Answer:
F = 4000 N
Explanation:
given,
mass of rocket (M)= 5000 Kg
10 Kg gas burns at speed (m)= 4000 m/s
time = 10 s
average force = ?
at the end the rocket is at rest
by conservation of momentum
M v + m v' = 0
5000 x v - 10 x 4000 = 0
5000 v = 40000
v = 8 m/s
speed of rocket = 8 m/s
now,
we know
change in momentum = F x Δ t
F = 4000 N
Hence, the average force applied to the rocket is equal to F = 4000 N
The question is incomplete. The complete question is :
The Rocket Club is planning to launch a pair of model rockets. To build the rocket, the club needs a rocket body paired with an engine. The table lists the mass of three possible rocket bodies and the force generated by three possible engines.
A 4-column table with 3 rows. The first column labeled Body has entries 1, 2, 3. The second column labeled Mass (grams) has entries 500, 1500, 750. The third column labeled Engine has entries 1, 2, 3. The fourth column labeled Force (Newtons) has entries 25, 20, 30.
Based on Newton’s laws of motion, which combination of rocket bodies and engines will result in the acceleration of 40 m/s2 at the start of the launch?
Body 3 + Engine 1
Body 2 + Engine 2
Body 1 + Engine 2
Body 1 + Engine 1
Solution :
Given :
Body Mass (gram) Engine Force (newtons)
1 500 1 25
2 1500 2 20
3 750 3 30
The body 1 has a mass of 500 gram which is equal to 0.5 kg
And engine 2 has a force of 20 newtons.
We know that according to Newton's laws of motion,
Force = mass x acceleration
20 = 0.5 x acceleration
Acceleration
Therefore, based on laws of motion of Newton, the Body 1 + Engine 2 combination of the rocket bodies and engines will result in an acceleration of at the start of the launch.
