EK 1001 physics Q

[engineeredout]

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430: "Two 1 kg lumps of clay sit in some mud. A boy picks up one lump of clay and rops it onto the second lump from a height of 1 m. The lumps stick together when they collide. If the second lump is driven into the mud a distance of 1cm, what is the average force exerted on the mud?"

Answer: B: 520N.

My answer: A: 500N


Okay so the stupid little **** drops the 1kg thing of clay, so vf^2 = 2*10*1 = 20, or vf ~ 4.5

Then, conserving momentum, m1v1 = (m1+m2)v2 == 4.5*1 = 2*v2, v2 = 2.25.

Then since it goes .01m into the mud, vi^2 = 2*a*.01 = 2.25^2 ~ 5, so solving, a ~ 250.

F=ma, so 2*250 = 500.

What am I doing wrong?

Thanks.

 

[minhaj]

KE = PE

.5mv^2 = mgh
.5(1)v^2 = 1*10*1
v = sqrt(20)

Now using conservation of momentum, m1v1 = m2v2
=> 1*sqrt(20) = 2*v2
=> v2 = sqrt5
since the lumps of clay end up 0.01m below the ground it had PE at the moment of impact.
PE at impact = 2*10*.01 = 0.2J
work done = PE + KE
=> W = .2 + 5 = 5.2J
since, W = Fd
5.2J = F(0.01)
F = 520N

hope that helps!

 

[engineeredout]

Okay I get your method, but I'm still not seeing why mine doesn't work.

 

[minhaj]

Okay I get your method, but I'm still not seeing why mine doesn't work.

The reason your method did not work was probably because you didnt take into account the acceleration due to gravity. If you do, then:

Fnet = m ( a + g )
Fnet = 2 (250 + 10)
Fnet = 520 N

 

[engineeredout]

The reason your method did not work was probably because you didnt take into account the acceleration due to gravity. If you do, then:

Fnet = m ( a + g )
Fnet = 2 (250 + 10)
Fnet = 520 N

😡😡 GAH WHO REMEMBERS GRAVITY WHEN THE OBJECT HAS ALREADY STRUCK THE GROUND.

Thanks

 

[physics junkie]

😡😡 GAH WHO REMEMBERS GRAVITY WHEN THE OBJECT HAS ALREADY STRUCK THE GROUND.

Thanks

Always draw a free body diagram.

 

[sodamnlucky]

My advice to you is that when you get hung up on a question like that, don't fret it. Chances are, that sort of calculation won't really show up...EK has TONS of questions like that...

 

[minhaj]

My advice to you is that when you get hung up on a question like that, don't fret it. Chances are, that sort of calculation won't really show up...EK has TONS of questions like that...

Well, if you understand how to work the problem the calculations aren't that complex. I dont know, I would consider this particular type of question fair game for the mcat. Thats just me though.

 

[engineeredout]

My advice to you is that when you get hung up on a question like that, don't fret it. Chances are, that sort of calculation won't really show up...EK has TONS of questions like that...

Aside from remembering gravity when the clay is being plowed into the ground, the calculations are really not THAT complex. Plus it was just pissing me off that I couldn't figure out why I wasn't getting it.

It could possibly show up as an independant Q.

 

[BerkReviewTeach]

KE = PE

.5mv^2 = mgh
.5(1)v^2 = 1*10*1
v = sqrt(20)

Now using conservation of momentum, m1v1 = m2v2
=> 1*sqrt(20) = 2*v2
=> v2 = sqrt5
since the lumps of clay end up 0.01m below the ground it had PE at the moment of impact.
PE at impact = 2*10*.01 = 0.2J
work done = PE + KE
=> W = .2 + 5 = 5.2J
since, W = Fd
5.2J = F(0.01)
F = 520N

hope that helps!

Excellent math and MCAT problem solving, but for what it's worth, you cannot use conservation of energy for an inelastic collision. The clay balls deform, so energy was not conserved during the overall process. Getting v at impact is valid your way (just like it was for Engineeredout). Using conservation of momentum is also not necessarily valid (because we don't know for sure if there is an absence of a net external force: Fgrav does not necessarily equal Ffriction).

If it helps with your calm, you really can't answer this question without making some invalid assumptions.

But this is a great question for sending a message to test takers. It appears that the test writer wants us to use conservation of energy, so in an MCAT persepctive it seems like we should do it. What this question tells us is that sometimes we need to think more about what the test writer wants us to answer more so than what the correct or best answer is.

 

[Heyeon]

Why is the Fnet not mg-ma?

 

[engineeredout]

Why is the Fnet not mg-ma?

Because the original problem is referring to collisions? I'm not sure what you're referencing exactly.

 

[minhaj]

Why is the Fnet not mg-ma?

because both g and a are in the same direction.

