HELP WITH AP PHYSICS- 5 min drill Multiple choice easy?

[girlphysics]

I need help with all of these. I am so lost on all of them and have been working for hours. Anything is so appreciated.


_____1. A rock is dropped off a cliff and falls the first half of the distance to the ground in t1 seconds. It falls the second half of the distance in t2 seconds. What is the value of t1/t2?
A) 1/2√2 B) 1/√2 C) ½
D) 1-(1/√2) E) √2-1

______2. A box of mass m slides on a horizontal surface with initial speed v0. It feels no forces other than it weight and the force from the surface. If the coefficient of kinetic friction between the box and surface is μ, how far does the box slide before coming to rest?
A) v02/2μg B) v02/μg C) 2v02/μg
D) mv02/μg E) 2 mv02/μg

_____3. An object initially at rest experiences a time-varying acceleration given by a = ( 2 m/s3 t) for t ≥ 0. How far does the oject travel in the first 3 seconds?
A) 9 m B) 12 m C) 18 m D) 24 m E) 27 m

______4. An engine provides 10kW of power to lift a heavy load at constant velocity a distance of 20 meters in 5 seconds. What is the mass of the load?
A) 100 kg B) 150 kg C) 200 kg D) 250 kg E)300 kg

______5. A ball (mass= .08 kg) is dropped onto the floor from a height of 2 m, and after bouncing off the floor rises almost to its original height. If the contact time with the floor was .04 s, what average force did the floor exert on the ball?
A) 0.16 N B) 16 N C) 25 N D) 36 N E) 64 N

 

[girlphysics]

For #4, I said Work/Time=Power so Fd/t=P so Ma*d/t=P because W=fd=ma*d so then I plugged in all the numbers and got .25 which leads me to believe its 250. ? I don't know I picked D

 

[cepheid]

Welcome to PF girlphysics!

I need help with all of these. I am so lost on all of them and have been working for hours. Anything is so appreciated. _____1. A rock is dropped off a cliff and falls the first half of the distance to the ground in t1 seconds. It falls the second half of the distance in t2 seconds. What is the value of t1/t2?
A) 1/2√2 B) 1/√2 C) ½
D) 1-(1/√2) E) √2-1

For this first problem, you can just use the relevant kinematics equation. What you want is the equation that tells you how distance varies with time for motion with constant acceleration. Start with that. Do you see any equations from your notes or your book that give distance vs. time for the case of a constant acceleration?

 

[girlphysics]

v0t+1/2at^2 so because initial is 0, just 1/2at^2??

 

[genericusrnme]

1)
How do you suppose you would find out what t1 is?
2)
Since I see you know what work is, how much work must be done on the box to make it come to rest?
3)
Do you know any calculus?
4)
You may be missing a factor of 1000 because the question uses kW = 1000 W
5)
What is the definition of force?

 

[cepheid]

For #4, I said Work/Time=Power so Fd/t=P so Ma*d/t=P because W=fd=ma*d so then I plugged in all the numbers and got .25 which leads me to believe its 250. ? I don't know I picked D

Your understanding of the physics seems fine. power = work/time = force*distance/time. Hence: P = mg*d/t

Solve for m.

I suspect that you simply made a mistake with the units. Be careful: that's 10 kilowatts, not 10 watts.

 

[girlphysics]

Let me just say, I have a terrible teacher and probably shouldn't be in AP physics, its halfway through the year and I know nothing. THANK YOU for your help. I also have not taken calculus and I have to turn this in tomorrow.
1.) t=√1/2ax ? I am not sure. I feel really stupid

 

[cepheid]

v0t+1/2at^2 so because initial is 0, just 1/2at^2??

Exactly. Now what is "a" in this situation? Hint: what is the only force that is present here?

So, you can solve for t1 using this equation. Call the initial height above the ground "h".

Then:

h/2 = 1/2*(a)*t₁²

Solving for t2 is trickier. You consider the halfway point to be the start of a new problem where you apply the same equation again. Only now there IS an initial velocity: it's equat to the velocity the object has reached at the half-way point. To compute this, you need another kinematics equation that gives you velocity vs. time.

 

[girlphysics]

OK So I got 4! yay. 1 down 4 to go.

 

[girlphysics]

a is 10m/s/s gravity...
so h= 5t^2 so t=√h/5

 

[girlphysics]

I don't know what equation..?

 

[girlphysics]

1)
How do you suppose you would find out what t1 is?
2)
Since I see you know what work is, how much work must be done on the box to make it come to rest?
3)
Do you know any calculus?
4)
You may be missing a factor of 1000 because the question uses kW = 1000 W
5)
What is the definition of force?

