# Mechanics of a block of mass homework

1. Mar 5, 2005

### Gughanath

A block of mass m = 2.30 kg slides down a 30.0° incline which is 3.60 m high. At the bottom, it strikes a block of mass M = 7.10 kg which is at rest on a horizontal surface, Fig. 7-41. (Assume a smooth transition at the bottom of the incline, an elastic collision, and ignore friction.)

2. Mar 5, 2005

### dextercioby

What's the question...?You didn't post the question..

Daniel.

3. Mar 5, 2005

### Gughanath

Sorry. Determine the speed of the smaller mass and the larger mass after collision?

4. Mar 5, 2005

### dextercioby

Okay then.The advice is simple:use the conservation of KE and momentum (in vector form).Then u'll have to solve a simple system.

Daniel.

5. Mar 5, 2005

### Gughanath

I did. I end up with two unknown values for a simultaneuos equation. But I keep getting the wrong answer. Do you have an answer to this question?

6. Mar 5, 2005

### dextercioby

Nope.I'm just trying to guide through the correct solving.It's your problem.

So post the equations & the reasoning for reaching them in the first place;

Daniel.

7. Mar 5, 2005

### Gughanath

Ok. I use the Gravitation potential Energy lost=Kinetic energy lost rule to determine the speed of the smaller mass before the collision. I get 8.40m/s. Is this method acceptable to work out the initial speed of the smaller mass?

8. Mar 5, 2005

### dextercioby

Yes,the initial value for the speed of the small mass is 8.40m/s...

Daniel.

9. Mar 5, 2005

### Gughanath

Good. Now i can build a simultaneous equation. Equation one deals witht the momentum of the system. Equation two deals with the kinetic energy of the system.

2.30x8.40 + 0 = 2.30v + 7.10V ..........(1)
0.5x2.30x8.40^2 + 0 = 0.5x2.30xv^2 + 7.10V^2..........(2)
Is this correct so far?

10. Mar 5, 2005

### Gughanath

sorry i have to correct equation (2)
0.5x2.30x8.40 + 0 = 0.5x2.30xv^2 + 0.5x7.1xV^2

11. Mar 5, 2005

### Gughanath

is this correct so far then?

12. Mar 5, 2005

### dextercioby

Assuming that the big mass is not exactly at the botton of the incline,then yes.

Daniel.

13. Mar 5, 2005

### Gughanath

Well from rearranging equation (2) to give V in terms of v, I end up with the answer
v=8.00m/s.
This is an electronic Physics homework that my school sets every week. After submitting it, the program tells me that my answer is wrong.

14. Mar 5, 2005

### dextercioby

Well,as u see,since a smaller object is collided with a bigger one,it will bounce back,which means that the velocity vector changes the sense...Which means that the final aswer should e negative...

Daniel.

P.S.Therefore,take "-v" in the momentum conservation law.

15. Mar 5, 2005

### Gughanath

But the question doesnt talk about velocity. The question asks for speed, which is not a vecor quantity. So just 8.00m/s should be fine. Maybe I am giving my answers to the wrong numbers of decimal places or significant figures?
Is my my speed correct though?

16. Mar 5, 2005

### dextercioby

Choosing "-v" in the component form of the momentum conservation eq.you'll end up with v~5.33ms^{-1}...

Daniel.