Physics Question Thread 2

[Pembleton]

Hi guys,

I don't have my physics textbook nearby and I was wondering if someone could remind me how to figure out the center of mass of two objects on a plank, for instance.

For example a 10 kg box is 1 m from the edge of a 10 m plank and a 20 kg box is 2 m from the other edge of the plank. Where is the center of mass on the plank?

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[RSAgator]

When an object is moved from within the focal length ofa diverging lens to a distance beyond the focal length, the image is observed to:

a) change from upright to inverted, and move away from the lens.
B) change from inverted to upright and move closer to the lens
c) remain upright, and move away from the lens
d) remain inverted and move closer to the lens

I drew this out about five times and still am getting the wrong answer. My answer was remained upright and closer to the lens but this isn't even an answer choice.

I'm pretty poor at optics so someone please correct me if I'm wrong, but without even drawing anything you should be able to get the answer. Firstly, only within the focal length of a converging lens and a concave mirror does the image switch. With this in mind you can immediately eliminate a and b. If you know that a diverging lens creates an image that is negative, imaginary, and upright you should be able to eliminate d, leaving you with C as the answer.

 

[iA-MD2013]

I'm pretty poor at optics so someone please correct me if I'm wrong, but without even drawing anything you should be able to get the answer. Firstly, only within the focal length of a converging lens and a concave mirror does the image switch. With this in mind you can immediately eliminate a and b. If you know that a diverging lens creates an image that is negative, imaginary, and upright you should be able to eliminate d, leaving you with C as the answer.

yep, diverging lens always produces upright image, no matter where the object is placed. C

 

[Kaustikos]

I'm pretty poor at optics so someone please correct me if I'm wrong, but without even drawing anything you should be able to get the answer. Firstly, only within the focal length of a converging lens and a concave mirror does the image switch. With this in mind you can immediately eliminate a and b. If you know that a diverging lens creates an image that is negative, imaginary, and upright you should be able to eliminate d, leaving you with C as the answer.

Seriously. I just stopped trying to understand EVERYTHING about optics and it's made answering questions so much easier by just using

m = -i/0 and 1/i + 1/o = 1/f. It answers almost everything you need, that and knowing just what divering/converging and concave/convex means.

 

[olygt]

Great technical explanation. However, to "simplify" it to a more intuitive answer, the bigger the disparity in the indecies of refraction, the larger the "bend". For a TIR problem you're thinking about bending away from the normal line. So, you want to go from low index of refraction to high index of refraction because that will cause a bend away from the normal line. And if you go from a very low index of refraction to a very high index of refraction, it will bend a great deal away from the normal.

If going from a low index of refraction to a higher index of refraction wouldn't the first medium (with the lower index) be at a bigger angle to the normal than the second medium?

 

[tncekm]

If going from a low index of refraction to a higher index of refraction wouldn't the first medium (with the lower index) be at a bigger angle to the normal than the second medium?

Looks like you're getting this stuff pretty well now, you caught that error Yes, you have to go from a slower medium (high index of refraction) to a faster medium (low index of refraction) to get TIR.

Just as an FYI another way to see this is to know that the critical angle = arcsin (n2/n1) and if n2 > n1 the arcsin function is undefined (because the sine of an angle has to be less than or equal to plus or minus one). And remember, the equation above is just a rearranged version of snells law.

 

[reminisce215]

Hi, how would you go about solving this photoelectric effect problem?

What power and frequency of incident radiation must be used to strike a metal (of work function 1x10^-18 J) to produce 10,000 electrons per second?

Thanks, I'll really appreciate your help!

 

[Kaustikos]

Hi, how would you go about solving this photoelectric effect problem?

What power and frequency of incident radiation must be used to strike a metal (of work function 1x10^-18 J) to produce 10,000 electrons per second?

Thanks, I'll really appreciate your help!

Sorry. I meant to answer this question so many times before.

