M1 Dynamics of a particle

For c) the impulse is the change in momentum. So calculate the velocity as it returns to the initial position and then the (change in) momentum. That will be the impulse necessary to bring the lift to rest.
d) should just require a bit of thinking about air resistance on the lift falling. Would air resistance affect the solution, if so, how?

I'll make my questions clearer , for c) Impulse is m(v-u) if I'm considering the lift only ( according to the question ) why is the mass used in this calculation (in the solution bank) including both the passenger and the lift? The impulse acts on the lift , I don't understand this part . For d)I thought of 2 explanations for the solution and Idk which ( if any ) is correct , I first thought that if air resistance was present , the impulse needed to cause the lift to come to rest would be less because air resistance helps oppose the motion .2nd possible explanation , if Impulse =m(v-u), air resistance should affect both v and u equally and decrease them , Impulse would decrease.

The safety string will act on both the lift and the passenger, hence both need to stop. Imagine there was a 10^6 kg (or something very large) mass inside the lift. The safetly line would break. The mass (person) inside the lift needs to be included.

For di), I'd agree, the height (release point) obained by the lift is the same. Depending on the velocity of the lift in free fall, it may not have that much effect in slowing down the lift, but there would be some effect. Hence the required impulse would be less. Imagine a lift falling in treacle (very high "air" resistance).
If the safety cord stops the lift, the final velocity "v"? must be 0? So the 2nd explanation is similar / same as the first?

Alright,thanks . May I ask another question ?

sure

For q1), when you do the calculation, you should get positive or negative velocities which give you the direction. So just assume one of the particles moves in a positive direction at the start of the question. Have a think about whether it matters which one you choose. The question seems to be deliberately vague in this respect to check whether you understand how to do it, so you'll need to practice / get used to it.
For q9), the truck is brakeing, which direction is the force acting on the truck? What will happen to the car, which direction will the force be acting through the towbar? Just think about whether the vehicles will have a positive or negative acceleration, the direction should be obvious. The towbar is rigid, so it directly transmits the force, you don't have to try and give it another name.