Mechanics of a block of mass homework

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Discussion Overview

The discussion revolves around a physics homework problem involving the mechanics of a block sliding down an incline and colliding with another block. Participants explore the application of conservation laws to determine the speeds of the blocks after an elastic collision, addressing both kinetic energy and momentum conservation.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents a scenario involving a block sliding down an incline and colliding with a stationary block, seeking help to solve the problem.
  • Another participant requests clarification on the specific question being asked.
  • A participant states that the speed of the smaller mass before the collision is calculated to be 8.40 m/s using gravitational potential energy and kinetic energy principles.
  • There is a discussion about setting up simultaneous equations for momentum and kinetic energy conservation, with one participant correcting an earlier equation regarding kinetic energy.
  • One participant expresses difficulty in obtaining the correct answer after rearranging the equations, leading to a calculated speed of 8.00 m/s for the smaller mass.
  • Another participant suggests that the direction of the velocity vector should be considered, indicating that the final speed should be negative due to the collision dynamics.
  • A participant argues that since the question asks for speed, which is a scalar quantity, the sign should not affect the answer, and questions whether significant figures might be the issue with the provided answer.
  • There is a suggestion that using a negative value for the velocity in the momentum conservation equation could lead to a different speed calculation of approximately 5.33 m/s.

Areas of Agreement / Disagreement

Participants express differing views on the treatment of velocity versus speed, with some emphasizing the importance of direction in the context of momentum conservation, while others maintain that the question's focus on speed should allow for a positive value. The discussion remains unresolved regarding the correct final speed of the smaller mass after the collision.

Contextual Notes

Participants have not reached a consensus on the correct approach to the problem, particularly regarding the treatment of velocity and speed in the context of the collision. There are also unresolved issues related to the accuracy of significant figures in the answers provided.

Gughanath
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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.)

:confused: Please help to solve this problem.
 
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What's the question...?You didn't post the question..

Daniel.
 
Sorry. Determine the speed of the smaller mass and the larger mass after collision?
 
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.
 
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?
 
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.
 
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?
 
Yes,the initial value for the speed of the small mass is 8.40m/s...

Daniel.
 
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
sorry i have to correct equation (2)
0.5x2.30x8.40 + 0 = 0.5x2.30xv^2 + 0.5x7.1xV^2
 
  • #11
is this correct so far then?
 
  • #12
Assuming that the big mass is not exactly at the botton of the incline,then yes.

Daniel.
 
  • #13
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
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
But the question doesn't 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
Choosing "-v" in the component form of the momentum conservation eq.you'll end up with v~5.33ms^{-1}...

Daniel.
 

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