Anyways, the current produced is independent of the work function equation.
The frequency of incident is equal to the work function x plancks constant.
The frequency of incidence dictates whether or not you have ejection of electrons, not the current. From the frequency, we have the energy of incidence. From there you have to find the power which is J/s. I think I remember the frequency was 1.5 x 10 ^-15?

Anyways, Convert the energy from that into power and there you go.

 

[reminisce215]

The frequency of incident is equal to the work function x plancks constant...
... I think I remember the frequency was 1.5 x 10 ^-15?

Thanks! I don't know why I didn't realize to use the definition of work function: the minimum energy with which an electron is bound to the metal. (So I think you meant frequency of incident = work function divided by planck's constant?)

 

[Kaustikos]

Thanks! I don't know why I didn't realize to use the definition of work function: the minimum energy with which an electron is bound to the metal. (So I think you meant frequency of incident = work function divided by planck's constant?)

Sorry, yeah.

But you understand everything else, right? I'm just making sure that what I say comes through because I stared at that 10,000 e/s for 1 minute wondering "what the hell".

 

[reminisce215]

Sorry, yeah.

But you understand everything else, right? I'm just making sure that what I say comes through because I stared at that 10,000 e/s for 1 minute wondering "what the hell".

Yeah, all my answers matched the answers given.

 

[TomSawyer]

A 50kg woman stands on a massless board. The board rests on a frictionless frozen lake 45 m from a rock. Ten 5kg blocks are stacked on the board next to the woman.

418) The woman slides 5 blocks to the right at 6 m/s, and 2s later slides the remaining 5 blocks to the left at 6m/s. At t=2s she is:

A. 49 m from the rock and stationary
B. 57 m from the rock and stationary
C. 57 m from the rock and moving at 1m/s to the right.
D. 57 m from the rock and moving at 2m/s to the right

The answer is A. 49 m from the rock and stationary, and i undersatnd i have to use the conservation of momentum to get that her momentum after sliding the first 5 blocks is 150 kg*m/s but i cant understand how to get the distance of 2m to the left in 2s to add the the 45 m she was away from the rock in the beginning. I'd really appreciate any help, thanks!

 

[RSAgator]

A 50kg woman stands on a massless board. The board rests on a frictionless frozen lake 45 m from a rock. Ten 5kg blocks are stacked on the board next to the woman.

418) The woman slides 5 blocks to the right at 6 m/s, and 2s later slides the remaining 5 blocks to the left at 6m/s. At t=2s she is:

A. 49 m from the rock and stationary
B. 57 m from the rock and stationary
C. 57 m from the rock and moving at 1m/s to the right.
D. 57 m from the rock and moving at 2m/s to the right

The answer is A. 49 m from the rock and stationary, and i undersatnd i have to use the conservation of momentum to get that her momentum after sliding the first 5 blocks is 150 kg*m/s but i cant understand how to get the distance of 2m to the left in 2s to add the the 45 m she was away from the rock in the beginning. I'd really appreciate any help, thanks!

initially at rest, so 0 = (50+5*5)*V_woman + (5*5)(V_block)

V_block is given as 6m/s so

75*V_woman = -25*6 = - 150

V_woman = -2

after 2 seconds she has moved a total distance x = t*V_woman = -4m

She then moves the other blocks to the left, which is the opposite side as the initial movement so the direction is negative.

(75)*(-2) = (50)V_woman + (25)(-6)

-150 = 50*V_woman -150
50*V_woman = 0
V_woman = 0

This means that after 2 seconds the woman stops, so she has moved a distance of -4m total meaning she is now 49 meters from the rock. Sorry if the math is hard to read.

edit: cleaned it up a little

I'm not sure if you're given a picture in this question but it doesn't seem like there's a way of knowing which direction, the right or the left, is towards the rock aside from looking at the answer choices.

 

[TypeSH07]

I think it was from a practice test answer explanation where I wrote this down. It said the higher the frequency the greater the index of refraction. Isn't that incorrect and shouldn't it be the opposite? (if frequency refers to speed that is...)

