Tricky problem worth 2 points on my final grade

[AryRezvani]

Tricky problem worth 2 points on my final grade :)

Homework Statement

Homework Equations

I honestly don't know how to tackle this.

I think it's going to contain ƩF=mg, and some basic kinematic equations.

The Attempt at a Solution

Where do is start?

 

[Nytik]

You are right that some kinematic equations will be needed.

But the first part of the problem is fundamentally about conversion of energy. What equations do you know for different kinds of mechanical energy?

 

[AryRezvani]

You are right that some kinematic equations will be needed.

But the first part of the problem is fundamentally about conversion of energy. What equations do you know for different kinds of mechanical energy?

Err

mgh=PE
1/2kx²=Springs.
1/2m²=KE

 

[Nytik]

Right, so at the beginning of the problem, the bigger block has some PE that is converted into KE.
When the blocks collide, the collision is elastic meaning KE is conserved. Of course momentum is conserved too (as always).
Using this information you should be able to find the velocities of the blocks as they are leaving the table. From there it's basic kinematics.

 

[AryRezvani]

Right, so at the beginning of the problem, the bigger block has some PE that is converted into KE.
When the blocks collide, the collision is elastic meaning KE is conserved. Of course momentum is conserved too (as always).
Using this information you should be able to find the velocities of the blocks as they are leaving the table. From there it's basic kinematics.

Mgh = 1/2mv², right?

How would you know what your mass is in this problem?

 

[Nytik]

Don't worry, the mass will cancel in all the equations you do.

So you have Mgh = $\frac{1}{2}$Mv², that's a good start.

You can rearrange that equation to find v. Next you'll need to write two equations for the collision, one for conservation of KE and one for conservation of momentum. These can be solved simultaneously.

 

[AryRezvani]

Don't worry, the mass will cancel in all the equations you do.

So you have Mgh = $\frac{1}{2}$Mv², that's a good start.

You can rearrange that equation to find v. Next you'll need to write two equations for the collision, one for conservation of KE and one for conservation of momentum. These can be solved simultaneously.

Okay, well rearranging that equation, you get 2mghv²/m.

O... Then the masses cancel out and you're left with 2gh=v².
Sqr[(2)(9.8)(.3)] = 2.619
So M hits m with a velocity of 2.619 m/s.
_______________________________________

Okay, now for conservation of momentum and KE:

I'm a little lost on this part.

 

[Nytik]

The momentum of the big block before they collide must be equal to the combined momentums of the big and little block after the collision. Do you know how to write an equation for this?

Similarly, the KE of the big block before the collision equals the combined KE of the big block and little block after the collision. Can you write this equation?

 

[AryRezvani]

The momentum of the big block before they collide must be equal to the combined momentums of the big and little block after the collision. Do you know how to write an equation for this?

Similarly, the KE of the big block before the collision equals the combined KE of the big block and little block after the collision. Can you write this equation?

Hmm I could give it a try. In regards to KE,

1/2Mv²= 1/2mv²+1/2Mv²

Not sure how to signify a collision above. Sorry for giving you a hard time. My physics professor isn't much a professor.

 

[Nytik]

That equation is perfect, I'm just going to add some subscript to make things clearer:

$\frac{1}{2}$Mv$_{i}^{2}$ = $\frac{1}{2}$Mv$^{2}_{1}$ + $\frac{1}{4}$Mv$^{2}_{2}$

So v$_{1}$ is the resulting velocity of the big block, and v$_{2}$ for the small block. Also notice I have substituted in the mass of the small block (m = 1/2M) to make it easier for you to cancel.

The equation for momentum is very similar,
Mv$_{i}$ = Mv$_{1}$ + $\frac{1}{2}$Mv$_{2}$

Now that you have two simultaneous equations for v$_{1}$ and v$_{2}$, you can solve them. (e.g. rearrange the momentum equation for v$_{1}$, then substitute the result into the KE equation). It might get a bit fiddly. Don't forget to cancel all the M's first!

 

[Nytik]

Hopefully you read this before solving, I accidentally missed out a 1/2 in the momentum equation (which I have now edited in). Sorry about that!

 

[AryRezvani]

Hopefully you read this before solving, I accidentally missed out a 1/2 in the momentum equation (which I have now edited in). Sorry about that!

That equation was ugly as hell hahaha.

Okay, the velocity of the smaller block was 1.75 m/s
the larger .8 m/s

 

[lendav_rott]

The 1st thing that comes to mind when I see this assignment is freefalling.

The blocks get an acceleration which is computable by the given information - once knocked off the table they will take exactly the same amount of time to reach the floor as they would have if just let fall to the floor from the same height. Once the time is known, everything else is a cinch.

I'm going to test this out,cuz technically, it should work

 

[Nytik]

Sorry it's taken me so long to get back to this thread. Did you check those numbers with the equations (sub them back into make sure they work)? Because they don't look correct to me.

By the way, when solving the two equations the preferred method is to keep all symbols in until the final step (I don't know if you solved them this way or not). Using this method the velocities I get are:

v$_{1}$ = $\frac{1}{3}$$\sqrt{2gh}$

v$_{2}$ = $\frac{4}{3}$$\sqrt{2gh}$

And these answers solve both equations so I believe them to be correct. Can you try again and see if you come up with these answers? (Perhaps you didn't put the 1/2 in that I missed first time around.)