Find the velocity of each object after the collision

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Homework Help Overview

The problem involves a collision between two objects of different masses, specifically a 28.0 g object moving at 19.0 cm/s and a 5.0 g object moving at 15.0 cm/s. The context is centered around elastic collisions and the application of conservation laws to determine the final velocities of both objects after the collision.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of conservation of momentum and kinetic energy to solve for the final velocities. There are attempts to set up equations based on these principles, but confusion arises regarding the treatment of unknowns and unit conversions. Some participants express uncertainty about the validity of their calculations and the methods used to derive the equations.

Discussion Status

There is an ongoing exploration of the problem with various participants attempting to clarify their approaches and reasoning. Some guidance has been offered regarding the need for two equations to solve for the two unknowns, but there is no explicit consensus on the correct method or resolution of the confusion surrounding the calculations.

Contextual Notes

Participants note that the problem specifies the need for velocities in cm/s, which has led to discussions about unit conversions. There is also mention of a previous class example that may have influenced their understanding of the problem, but discrepancies in the outcomes have led to further confusion.

mandy9008
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Homework Statement


A 28.0 g object moving to the right at 19.0 cm/s overtakes and collides elastically with a 5.0 g object moving in the same direction at 15.0 cm/s. Find the velocity of each object after the collision.

Homework Equations


mv + mv = mv + mv

The Attempt at a Solution


(28.0 g)(19.0 cm/s) + (5.0 g)(15.0 cm/s) = (28.0 g)v1f + (5.0 g)v2f
607 g cm/s = 28 g v1f + 5 g v2f

607 cm/s = 28 v1f + 5 v2f
(574 cm/s = v1f + v2f )-5

607 cm/s = 28 v1f + 5 v2f
-2870 cm/s = 5 v1f - 5 v2f

-2263 cm/s = 23 v1f
v1f = -98.4 cm/s

v2f = -5.83 cm/s

I know that both velocities have to be positive, so I know these values cannot be right.
 
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mandy9008 said:

The Attempt at a Solution


(28.0 g)(19.0 cm/s) + (5.0 g)(15.0 cm/s) = (28.0 g)v1f + (5.0 g)v2f

Good, though you should convert your units.

mandy9008 said:
607 g cm/s = 28 g v1f + 5 g v2f

607 cm/s = 28 v1f + 5 v2f
(574 cm/s = v1f + v2f )-5

607 cm/s = 28 v1f + 5 v2f
-2870 cm/s = 5 v1f - 5 v2f

-2263 cm/s = 23 v1f
v1f = -98.4 cm/s

v2f = -5.83 cm/s

I know that both velocities have to be positive, so I know these values cannot be right.

Where did you get these numbers from?
 
the reason that I did not convert my units was because the answer asks for velocities in cm/s.

607 g cm/s = 28 g v1f + 5 g v2f I solved the mv + mv = mv + mv equation for with the 2 unknowns.

then I subtracted 33 from each side to set v1f and v2f equal to 1, and i got 574 cm/s = v1f + v2f

then I multiplied the bottom equation by 5 so I could cancel out the vf2 so I could solve for vf1, thus getting -2870 cm/s = 5 v1f - 5 v2f

after that, I added the 2 equations together and got -2263 cm/s = 23 v1f, then i solved for v1f: v1f = -98.4 cm/s

then I took the v1f that I got and plugged it into the original equation and got v2f: v2f = -5.83 cm/s
 
mandy9008 said:
the reason that I did not convert my units was because the answer asks for velocities in cm/s.