 

[BloodySurgeon]

I think it was from a practice test answer explanation where I wrote this down. It said the higher the frequency the greater the index of refraction. Isn't that incorrect and shouldn't it be the opposite? (if frequency refers to speed that is...)

I just read your statement out of context so I could be wrong but i think you are talking about defracted light are you not? Well anyways light does not change frequency from one medium to the next. Only wavelength and speed does. n = c/v ; v=f x wavelength ; if wavelength or velocity increases, n decreases. This is why red light in a rainbow is on top.

 

[tncekm]

I think it was from a practice test answer explanation where I wrote this down. It said the higher the frequency the greater the index of refraction. Isn't that incorrect and shouldn't it be the opposite? (if frequency refers to speed that is...)

Incorrect
I know where you're coming from, that's my first inclination. I think to myself "Higher frequency = higher energy, therefore traveling through a medium it should be traveling slightly "faster" & n=c/v, so a higher frequency / higher energy, photon for example, will have a lower index of refraction. " BUT THAT'S WRONG

Correct
Higher frequency materials will go SLOWER through a medium (I don't know why), so a higher frequency material has a higher index of refraction!

 

[tncekm]

I just read your statement out of context so I could be wrong but i think you are talking about defracted light are you not? Well anyways light does not change frequency from one medium to the next. Only wavelength and speed does. n = c/v ; v=f x wavelength ; if wavelength or velocity increases, n decreases. This is why red light in a rainbow is on top.

Okay, that explains "why"

 

[Engineer2Doctor]

A softball pitcher swings a ball of mass 0.25kg around a vertical circular path of radius 0.60m before releasing it from hand. The pitcher maintains a component force on the ball of constitute magnitude of 30N in the direction of motion around the complete path. The ball's speed at the top of the cicle is 15m/s. If the ball is released at the bottom of the circle, what is its spped upon release?

This question is from the section of conservatinve and nonconservative forces.

I tried using the following way to solve it but did not come up with the correct answer of 26.5m/s.

30N + KEi + PEi = KEf + PEf
KEi =15m/s
PEi = mgh = .25*9.8*1.2
PEf = 0

I solved for the velocity by solving for KEf but like I said, I did not come up with 26.5m/s.

Any thoughts?

 

[tncekm]

circumference = 2πr = 2π(0.6m)=1.2πm. The distance covered from the top to the bottom of the circle is ½ x 1.2πm, or 0.6πm which is about equal to 2m.

F=ma therefore a=F/m = 30N/0.25kg = 120m/s²

v² = v₀² + 2ad = (15m/s)² + 2(120)2 = 225 + 480 = 705

v=√(705) = 26.5m/s

So, it sounds complicated, but it ended up being simply that a force was applied over a distance (a semicircle of about 2m) which means that the ball accelerated at a rate proportional to the force applied by the pitcher. Tricky wording... welcome to the MCAT BTW, what prep book is this from?

 

[tncekm]

Just realized you said it was covered in the conservative / non conservative / work section. So, here is a different way to solve it using the work energy theorem.

W = Fd = ΔKE = 1/2m(fv²-vi&#178

vf² = 2Fd/m + vi²
vf² = 2(30)2/0.25 + 225 = 705
vf = 26.5m/s

 

[Engineer2Doctor]

Thanks! Your 2nd explanation makes sense. The work performed in moving results in the increase in kinetic energy.

This prob was from my textbook 'Pricinciples of Physics' by Serway/Jewett.

 

[tncekm]

Thanks! Your 2nd explanation makes sense. The work performed in moving results in the increase in kinetic energy.

This prob was from my textbook 'Pricinciples of Physics' by Serway/Jewett.

Glad to help; but also make sure you can understand it the 1st way, too. You want to be able to attack a problem from as many different angles as possible!