 

[TheBoondocks]

The reason your method did not work was probably because you didnt take into account the acceleration due to gravity. If you do, then:

Fnet = m ( a + g )
Fnet = 2 (250 + 10)
Fnet = 520 N

I have a question on this. I think it's a poor question in general. When you calculate v2, the logic and physics makes sense. However, when you calculate a, that is the overal acceleration of the net force. So why then, can you add 10 to a. This doesn't make physical sense to me. I'm aware of doing this with tension.

Ex. An 800 kg elevator is falling at 10m/s and then is slowed down at a constant rate in 25 meters. Find the tension. We can calculate the a by using the equation you guys used, vof^2 -vo^2=2ax. X here is 25 and vof is 0. When you solve for a it is equal to 2m/s. Now, the Fnet=m*a(just calculated). Fnet= T-mg. ma +mg=T. m(a+g)=T=9440N. This where I have seen this. However, never have I seen a calculated from the aforementioned equation and then added g to it. Please tell me what is wrong in my analysis. I see that it asked for AVERAGE force. Which leads me to think they wanted W=FDcos(theta) as you solved it. This makes sense. However, I don't follow the logic of the other method.

I know that (m1+m2)g +force(from clay 1)=ma. So how is it that once you calculate a to be 250, can you say the actual acceleration is (a+g).

 

[TheBoondocks]

I have a question on this. I think it's a poor question in general. When you calculate v2, the logic and physics makes sense. However, when you calculate a, that is the overal acceleration of the net force. So why then, can you add 10 to a. This doesn't make physical sense to me. I'm aware of doing this with tension.

Ex. An 800 kg elevator is falling at 10m/s and then is slowed down at a constant rate in 25 meters. Find the tension. We can calculate the a by using the equation you guys used, vof^2 -vo^2=2ax. X here is 25 and vof is 0. When you solve for a it is equal to 2m/s. Now, the Fnet=m*a(just calculated). Fnet= T-mg. ma +mg=T. m(a+g)=T=9440N. This where I have seen this. However, never have I seen a calculated from the aforementioned equation and then added g to it. Please tell me what is wrong in my analysis. I see that it asked for AVERAGE force. Which leads me to think they wanted W=FDcos(theta) as you solved it. This makes sense. However, I don't follow the logic of the other method.

I know that (m1+m2)g +force(from clay 1)=ma. So how is it that once you calculate a to be 250, can you say the actual acceleration is (a+g).

bump

 

[spyderracing32]

Excellent math and MCAT problem solving, but for what it's worth, you cannot use conservation of energy for an inelastic collision. The clay balls deform, so energy was not conserved during the overall process. Getting v at impact is valid your way (just like it was for Engineeredout). Using conservation of momentum is also not necessarily valid (because we don't know for sure if there is an absence of a net external force: Fgrav does not necessarily equal Ffriction).

Well generally these types of questions designate whether or not friction is negligible (which it normally is), and there is no reason that you can't use conservation of energy to determine the impact speed if you designate the point of impact to be where PE=0 because energy is conserved in the falling (obviously you can't do (1/2)(m1)(vi)^2 = (1/2)(m1+m2)(vf)^2). Even if there is friction present you simply use PE + Wext = KE where the external work (Wext) done is attributed to the force of friction.

 

[TheBoondocks]

Well generally these types of questions designate whether or not friction is negligible (which it normally is), and there is no reason that you can't use conservation of energy to determine the impact speed if you designate the point of impact to be where PE=0 because energy is conserved in the falling (obviously you can't do (1/2)(m1)(vi)^2 = (1/2)(m1+m2)(vf)^2). Even if there is friction present you simply use PE + Wext = KE where the external work (Wext) done is attributed to the force of friction.

Yea, i was right. You're supposed to solve via W=Fd. The average force indicates this. Well done guys.

 

[BerkReviewTeach]

Well generally these types of questions designate whether or not friction is negligible (which it normally is), and there is no reason that you can't use conservation of energy to determine the impact speed if you designate the point of impact to be where PE=0 because energy is conserved in the falling (obviously you can't do (1/2)(m1)(vi)^2 = (1/2)(m1+m2)(vf)^2). Even if there is friction present you simply use PE + Wext = KE where the external work (Wext) done is attributed to the force of friction.

The question is asking for average force, not speed at impact. Unfortunately, you have to know more information to determing the energy loss during an inelastic (deforming) collision.

Boondocks is right on the money.

 

[premed1001]

this question involves a ridiculous amount of time to solve, is this type possible on the real test?

 

[premed1001]

The reason your method did not work was probably because you didnt take into account the acceleration due to gravity. If you do, then:

Fnet = m ( a + g )
Fnet = 2 (250 + 10)
Fnet = 520 N

furthermore this doesn't make any sense.
force of mg = 2*10 is pointing down
force of ma(due to the mud stopping) = 250*2 is pointing UP, not down
so why then are these accelerations added instead of subtracted?