Quite a few phys 12 questions, on unit 3 dynamics,energy,work,power

[gdhillon]

Fist a little bit of background on why I am doing physics 12, I am currently a first year electrician and recently found out in order to get a job with BCHydro physics 12 is a pre req. Unfortunately this is the only science I didn't take in highscool. The job posting for apprentice ships will be up early/mid september and run about 6 weeks so I have about 3 months to finish this course. It is all online (no textbook) I have been using/abusing Salmon Khan's tutorials. I am now on unit 4 of 8. There is a test every two units and no final :D. I really would like to do the best I possibly could for this course to make my resume look better. So for this unit (dynamics,energy,work,power,efficiency) I have done all the self marking assignment along with the lessons and I actually found it easy. Now for the send-in worth marks I am having quite a few troubles. I am not sure if I can post all my questions in one thread but I am going to anyway (mods I apologize in advance if I am not allowed to do this please email me if I am in the wrong) I have completed the send in and I sent my answers to the teacher and these are the ones he told me I need to relook, I have relooked them over and over and am on the brink of tearing my hair out! haha. I am not looking for answers I am looking for what i am doing wrong when I approach these questions. Thanks in advance :)


A 1.2 x 103 kg car is traveling at a velocity of 20.0 m/s East when its brakes are locked.
Assuming a force of friction of 2.5 x 104
N, what is the velocity of the car after 0.50 sec?



I used mv(before)=mv(afterbrakes)



3.I got 30 m/s for the velocity of the car after 30 seconds


Q:Two blocks are tied together on an inlcline The first box is on a 30 degree incline witht he horizontal and is 1 KG, the wire from it runs over a pulley that goes to a mass of 2KG . If both the pulley and the incline are frictionless:
a) What is the acceleration of the 1.0 kg block up the incline?
b) What is the tension in the string joining the two blocks.

I first found the FG of the 1KG block and from that used sohcahtoa to get my normal force and my fapp (-9.8sin30) negative because it is 'wanting' to go down the incline away from the 2kg mass. For a) I used Ʃf=ma, 19.6-(9.8sin30)=3a acceleration is 4.9m/s^2
for b) I used Ʃf=ma and had T-19.6=2(4.9) and got tension=29N


Q:A boy of mass 45 kg is traveling without a seat belt, in a car with a velocity of 14 m/s.
When the car stops suddenly, he is stopped by the windshield in a distance of 0.024m.
What force does he exert on the windshield?

i set the impulse of the vehicle (mv)= to FD (because (delta)E=W=Fd) and got a force of 2.6*10^4N. That seems to high of a force to me.


Q:A sandbag stops a 12 g bullet which was traveling at 860 m/s. The bullet penetrated a
distance of 24 cm into the sand?
a) How long did it take the bullet to stop?
b) What average force acted on the bullet?

I first found the force by using (delta E)=Fd and got 18490N then I put that force into (delta)p=Ft and got a time of 5.6*10^-3 sec.


Q:A 46 g steel marble collides obliquely with an identical stationary marble, and continues
at 55 to its original direction. The collision is perfectly elastic. What is the angle between
the direction taken by the target ball and the original direction of the incident ball?

I know the collision being elastic means the kinetic energy is conserved, but I am completely lost on how to solve this question I tried drawing the vectors tip to tail but I didn't know how I would find theta because I don't know any of the sides.


Q:A stationary lift raft of mass 160 kg is carrying two survivors with masses of 55 kg and
72 kg. They dive off the raft at the same instant, the 55kg person East at 4.4 m/s and the 72
kg person North at 4.2 m/s. AT what speed and in what direction does the raft start to move?

What i did here was used p=mv to find both peoples impulses then I said p(east person)+p(north person)=P(raft) So I have 242+ 302.4=160V and I got a velocity of 3.4m/s @15degrees S OF W My teacher said the direction was correct on this one


Q:A 40000N truck moving West at a velocity of 8.0 m/s collides with a 30000N truck
moving South at a velocity of 5.0 m/s. If these two vehicles lock together upon impact, what
is the initial velocity of the vehicles after the collision?


I added the (momentum) vectors tip to tail, i then used Pythagoras with the momentum's i calculated and solved for velocity and got .52m/s @25degrees S of W


Q:An object explodes into three equal masses. One mass moves East at a velocity of 15.0
m/s. If a second mass moves at a velocity of 10.0 m/s 45.0 South of East, what is the
velocity of the third mass?

I drew the vectors tip to tail then used cos law to find the velocity of the third mass. c^2=15^2+10^2-2(15)(10)cos 45 and got a velocity of 10.623m/s=) 11m/s I then used sin law to get theta and had sin45/10=sinc\10.6239 and got 49 dgrees S Of E


Qow much work did a horse do that pulled a 200 kg wagon 80 km along a level road if
the effective coefficient of friction was 0.060?