 

[Engineer2Doctor]

The objects are connected by a light string passing over a light frictionless pulley. I did my best to depict the graphic. The 5Kg object is released from rest. Using the principle of conservation of energy, (a) determine the speed of the 3Kg object just as the 5Kg object hits the ground, and (b) find the maximum height to which hte 3Kd object rises

The answer for (a) is: 4.43m/s and (b) 5m

---------------------------
|
|
|
|
|
|
O
| |
| |
| |
| |
| M1=5kg
|
|
|
|
M2=3kg
-----------------------------------------

 

[tncekm]

What heights do they start from? We need to know this to determine speed at impact.

But, this is a classic ∑F=ma problem.

 

[Engineer2Doctor]

Oops, forgot that.

M2 is on the ground. And M1 is 4m above the ground.

 

[tncekm]

Find acceleration
============
∑F = ma; m = Ma + Mb
ma = Fa - Fb; where Fa = force felt by 5kg weight and Fb = force felt by 3kg weight.
a = [Fa-Fb]/(Ma + Mb) = (50-30)/8 = 2.5m/s²

Find Velocity at Impact
==============
v=√(2aD) = √(2(2.5)4) = 4.47m/s

This means that the 5kg weight is moving 4.47m/s toward the ground as it strikes the ground (it has traveled 4m down) and the 3kg weight is moving 4.47m/s in the opposite direction (as it hits a position 4m above the ground).

Find max height
============
v²=v₀² + 2aD --> v₀² = - 2aD
D = v₀²/2a = 20/(2*10) = 1m

So, it will move 1m above the point where it is at its maximum velocity while moving against the direction of gravity (which occurs at 4m, the distance the 5km weight traveled).

4m + 1m = 5m.

 

[Engineer2Doctor]

Thanks with all the good responses!! I've got one (at least) more question coming up on work/energy.

 

[Engineer2Doctor]

A toy cannon uses a spring to project a 0.0053Kg soft rubber ball. The spring is originally compressed by 0.05m and has a force constant of 8N/m. When it is fired, the ball moves 0.15m through the horizontal barrel of the cannon and the barrel exerts a constant friction force of 0.032N on the ball.

(a) With what speed does the projectile leave the arrel of the cannon? ANS: 1.40m/s
(b) At what point does the ball have maximum speed? ANS: 0.046m after release
(c) What is this maximum speed? ANS: 1.79m/s


I was able to solve (a).

KEf + PEf = KEi + PEi - Fd
½ MVf2 + 0 = 0 + ½ KX2 – Fd
Vf = 1.40m/s

For (b), I would think that the ball would have max speed right after leaving the spring when all of the potential energy from the spring is transferred to the ball and before the friction force can subtract from it. So, d = 0m. Accordingly, this would result in Vf = 1.94m/s for (c).

However, that’s not the case. Can someone explain?

 

[phospho]

if we put a 60 W bulb in its own circuit and a 100 W bulb in its own circuit, obviously, the 100 W bulb will be brighter.

Why is it that if we put them in series, the 60 W one will be brighter? It says here that it's because the 60 W bulb has higher resistance. I'm not getting this at all. Any help would be appreciated

 

[tncekm]

Equations
P = I²R = IV = V²/R
V=IR

Hypothetical Conditions
First, lets define some conditions. Lets say we've got a 10V power source.

Determining 60W and 100W resistances by putting them in their own circuits
For the 60W bulb we've got a current of 60W/10V = 6A. This means we've got a resistance of 10V/6A = 5/3 Ohms.

For the 100W bulb we've got a current of 100W/10V = 10A. This means we've got a resistance of 10V/10A = 1 Ohm

Put them in series
Put them in series and your total resistance becomes 5/3 + 3/3 Ohms = 8/3 Ohms. The new current for the series is I = 10V/(8/3 Ohms)= 15/4 Amps

Result
Since P=I²R you can see that with a fixed current (resistors in series have the same current) the one with the greater resistance will put out more power, that is, more light.

