Use the equations below:
P=mv
Ek=1/2mv^2
1) During a docking manoeuvre, a supply pod of mass 4000kg accidentally collides with a small space station of mass 100,000 kg, at a relative velocity of 2.00ms-1. The pod attaches to the station.
You can assume that the space station is stationary before the collision.
A) Calculate the momentum of the system before the collision.
B) State the momentum of the system after the collision.
C) Calculate the change in velocity of the space station
D) Calculate the kinetic energy of the objects before the impact.
E) Explain whether the collision is inelastic or elastic and why.
12marks
Your problems are all the same. You do not seem to have tried any. I will outline this one, period.
Initial momentum:
P = 4,000 * 2 + 100,000 * 0 = 8,000
That is of course the final momentum as well.
8,000 = (4,000+100,000) v
Since they stick together, v is the final speed of both and is the change of speed of the station.
Ke before = (1/2)(4,000)(4) + 0
Ke after = (1/2)(104,000)(v^2)
the final Ke will be less than the initial Ke because the collision is inelastic (no springs, stick together, get warm)
Please help with the rest too as each question is the same equations involved but very different and i need help with them.
I think its 0.077ms-1
so is ke before half 4000 times zero?
the answer worlb be zero ek
you square velocity so what is beingquare for velocity before and wat is with the extra number of adding zero
oh= i get 2 squared is 4 ohh
for the second final ek the 0.5 x 104000x2 squared is 208000 which is more
and if the velocity is wrong then how do you do it
if yuo cantwork ou the velocity then how do u know that it is less