2.) I don't know it must only be effected by the normal force and coefficient of friction, and F=μFn. I don't know what to do.
5.) F=ma. I worked on this for like an hour. Dont I need the change in velocity? to find the impulse?

 

[girlphysics]

I just need the letter answers if its easier to work backwards...

 

[cepheid]

a is 10m/s/s gravity...
so h= 5t^2 so t=√h/5

Yeah exactly. The answer I was going for was that a = g, where g is defined as 9.81 m/s/s (or 10 m/s/s if you're not too concerned about accuracy).

Your work is pretty close. Be a little more careful with your algebra. The equation was

h/2 = (1/2)gt₁²

multiply both sides by 2:

h = gt₁²

t₁ = √(h/g)

I don't know what equation..?

You don't know how velocity varies with time during constant acceleration? Hint: what is the definition of acceleration?

The question you're trying to answer is: how fast is it going when it reaches the halfway point?

 

[SammyS]

I need help with all of these. I am so lost on all of them and have been working for hours. Anything is so appreciated.


_____1. A rock is dropped off a cliff and falls the first half of the distance to the ground in t1 seconds. It falls the second half of the distance in t2 seconds. What is the value of t1/t2?
A) 1/2√2 B) 1/√2 C) ½
D) 1-(1/√2) E) √2-1

The wording of Question #1. must be in error. The time, t₁ to go the first half of the distance must be greater than time, t₂, the time to go the second half of the distance.

Therefore t₁/t₂ > 1 . But ALL of the answers are less than 1.

Perhaps if the time t₂ represented the time to go the total distance, then one of the listed answers might be correct.

 

[girlphysics]

oh that makes sense!

well acceleration is the rate at which velocity changes...

v^2=v0+at

 

[girlphysics]

but there's also v^2=v0^2+a(x-x0)

 

[girlphysics]

The wording of Question #1. must be in error. The time, t₁ to go the first half of the distance must be greater than time, t₂, the time to go the second half of the distance.

Therefore t₁/t₂ > 1 . But ALL of the answers are less than 1.

Perhaps if the time t₂ represented the time to go the total distance, then one of the listed answers might be correct.

THATS exactly what I thought at first, but I know this question is right...

 

[cepheid]

oh that makes sense!

well acceleration is the rate at which velocity changes...

v^2=v0+at

It's actually v = v0 + at. (Remember that equations have to be dimensionally consistent: if the right-hand side has dimensions of velocity, the left-hand side must as well, otherwise it is nonsense).

Can you see how to use this equation (velocity vs. time) to figure out how fast the object was going at time t = t1 when it reaches the half-way point?

THATS exactly what I thought at first, but I know this question is right...

How do you know? It could easily have a typo...

 

[girlphysics]

It's actually v = v0 + at. (Remember that equations have to be dimensionally consistent: if the right-hand side has dimensions of velocity, the left-hand side must as well, otherwise it is nonsense).

Can you see how to use this equation (velocity vs. time) to figure out how fast the object was going at time t = t1 when it reaches the half-way point?



How do you know? It could easily have a typo...

Well plug in t1 for t?
I don't know for sure, but I asked two friends who have the same problem if they were able to find the right answer and they both said yes, and I am pretty sure my teacher has been using the same problems for like his 30 years in teaching- not saying that everyone could have missed the typo.

 

[girlphysics]

how fast is it going when it reaches the halfway point? -so If I plug in t1 and use gravity, but it does have an intial velocity now right? how do I get that?

 

[girlphysics]

thank you so much cepheid for helping me- I am a dumb blonde when it comes to physics

 

[cepheid]

how fast is it going when it reaches the halfway point? -so If I plug in t1 and use gravity, but it does have an intial velocity now right? how do I get that?

You seem to be confusing the two stages of the problem. In the first stage of the problem, we consider it accelerating from rest at t = 0 to t = t1. We know that it moves a distance h/2 in this time interval, therefore we are able to solve for how long the time interval is by using the equation for distance vs. time. That's what we've done so far.

Now we want to consider the second stage, which is the stage of the motion between t = t1 and t = t1 + t2 [I EDITED THIS]. We know it moves down by another h/2 distance in this stage. So in principle, we can use the distance vs. time equation AGAIN to solve for the time interval t2 [I EDITED THIS]. The problem is, that during THIS stage, we don't know what the INITIAL velocity is (i.e. the velocity at t1).