Expansion on the Result
Okay, the reason that the 100W bulb puts out more light by itself is b/c it draws more current than the 60W bulb when the 60W bulb is by itself. And, remember, power is proportional to the square of current! If you put them in series you've got both bulbs pulling current, so both experience the same current, and therefore the one with greater resistance is going to put out more power.

 

[Kaustikos]

if we put a 60 W bulb in its own circuit and a 100 W bulb in its own circuit, obviously, the 100 W bulb will be brighter.

Why is it that if we put them in series, the 60 W one will be brighter? It says here that it's because the 60 W bulb has higher resistance. I'm not getting this at all. Any help would be appreciated

From what I remember, you use P = V^2/R. Setting the V the same for both allows us to determine the Resistance. Putting it at,say, 120 makes R = 120^2/P. The higher the power, the lower the resistance. The inverse is true for the 60 watt bulb.
The 60 watt bulb, thus, has more resistance and will incurr a larger power loss. This all boils down to the resistance and how the current is the same for both. If you run a current through a high-resistance wire, it will lose some of its power in heat. Since the wires share the same current, the brightness is correlated to the resistance, not the current.

If the bulbs were wired in parallel, the opposite would occur and cause the 100 W to glow brighter since they share the same Voltage drop, not the same current. V = IR is used in this circumstance to help explain it. The Voltage is the same and, knowing the resistance differences, the lower resistance bulb will receive more current and thus glow brighter than the higher resistance bulb. The brightness, in this case, is correlated with the current, not the resistance (so to speak).


I KNEW IT - beat me to it!

Good explanation, too.

 

[phospho]

Equations
P = I²R = IV = V²/R
V=IR

Hypothetical Conditions
First, lets define some conditions. Lets say we've got a 10V power source.

Determining 60W and 100W resistances by putting them in their own circuits
For the 60W bulb we've got a current of 60W/10V = 6A. This means we've got a resistance of 10V/6A = 5/3 Ohms.

For the 100W bulb we've got a current of 100W/10V = 10A. This means we've got a resistance of 10V/10A = 1 Ohm

Put them in series
Put them in series and your total resistance becomes 5/3 + 3/3 Ohms = 8/3 Ohms. The new current for the series is I = 10V/(8/3 Ohms)= 15/4 Amps

Result
Since P=I²R you can see that with a fixed current (resistors in series have the same current) the one with the greater resistance will put out more power, that is, more light.

Expansion on the Result
Okay, the reason that the 100W bulb puts out more light by itself is b/c it draws more current than the 60W bulb when the 60W bulb is by itself. And, remember, power is proportional to the square of current! If you put them in series you've got both bulbs pulling current, so both experience the same current, and therefore the one with greater resistance is going to put out more power.

From what I remember, you use P = V^2/R. Setting the V the same for both allows us to determine the Resistance. Putting it at,say, 120 makes R = 120^2/P. The higher the power, the lower the resistance. The inverse is true for the 60 watt bulb.
The 60 watt bulb, thus, has more resistance and will incurr a larger power loss. This all boils down to the resistance and how the current is the same for both. If you run a current through a high-resistance wire, it will lose some of its power in heat. Since the wires share the same current, the brightness is correlated to the resistance, not the current.

If the bulbs were wired in parallel, the opposite would occur and cause the 100 W to glow brighter since they share the same Voltage drop, not the same current. V = IR is used in this circumstance to help explain it. The Voltage is the same and, knowing the resistance differences, the lower resistance bulb will receive more current and thus glow brighter than the higher resistance bulb. The brightness, in this case, is correlated with the current, not the resistance (so to speak).


I KNEW IT - beat me to it!

Good explanation, too.

thank you both very much. I completely forgot about that p=i^2r. It makes much more sense now.

 

[nikeshp]

A toy cannon uses a spring to project a 0.0053Kg soft rubber ball. The spring is originally compressed by 0.05m and has a force constant of 8N/m. When it is fired, the ball moves 0.15m through the horizontal barrel of the cannon and the barrel exerts a constant friction force of 0.032N on the ball.

