Rod's Physics 3/4 Questions Thread

[Rod]

Hi guys, just stuck on a question I need to do for holiday homework.

A green dodgem car of mass 400kg has a head-on-collision with a red dodgem car of mass 300kg, Both dodgem cars were travelling at a speed of 2 m/s before the collision. What is the rebound speed of the green dodgem car if the red dodgem car rebounds at a speed of 1 m/s.

This is what I did:

Total momentum before collision = P=mv
so (400x2) + (300x-2)
so total momentum before collision = 200 kg m/s.

Momentum of red car is 300x1=300 kg m/s

So total momentum after collision is 200=400v + 300
                                                     200-300 =400v
                                                     -100=400v
                                                     v=0.25 m/s

The booklet's answer is 1 m/s. I think this was my teacher's answer and I have no idea how she gets 1 m/s.

Thank you

[Ya Habibi]

Just going to post here so I remind myself to do it in the morning

[Rod]

Rightyo thanks buddy 

I actually think my answer is right, the teacher's suggested solution must have been a mistake. So it would be great if you could just 'proof read' or something to see if I am correct tomorrow morning.

[Rod]

Hi guys I don't really get this:

So the sole purpose of air bags and padded dashboards is to 'increase the time interval during which the momentum of the vehicle’s occupants changes during a collision.' I always thought it was to reduce the individual's kinetic energy and momentum, so upon collision small amounts of energy is transferred and therefore the person does not get hurt as much. Can someone please explain what 'increasing the time interval' does?

Thanks

[lzxnl]

Hi guys, just stuck on a question I need to do for holiday homework.

A green dodgem car of mass 400kg has a head-on-collision with a red dodgem car of mass 300kg, Both dodgem cars were travelling at a speed of 2 m/s before the collision. What is the rebound speed of the green dodgem car if the red dodgem car rebounds at a speed of 1 m/s.

This is what I did:

Total momentum before collision = P=mv
so (400x2) + (300x-2)
so total momentum before collision = 200 kg m/s.

Momentum of red car is 300x1=300 kg m/s

So total momentum after collision is 200=400v + 300
                                                     200-300 =400v
                                                     -100=400v
                                                     v=0.25 m/s

The booklet's answer is 1 m/s. I think this was my teacher's answer and I have no idea how she gets 1 m/s.

Thank you

Taking the direction of the 400 kg car's motion to be positive, initial momentum is (400-300)*2 = 200 N s
The red 300 kg rebounds with a speed of 1 m/s; this is positive, so the final momentum = 200 N s = 400v + 300 = 200
400v = -100
I agree with your answer

Hi guys I don't really get this:

So the sole purpose of air bags and padded dashboards is to 'increase the time interval during which the momentum of the vehicle’s occupants changes during a collision.' I always thought it was to reduce the individual's kinetic energy and momentum, so upon collision small amounts of energy is transferred and therefore the person does not get hurt as much. Can someone please explain what 'increasing the time interval' does?

Thanks

During a collision, the person's momentum and KE will always be reduced to zero as they always end up stationary (that is an assumption I'm going to be making). It doesn't matter if you have air bags or not; the final velocity, and hence KE and momentum, is zero. Adding airbags, however, increases the time of the collision. It's like jumping onto a massive sponge mattress from 10 m high and jumping onto concrete. As the sponge mattress is massively deformable, the impact time is much larger, which means the average force = decreases. Now in the case of a car collision, if the average force on the car decreases, the average acceleration of the passenger decreases. The force experienced by the passenger is ma; thus, the passenger experiences a weaker average force and is thus hurt less.

[Ya Habibi]

Pretty much what I did was what the guy above wrote. Consult your teacher.

And with the second question.
By increasing the time taken, it lowers the impulse applied. A high impulse is achieved when there is a high force applied within a small time period. So, by adding protective gear it will be increase the time that the force is applied and thus reduce the force felt by the person.

[Aurelian]

And with the second question.
By increasing the time taken, it lowers the impulse applied. A high impulse is achieved when there is a high force applied within a small time period. So, by adding protective gear it will be increase the time that the force is applied and thus reduce the force felt by the person.

No this isn't quite accurate, unfortunately. The impulse is simply the change on momentum ∆p associated with the collision. Since the initial and final momentum of an occupant is the same whether or not airbags are present (start off with some momentum p, end up stationary with p = 0), then ∆p/the impulse associated with the collision is also not going to change.

The key, as lzxnl pointed out, is just that airbags increase collision time ∆t so that the average force applied to an occupant is reduced (since F_avg = ∆p/∆t and ∆p is constant, as explained above).

[Rod]

Taking the direction of the 400 kg car's motion to be positive, initial momentum is (400-300)*2 = 200 N s
The red 300 kg rebounds with a speed of 1 m/s; this is positive, so the final momentum = 200 N s = 400v + 300 = 200
400v = -100
I agree with your answer
During a collision, the person's momentum and KE will always be reduced to zero as they always end up stationary (that is an assumption I'm going to be making). It doesn't matter if you have air bags or not; the final velocity, and hence KE and momentum, is zero. Adding airbags, however, increases the time of the collision. It's like jumping onto a massive sponge mattress from 10 m high and jumping onto concrete. As the sponge mattress is massively deformable, the impact time is much larger, which means the average force = decreases. Now in the case of a car collision, if the average force on the car decreases, the average acceleration of the passenger decreases. The force experienced by the passenger is ma; thus, the passenger experiences a weaker average force and is thus hurt less.

Thank you I think I get it now. So by increasing the time the force of the person decreases. By deceasing the force, the acceleration decreases so the passenger wouldn't feel as much force?

[Aurelian]

Thank you I think I get it now. So by increasing the time the force of the person decreases. By deceasing the force, the acceleration decreases so the passenger wouldn't feel as much force?

