V
m= 100 kg
v=5000 m/s
t= 7min = 420 secs

500N 45 degrees NoW

The Attempt at a Solution

a) m$$\Delta$$V = 100kg(5000m/s) = 500000 kg m/s
b) I am stuck afterwards I am guessing its cos(45)500N * 420 seconds
c) answer from b divide by 100kg (F$$\Delta$$t / m = V)
d) I don't 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.

 

[thomate1]

how to use the fuel speed

a) The impulse is equal to the change in momentum, which is 500000 kg m/s in the direction 45 degrees North of West.
b) To find the new momentum vector, we need to use vector addition. The initial momentum vector is 100kg * 5000m/s in the direction due East. The impulse vector is 500000 kg m/s in the direction 45 degrees North of West. Adding these vectors results in a new momentum vector of approximately 500000 kg m/s in the direction 30 degrees North of East.
c) To find the new speed, we can use the Pythagorean theorem: v = √(500000^2 + 5000^2) = 500050 m/s. The direction can be found using trigonometry: tanθ = 5000/500000, θ = 0.572 degrees North of East.
d) To find the mass of the propellant used, we can use the formula m = FΔt/Δv, where F is the force produced by the thruster (500N), Δt is the time the thruster is fired (420 seconds), and Δv is the speed of the fuel ejected (3000m/s). Plugging in these values gives us a mass of approximately 70 kg.