(a) With what speed does the projectile leave the arrel of the cannon? ANS: 1.40m/s
(b) At what point does the ball have maximum speed? ANS: 0.046m after release
(c) What is this maximum speed? ANS: 1.79m/s


I was able to solve (a).

KEf + PEf = KEi + PEi - Fd
½ MVf2 + 0 = 0 + ½ KX2 – Fd
Vf = 1.40m/s

For (b), I would think that the ball would have max speed right after leaving the spring when all of the potential energy from the spring is transferred to the ball and before the friction force can subtract from it. So, d = 0m. Accordingly, this would result in Vf = 1.94m/s for (c).

However, that’s not the case. Can someone explain?

im going to bump this for you, tried to figure it out but didnt come to me

 

[tncekm]

A toy cannon uses a spring to project a 0.0053Kg soft rubber ball. The spring is originally compressed by 0.05m and has a force constant of 8N/m. When it is fired, the ball moves 0.15m through the horizontal barrel of the cannon and the barrel exerts a constant friction force of 0.032N on the ball.

(a) With what speed does the projectile leave the arrel of the cannon? ANS: 1.40m/s
(b) At what point does the ball have maximum speed? ANS: 0.046m after release
(c) What is this maximum speed? ANS: 1.79m/s


I was able to solve (a).

KEf + PEf = KEi + PEi - Fd
½ MVf2 + 0 = 0 + ½ KX2 – Fd
Vf = 1.40m/s

For (b), I would think that the ball would have max speed right after leaving the spring when all of the potential energy from the spring is transferred to the ball and before the friction force can subtract from it. So, d = 0m. Accordingly, this would result in Vf = 1.94m/s for (c).

However, that’s not the case. Can someone explain?

I think its an oversight / typo made by the author. You're definitely correct.

When the spring is compressed maximally (0.05m) all energy is stored as potential energy. When the spring has been released from compression and is at is equilibrium length it is no longer exerting a force on the ball and all of the PE has been converted to KE and KEmax is where Vmax occurs.

 

[da8s0859q]

Y'know, it's funny, we bust our asses to handle questions like that, but then some of us never even see it on the MCAT.

I know I didn't. Hell, I never even used one kinematics equation.

Okay, you probably didn't want to hear that. Movin' right along...

 

[Kaustikos]

Y'know, it's funny, we bust our asses to handle questions like that, but then some of us never even see it on the MCAT.

I know I didn't. Hell, I never even used one kinematics equation.

Okay, you probably didn't want to hear that. Movin' right along...

That's the general, unfortunate, consensus with myself so far. This past practice mcat was extremely disappointing. I was hoping for a lot more than what I got.

 

[da8s0859q]

That's the general, unfortunate, consensus with myself so far. This past practice mcat was extremely disappointing. I was hoping for a lot more than what I got.

That's been exactly the case with me on the real one(s), too. It sucks because I was really hoping for some optics goodness, or some kinematics, or some of what's been in the past few pages of this thread, but nooo. Of course the AAMC wouldn't give me that. I could've actually answered it.

 

[phospho]

Equations
P = I²R = IV = V²/R
V=IR

Hypothetical Conditions
First, lets define some conditions. Lets say we've got a 10V power source.

Determining 60W and 100W resistances by putting them in their own circuits
For the 60W bulb we've got a current of 60W/10V = 6A. This means we've got a resistance of 10V/6A = 5/3 Ohms.

For the 100W bulb we've got a current of 100W/10V = 10A. This means we've got a resistance of 10V/10A = 1 Ohm

Put them in series
Put them in series and your total resistance becomes 5/3 + 3/3 Ohms = 8/3 Ohms. The new current for the series is I = 10V/(8/3 Ohms)= 15/4 Amps

Result
Since P=I²R you can see that with a fixed current (resistors in series have the same current) the one with the greater resistance will put out more power, that is, more light.