There's a bit of extraneous reasoning in what lzxnl said. Just keep it simple: F_avg = ∆p/∆t (by definition); ∆t increases while ∆p stays the same, therefore the average force F_avg exerted on an occupant in a collision is reduced. Since the occupant experiences a smaller magnitude force, they are less likely to be harmed.

Hope that helps =)

[PB]

^Thats right. It is only the force that hurts the person, not necessarily the acceleration! Just watch out for that wording
The reason:
- an oxygen molecule might be accelerating towards you at 200m/s^-2 but it obviously would do absolutely nothing to you
-this time, a car is accelerating towards you at 200m/s^-2...now that might hurt a little. This is because the car's mass is so much larger than an oxygen's mass, hence the force exerted on you (F=ma) is larger too.

So yeh, my point is that the damage done to a person is dependant on force, not acceleration.

[Rod]

There's a bit of extraneous reasoning in what lzxnl said. Just keep it simple: F_avg = ∆p/∆t (by definition); ∆t increases while ∆p stays the same, therefore the average force F_avg exerted on an occupant in a collision is reduced. Since the occupant experiences a smaller magnitude force, they are less likely to be harmed.

Hope that helps =)

Ohh I fully get it now; F=ma and a=v/t, as t increases a decreases, and as a decreases f decreases. So if the force exerted on an occupant in a collision is weaker, obviously the passenger would be more likely to survive.

Thanks for the help everyone! I might ask more questions in the future if that is okay, really appreciate it.

I also wanted to ask all you guys what you did during the holidays for physics, chem and english. You all have seem to have done tremendously, so it would be great if you guys could give me some tips on what to do on the holidays. I've started from day one of the holidays, and am happy with my progress. Although at times I just can't be stuffed and I procrastinate or just go outside. I'm playing cricket as well so that is another distraction. What do you guys recommend me doing to keep out procrastination and balance my life a bit more? Did you guys procrastinate or do any extra-curricular activities during the holidays?

Thanks

PS - Should I also do a bit of methods and health? I'm not doing health because I don't think it would be that useful as it is not that hard. And I find it hard to self-learn maths.

[PB]

Hey rod,
For Physics ( I can only comment on Physics as I didn't do too well in Eng/Chem). My recommendation is to try read through and briefly graspe all the relevant concepts in your textbook. Try to get a feel of the whole course in its entirety so that you can enter your physics classroom always having a big picture of the subject in mind!
I personally did not use this strategy but on hindsight - this is what I would definitely have changed in my VCE year. I made the mistake of trying to tackle questions during the holidays, which is highly inefficient as I wouldn't need to know how to answer the questions until the next year. It also slowed my progress down considerably (questions took the bulk of my studytime) and I only managed to reach the end of the 3rd chapter before school started...not good.
This is advice I got from 99.95ers late in the year (which wasn't much help anymore) and I believe the great nliu would give this same advice
As for your dilema between Methods/Health  I think it is purely your decision to make. Though I might advice you to look at your school's performance in both subjects to make a judgement (should it come down to that stage). For example, your school might do better in Health overall each year, which may testify to a better Health teaching faculty.
Kind Regards,
PB

P.S. don't hesistate to ask any other conceptual problems for Physics that is what the AN community is here for!

[Rod]

Hey rod,
For Physics ( I can only comment on Physics as I didn't do too well in Eng/Chem). My recommendation is to try read through and briefly graspe all the relevant concepts in your textbook. Try to get a feel of the whole course in its entirety so that you can enter your physics classroom always having a big picture of the subject in mind!
I personally did not use this strategy but on hindsight - this is what I would definitely have changed in my VCE year. I made the mistake of trying to tackle questions during the holidays, which is highly inefficient as I wouldn't need to know how to answer the questions until the next year. It also slowed my progress down considerably (questions took the bulk of my studytime) and I only managed to reach the end of the 3rd chapter before school started...not good.
This is advice I got from 99.95ers late in the year (which wasn't much help anymore) and I believe the great nliu would give this same advice
As for your dilema between Methods/Health  I think it is purely your decision to make. Though I might advice you to look at your school's performance in both subjects to make a judgement (should it come down to that stage). For example, your school might do better in Health overall each year, which may testify to a better Health teaching faculty.
Kind Regards,
PB

P.S. don't hesistate to ask any other conceptual problems for Physics that is what the AN community is here for!

Thank you so much PB!

I'm going to follow everything you have said apart from the questions part. I thought doing questions would consolidate my knowledge and prove that I've studied that chapter well and can move on to the next one. So I think I will be doing questions, but not as much. Initially, I've been doing the chapter review questions, chapter questions, A+ note questions and heinemann text book questions but might just do one, not four. I'm up to chapter 3 with 6.5 weeks remaining.

So this is what I'm doing:
- Reading the whole chapter
- Summarising it and taking notes
- Doing four lots of questions
- Then moving on

Is that okay? The only thing I'll change after reading your advice is probably do only one lot of questions, maybe the chapter review. Thanks for all your help, I joined Atarnotes yesterday and I can't believe how awesome it is. How VCE specialists like you are sparing their time and helping others in need.

[Rod]

Hi guys, another question here

So just imagine this, I'm bored so I go outside in the back yard. I take a 20 metre run up and charge a brick wall. After colliding, my momentum becomes zero as the brick wall submits a force to stop me. So since momentum is always conserved, where would the momentum go? I've been reading my text book and apparently, in this case some will go into the brick wall and some transferred to the Earth. What does being transferred to the Earth mean?

Thanks guys

[SocialRhubarb]

You have slightly changed the speed at which the earth moves through space, and also changed the rate of rotation of the earth.