Answer:
5398.4km/h
Explanation:
IN THIS CASE THE MOMENTUM IS CONSERVED. THE VALUE OF MOMENTUM OF ONE COMBINED ROCKET WILL BE SAME AS OF TWO COMBINED.
Let mass of module be m
then
mass of motor = 4m (four times the mass of rocket module)
total mass = m + 4m = 5m
combined velocity = V = 5320kph
Let
absolute (relative to earth)motor velocity after disengagement = v
then
rocket module velocity (relative to earth) after disengagement = v+98 (relative velocity = 98)
momentum conservation equation
combined momentum = module momentum + motor momentum
(m+4m)V = m(v+98) + 4m*v
5mV = 98m+mv + 4mv
5V = 98+v + 4v (m cancels out)
5V - 98 = 5v
((5*5320)-98)/5 = v
v = 5300.4 km/h
velocity of rocket module relative to earth = v +98
= 5300.4 + 98
= 5398.4km/h
Answer:
9.03 times 10times 23 calculate it and there ya go
