physics questions

In the first 0.05s, we stop at this time to calculate the average force acting on water by rocket. It is an instantaneous motion of the rocket. It is really important about the weight of water. By Newton's third law, action and reaction, the force from the weight of water acts to, the force pushing up the rocket gained.
Now, the case is the momentum of water is pointing downwards, so the momentum of the rocket by the water is pointing opposite, upwards. You don't need to mix up and misunderstand the rocket and water, they can be analysed individually.
Calculating momentum questions, you need to know the knowledge of Motion ( Newton's laws) and also (direction and magnitude).

2014-09-02 04:06:15 補充：
http://postimg.org/image/thsyxk6hr/

1. v^2=u^2+2as
v^2=0+2(10)(5)
v=10m/s
2. Change in momentum= mv-mu = 70(10)= 700 Ns
Impact force = 700 /0.2 = 3500N
3. 3500/70 = 50 times

2014-09-02 04:13:53 補充：
Alter: 2. Change in momentum= mv-mu = 0-70(10)= -700 Ns

1. The force acting on the water is vertically downward, the same direction as the weight of water. Hence, the weight doesn't concern with the acting force, as it is given by gravity, not by the rocket.

2.(a) Let x be the initial distance of A from the mirror. Hence, image of A is x behind the mirror.
Separation between A and its image A' = 2x

After 1 s, the new distance of A from the mirror = x-1
New distance of image A' from mirror = x-1
New separation between A and A' = 2(x-1)

Hence, in 1 s, as seen by A, A' moves towards A by a distance = 2(x-1) - 2x
= -2 m/s
i.e. velocity of A' as seen by A = 2 m/s (the -ve sign indicates it is towards A)

The external observer sees A' moves 1 m/s towards A.

(b) Let x be the initial distance of A from the mirror. Hence, image of A (A') is at distance x behind the mirror.
Initial separation between A and A' = 2x

After 1 s, new distance of A from mirror = x-1
New distance of A' from mirror = x-1
New separation of A and A' = 2(x-1)
i.e. in 1 s, as seen by A, A' has moved towards him = 2(x-1) - 2x = -2 m/s
Hence, velocity of A' as seen by A = 2 m/s towards him

For an external observer, let x' be the distance of A' from the initial position of A after time of 1 s. We have,
(x-1) + x' = 1 + x
i.e. x' = 2 m
Hence, velocity of A as seen by an observer = 2 m/s



2014-08-25 22:35:25 補充：
Your suppl question:
The force acting on the water by the rocket equals to the rate of change of momentum of the water. This simply follows from Newton's Second Law. As said, this doesn't concern with the weight of the water, which is given by gravity.

2014-08-25 22:37:31 補充：
(cont'd)... Before the rocket is launched (i.e. when it is still stationary), the weight (of water) has already existed.

2014-08-25 22:37:48 補充：
cont'd)... Before the rocket is launched (i.e. when it is still stationary), the weight (of water) has already existed.

2014-08-26 11:12:11 補充：
Pl see 意見 section.

2014-08-26 11:16:16 補充：
Newton's Second Law relates the force and the rate of change of momentum of an object. In this problem, the force so found is that acted on the water by the rocket. How come you have to substract the weight from the force? The two forces are independent from each other.

2014-08-27 00:14:47 補充：
Your additional question:
The air cushion problem is completely different from the rocket. In the former, there is an normal reaction acting on the man and the additional retarding force from the cushion. Both forces contributes to stop the falling man.

2014-08-27 00:18:43 補充：
In the rocket problem, the weight of water doesn't contribute to ejecting the water from the rocket. The only force that acts on ejecting the water comes from the gas pressure inside the rocket.