To solve for the velocity at t1, we can use the equation for velocity vs. time: v = v0 + at, and consider it accelerating from rest at t = 0 to some final speed at t = t1. Using this equation, can you solve for how fast it is going at t = t1? Hint: what is v in this case? What is v0 in this case? What is the time interval t in this case?

Once you know the velocity at t1, you use that in the distance vs. time equation for the second stage.

 

[cepheid]

On second thought, this will be algebraically tedious. A much simpler way is to compute the TOTAL time of the motion t_total, which is the time required to move the total distance h. Then you can compute t2 using t2 = t_Total - t1.

Sorry.

 

[cepheid]

Perhaps if the time t₂ represented the time to go the total distance, then one of the listed answers might be correct.

If you assume that they want the reciprocal of what she wrote: i.e. t2/t1, then things work out in the sense that you get one of the given answers.

Well plug in t1 for t?
I don't know for sure, but I asked two friends who have the same problem if they were able to find the right answer and they both said yes, and I am pretty sure my teacher has been using the same problems for like his 30 years in teaching- not saying that everyone could have missed the typo.

See above.

 

[SammyS]

If you assume that they want the reciprocal of what she wrote: i.e. t2/t1, then things work out in the sense that you get one of the given answers.

Right.

That makes good sense.

 

[girlphysics]

so how do I solve for the total time?

 

[girlphysics]

If you assume that they want the reciprocal of what she wrote: i.e. t2/t1, then things work out in the sense that you get one of the given answers.



See above.

So do you mean that the given answer will be total time/t1? I am so confused

 

[girlphysics]

or do you mean it will turn out to just be t2/t1? Because I didnotice immediatley all the answers are less than 1 and they shouldn't be.

 

[cepheid]

so how do I solve for the total time?

EXACTLY the same way you solved for t1, with the difference that the distance traveled is the full height h, now, instead of h/2. Think about it: it's an exactly equivalent situation in both cases. You have a ball that starts at rest from some height, and then it gets dropped, and you want to know how long it will take to fall a certain distance.

or do you mean it will turn out to just be t2/t1? Because I didnotice immediatley all the answers are less than 1 and they shouldn't be.

We think that the problem is asking for t2/t1 rather than t1/t2 (which is what you wrote), simply because the former makes more sense given the answer choices (which are all less than 1).

 

[girlphysics]

EXACTLY the same way you solved for t1, with the difference that the distance traveled is the full height h, now, instead of h/2. Think about it: it's an exactly equivalent situation in both cases. You have a ball that starts at rest from some height, and then it gets dropped, and you want to know how long it will take to fall a certain distance.



We think that the problem is asking for t2/t1 rather than t1/t2 (which is what you wrote), simply because the former makes more sense given the answer choices (which are all less than 1).

thank you! Now can you help me with number 2? I am not sure how to start it.

 

[cepheid]

thank you! Now can you help me with number 2? I am not sure how to start it.

To know how far the block will go, you need to know its acceleration. Then you can just use the kinematics equations like before.

To find the acceleration, you can use Newton's Second Law, which says that

a = F/m

where a is the acceleration of the body, F is the NET force acting on it, and m is the mass of the body.

The NET force is the sum of all forces acting on the body. You can do this summing separately for the horizontal and vertical directions (i.e. you can apply Newton's 2nd law separately for each direction).

The key to finding the net force is to draw a free body diagram. It's called that because it includes only the object in question and the forces acting on it (nothing from the surroundings). A free body diagram for the block is therefore an inventory of all the forces that are acting on it.

Newton's 2nd law for the y (vertical) direction will give you the normal force. Hint: what is the acceleration in the vertical direction?

Once you know the normal force, you know the frictional force. And since the friction force is the only force that acts in the horizontal direction, that completes the force inventory for that direction. So now you know the net horizontal force and hence the horizontal acceleration.

 

[girlphysics]

To know how far the block will go, you need to know its acceleration. Then you can just use the kinematics equations like before.

To find the acceleration, you can use Newton's Second Law, which says that

a = F/m

where a is the acceleration of the body, F is the NET force acting on it, and m is the mass of the body.

The NET force is the sum of all forces acting on the body. You can do this summing separately for the horizontal and vertical directions (i.e. you can apply Newton's 2nd law separately for each direction).

The key to finding the net force is to draw a free body diagram. It's called that because it includes only the object in question and the forces acting on it (nothing from the surroundings). A free body diagram for the block is therefore an inventory of all the forces that are acting on it.