I used W=Fd to find the work (W=1960*80000=156800000) then I got the force of friction by ff=(coeff)Fn and got (.06)*1960=117.6. I then got the W(friction)=117.6(80000)=-9408000 and by subtracting the 2 Work answers I got 14739200J


A heavy box of mass “n” kg slides 12.0 m along a straight frictionless 30 degree incline. If
the box starts from rest at the top of the incline, what is its speed at the bottom. (hint: think
energy)

I used Fd=1/2mv^2 which was msin30(12)=1-2mv^2 and I canceled the masses then came out with velocity=3.5m/s


A 5.0 kg object is accelerated from rest to 6.0 m/s while moving 2.0 m across a level
surface. If the force of friction is 4.0 N, calculate the power output.

I used d=vt to find the time which was .33sec then i put that into P=Fd/t and had (49-4)(2)/.35 and got 270 Watts


I know that's a lot of questions but any help would be greatly appreciated, again I'm not looking for answers just direction in where I am going wrong and what I need to change to be on track!

Thanks,

Gavin

 

[tiny-tim]

Welcome to PF!

Hello Gavin! Welcome to PF!

Please don't put so many questions together … there's no way anyone will want to answer all of them.

A 1.2 x 103 kg car is traveling at a velocity of 20.0 m/s East when its brakes are locked.
Assuming a force of friction of 2.5 x 104
N, what is the velocity of the car after 0.50 sec?

I used mv(before)=mv(afterbrakes)

3.I got 30 m/s for the velocity of the car after 30 seconds

i] how can the speed be greater after braking?

ii] mv(before)=mv(after) only applies when there's no external force (in that direction) …

the brakes are an external force

iii] always show your calculations … how did you get 30 ?

iv] look at the significant figures … if it was 30, wouldn't it be 30.0 ? (if not, why not?)

Try again. ​

(and please start some new threads for the remaining questions)

 

[gdhillon]

Thank you for the warm welcome, Okay ill put them all in separate threads.

I got that by first finding the momentum before. p=1200(20)=24000N*s
then i went epi+eki=epf+ekf...since there's no height I could take out the potential energies so i now have eki=ekf
I then found the impulse of the force of friction so p=Ft, p=25000(.5)=-12500N*s so then i had 24000=1200v-12500 and got 30m/s...two sig figs because that's the least amount of sig figs in the question. I am going to retry this question from scratch after i make separate threads for the other questions.

Thanks,

Gavin

 

[tiny-tim]

Hi Gavin!

then i went epi+eki=epf+ekf...since there's no height I could take out the potential energies so i now have eki=ekf

i don't understand why you're saying this … you haven't used it, and it's not true anyway

I then found the impulse of the force of friction so p=Ft, p=25000(.5)=-12500N*s

correct

so then i had 24000=1200v-12500 and got 30m/s

again, how can the speed be greater after braking?

isn't it obvious that you must have a minus in the wrong place?

(it might avoid mistakes if you were to state a reason, or description, before every equation)

...two sig figs because that's the least amount of sig figs in the question.

no, that only applies when there's only multiplication (and division)

where there's also addition (or subtraction), the rule is different, see significant figures

(it often gives the same result, but you must show that your result is correct)

 

[Nickie]

Hello Gavin,

First of all, I want to commend you for taking on the challenge of completing physics 12 in such a short amount of time. It's not an easy task, and it shows your determination and dedication to your career goals.

Now, let's take a look at the questions you have provided. It seems like you have a good understanding of the concepts and are using the correct equations to solve the problems. However, there are a few things that I noticed that may be causing some errors in your answers.

For the first question, you have correctly used the equation mv(before)=mv(after brakes) to solve for the velocity of the car after the brakes are locked. However, you have not taken into account the time that has passed (0.50 sec). Remember, the force of friction acts over a certain amount of time, which will cause the car to decelerate. So you will need to use the equation F=ma to find the deceleration, and then use that in your equation for velocity.

For the second question, you have correctly used the equations for finding the acceleration and tension in the string. However, I noticed a small error in your calculations for the tension. When you subtracted the weight of the 1kg block from the normal force, you should have used 9.8cos30 (not sin30) since the block is on an incline, not on a vertical surface.

For the third question, you have correctly used the concept of impulse to solve for the force exerted on the windshield. However, you have used the wrong units for the distance (0.024 m). It should be in meters, not centimeters, so your answer of 2.6*10^4 N is too high. Make sure to always check your units and use the correct conversion factors.

For the fourth question, you have correctly found the force of the bullet using the equation (delta E)=Fd. However, you have not taken into account the fact that the bullet stops in 0.24 m, not in 1 m. So you will need to divide your answer by 0.24 to get the correct force.

For the fifth question, you have correctly identified that the collision is elastic and that kinetic energy is conserved. However, you can use the concepts of conservation of momentum and conservation of energy to solve for the angle. Remember, the total momentum and total kinetic energy before and after the collision