Expansion on the Result
Okay, the reason that the 100W bulb puts out more light by itself is b/c it draws more current than the 60W bulb when the 60W bulb is by itself. And, remember, power is proportional to the square of current! If you put them in series you've got both bulbs pulling current, so both experience the same current, and therefore the one with greater resistance is going to put out more power.

Thank you so much for taking the time to explain this. The concept helped me a LOT on the real thing yesterday

 

[tncekm]

Glad to hear it! I hope the test went well for you

 

[secret25]

So I'm absolutely terrible at Magnetic Fields and my test is in two weeks so I'm hoping someone can help me just figure it out... my question is pretty basic I think...

In Kaplan, in one of their "Notes" they say "To determine sense of rotation (clockwise or counterclockwise) of a charged particle undergoing uniform circular motion, apply right hand rule for magnetic force on the particle."

So if all you know is direction of the magnetic field, how can you determine clockwise/counterclockwise direction of charge and the direction of magnetic force? Or, were they trying to say if you know one, you can find the other?

Also Kaplan also states "A charged particle moving perpendicularly to a constant, uniform magnetic field, undergoes uniform circular motion." Does that mean it ALWAYS undergoes uniform circular motion? In the example before that comment, it showed a charged particle moving directly upward through a uniform magnetic field, so I'm just completely lost now.

Hope that made sense and thanks for any help!

 

[tncekm]

Okay, if you know the direction of the magnetic field and if you know the direction of the charged particles travel you can predict how it will "turn".

Let say we've got a positive particle traveling at some speed 'v' entering perpindicular to a magnetic field that is going to into the page. Using right hand rule #2 (which is essentially an easy way to figure out F=q(v x B)sinθ you know that the force is going to be "mutually perpindicular" to the direction of 'v' and B at any given "instant". This means, the force will always be applied at a 90 degree angle to the direction of travel.

In the outlined case the force would be applied at a 90 degree angle to both its direction of travel and the magnetic field, so it would go counter clockwise. If it were an electron we were using, then it would go clockwise.

http://physicsed.buffalostate.edu/SeatExpts/resource/rhr/rhr.htm

Hope that helps, but its really hard to explain without "show and tell".

 

[Engineer2Doctor]

The chapter on linear momentum took some time for me to review. It wasn't the easiest for me to grasp so please bear with me for the next few days as I pose questions on this topic.

First question:

Two blocks with masses M and 3M are placed on a horizontal fricitonless surface. Al lifht spring is attached to one fo them, and hte blicks are pushed together with the spring between them. A cord initially holding the blocks together is burned; after this, the block of mass 3M moves to the right with a speed of 2m/s.

(a) What is the speed of the block of mass M?
This was easy. m1v1 = m2v2. v1 = 6m/s

(b) Find the original elastic potential energy in the spring, taking M = 0.35kg.

I thought about using conservation of energy here. PEi + KEi = PEf +KEf
with KEi and PEf being 0. But that did not give me the correct answer of 8.4J.

 

[gogopogo]

Two blocks with masses M and 3M are placed on a horizontal fricitonless surface. Al lifht spring is attached to one fo them, and hte blicks are pushed together with the spring between them. A cord initially holding the blocks together is burned; after this, the block of mass 3M moves to the right with a speed of 2m/s.

(a) What is the speed of the block of mass M?
This was easy. m1v1 = m2v2. v1 = 6m/s

(b) Find the original elastic potential energy in the spring, taking M = 0.35kg.

I thought about using conservation of energy here. PEi + KEi = PEf +KEf
with KEi and PEf being 0. But that did not give me the correct answer of 8.4J.

Here goes...

Each block moves away with kinetic energy 0.5m1v^2 and 0.53m2v^2, respectively, and that energy had to come from somewhere, hence a potential energy must have been stored up, with value P

P = 0.5(0.35kg)6m/s^2 + 0.5(1.05kg)2m/s^2

Roughly solving in my head gave about 7.7ish, which would probably be close enough...

Hope this makes sense