Newton's 2nd law for the y (vertical) direction will give you the normal force. Hint: what is the acceleration in the vertical direction?

Once you know the normal force, you know the frictional force. And since the friction force is the only force that acts in the horizontal direction, that completes the force inventory for that direction. So now you know the net horizontal force and hence the horizontal acceleration.

ahh thank you! My last query is about the ball dropping from 2m and bouncing to about the same height. The force of the floor on the ball should just be mg right? I think I have to find the impulse, the change in momentum or whatever, but I think I need the chage in velocity. I am not sure what to do.

 

[cepheid]

ahh thank you! My last query is about the ball dropping from 2m and bouncing to about the same height. The force of the floor on the ball should just be mg right? I think I have to find the impulse, the change in momentum or whatever, but I think I need the chage in velocity. I am not sure what to do.

The impulse-momentum theorem says that:

impulse = change in momentum

The impulse is equal to the average force on the object multiplied by the time interval over which the force acts. So:

FΔt = Δp = mΔv

where p is momentum, and assuming the mass doesn't change, the change in momentum is just equal to the mass times the change in velocity.

So you're right. In order to solve for F, you DO need to know the change in velocity Δv. To get that, you need to use another physics principle (a very important one) which is the conversation of energy. Think about it: the ball starts out a height h with gravitational potential energy mgh. The ball falls to height h = 0, which means its potential energy goes to 0. The conservation of energy says that all of that initial potential energy is converted into kinetic energy. That means you know the kinetic energy at the moment of impact, which means you know the speed at the moment of impact.

On the way back up, the reverse is true. Whatever initial kinetic energy the ball has just after the impact is entirely converted back into gravitational potential energy as the ball reaches its maximum height. Now, the fact that the ball reaches almost the same height as before tells you that the ball has almost the same amount of kinetic energy after the impact as it did just before the impact. In other words, the collision is almost perfectly elastic (no kinetic energy is lost). If you assume a perfectly elastic collision, then the kinetic energy before is the same as the kinetic energy after. This means that the speed "v" before is the same as the speed "v" after.

Now, remember that momentum is a vector. So it's not enough to consider speed. We have to consider velocity. Before the collision, the velocity was -v (downward). After the collision, it had changed to +v (upward). They have the same magnitude (speed) but opposite directions. Therefore Δv = v_after - v_before = v - (-v) = 2v.

You need to use the conservation of energy to compute what "v" actually is. I can't help you any further today because I have to go to sleep.

EDIT: By the way, looking at #3, I'm pretty sure you DO need calculus to solve it.

 

[girlphysics]

The impulse-momentum theorem says that:

impulse = change in momentum

The impulse is equal to the average force on the object multiplied by the time interval over which the force acts. So:

FΔt = Δp = mΔv

where p is momentum, and assuming the mass doesn't change, the change in momentum is just equal to the mass times the change in velocity.

So you're right. In order to solve for F, you DO need to know the change in velocity Δv. To get that, you need to use another physics principle (a very important one) which is the conversation of energy. Think about it: the ball starts out a height h with gravitational potential energy mgh. The ball falls to height h = 0, which means its potential energy goes to 0. The conservation of energy says that all of that initial potential energy is converted into kinetic energy. That means you know the kinetic energy at the moment of impact, which means you know the speed at the moment of impact.

On the way back up, the reverse is true. Whatever initial kinetic energy the ball has just after the impact is entirely converted back into gravitational potential energy as the ball reaches its maximum height. Now, the fact that the ball reaches almost the same height as before tells you that the ball has almost the same amount of kinetic energy after the impact as it did just before the impact. In other words, the collision is almost perfectly elastic (no kinetic energy is lost). If you assume a perfectly elastic collision, then the kinetic energy before is the same as the kinetic energy after. This means that the speed "v" before is the same as the speed "v" after.

Now, remember that momentum is a vector. So it's not enough to consider speed. We have to consider velocity. Before the collision, the velocity was -v (downward). After the collision, it had changed to +v (upward). They have the same magnitude (speed) but opposite directions. Therefore Δv = v_after - v_before = v - (-v) = 2v.

You need to use the conservation of energy to compute what "v" actually is. I can't help you any further today because I have to go to sleep.

EDIT: By the way, looking at #3, I'm pretty sure you DO need calculus to solve it.

Thank you for your help. I handed this in yesterday, Are you allowed to tell me what you would have chosen as the correct answers? I am still trying to fully understand these questions, and I ended up guessing. 2.A 3.B 4.D 5.E