607 g cm/s = 28 g v1f + 5 g v2f I solved the mv + mv = mv + mv equation for with the 2 unknowns.

then I subtracted 33 from each side to set v1f and v2f equal to 1, and i got 574 cm/s = v1f + v2f

then I multiplied the bottom equation by 5 so I could cancel out the vf2 so I could solve for vf1, thus getting -2870 cm/s = 5 v1f - 5 v2f

after that, I added the 2 equations together and got -2263 cm/s = 23 v1f, then i solved for v1f: v1f = -98.4 cm/s

then I took the v1f that I got and plugged it into the original equation and got v2f: v2f = -5.83 cm/s


I am still perplexed as to how you solved one equation with two unknowns.
 
okay, how would you have done this problem. It is hard to explain how I did it without talking in person
 
mandy9008 said:
okay, how would you have done this problem. It is hard to explain how I did it without talking in person

Collision: conservation of momentum
Elastic: Conservation of energy

Two equations, two unknowns.
 
but i have two unknowns. I am confused
 
mandy9008 said:
but i have two unknowns. I am confused

If you are talking about what you did, then I too am confused.

If you meant what I suggested; then each conservation law would give an equation

You end up with something like

ax+by=e1
cx+dy=e2

Which you can solve for both x and y.
 
mandy9008 said:
the reason that I did not convert my units was because the answer asks for velocities in cm/s.

607 g cm/s = 28 g v1f + 5 g v2f I solved the mv + mv = mv + mv equation for with the 2 unknowns.

then I subtracted 33 from each side to set v1f and v2f equal to 1, and i got 574 cm/s = v1f + v2f

then I multiplied the bottom equation by 5 so I could cancel out the vf2 so I could solve for vf1, thus getting -2870 cm/s = 5 v1f - 5 v2f

after that, I added the 2 equations together and got -2263 cm/s = 23 v1f, then i solved for v1f: v1f = -98.4 cm/s

then I took the v1f that I got and plugged it into the original equation and got v2f: v2f = -5.83 cm/s


Hello Mandy,

You said subtracted 33 from each side but v1f and v2f are not the same you cannot do like this.

You should use law of conservation of momentum and law of conservation of kinetic energy to solve 2 unknown problem.
 
  • #10
what are your a's and b's and such?
 
  • #11
mandy9008 said:
what are your a's and b's and such?

a,b,c,d,e1,e2 are just random constants, I am just showing you how you'd get two equations to solve. (One will not be linear though)
 
  • #12
I am way more confused then when I started. My prof did a problem similar to this one in class today and he did it the way that I did it, but my answer is not coming out right. UGH
 
  • #13
mandy9008 said:
I am way more confused then when I started. My prof did a problem similar to this one in class today and he did it the way that I did it, but my answer is not coming out right. UGH

Here your first equation, it is correct.
607 g cm/s = 28 g v1f + 5 g v2f

Second equation is
Before collision total KE= After collision total KE
 
  • #14
mandy9008 said:
I am way more confused then when I started. My prof did a problem similar to this one in class today and he did it the way that I did it, but my answer is not coming out right. UGH

Then let's start again, but without doing anything major.

mandy9008 said:

Homework Statement


A 28.0 g object moving to the right at 19.0 cm/s overtakes and collides elastically with a 5.0 g object moving in the same direction at 15.0 cm/s. Find the velocity of each object after the collision.

The problem says the masses collide elastically.

Collision means that you will need to apply conservation of linear momentum.
Elastic means you will need to apply conservation of kinetic energy.

mandy9008 said:

The Attempt at a Solution


(28.0 g)(19.0 cm/s) + (5.0 g)(15.0 cm/s) = (28.0 g)v1f + (5.0 g)v2f

This is what you get when you applied conservation of momentum.

Kinetic energy of a mass is given by 1/2mv2.

So kinetic energy before = kinetic energy after. So form this equation now.
 
  • #15
1/2 mv21i + 1/2 mv22i = 1/2 mv21f + 1/2 mv22f

1/2 (28g)(19 cm/s)2+ 1/2 (5 g)(15 cm/s)2 = 1/2 (28g)v21f + 1/2 (5g)v22f
 
  • #16
mandy9008 said:
1/2 mv21i + 1/2 mv22i = 1/2 mv21f + 1/2 mv22f

1/2 (28g)(19 cm/s)2+ 1/2 (5 g)(15 cm/s)2 = 1/2 (28g)v21f + 1/2 (5g)v22f

Yes, right.
You have previous for momentum. You can easy to get the answer.
 

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