- #1

## Homework Statement

There is a 4 kg mass that has a speed of 6 m/sec on a horizontal frictionless surface. The mass collides head-on and elastically with an identical 4 kg mass initially at rest.

The final speed of the first 4 kg mass is:

(a) 0 m/s (b) 2 m/s (c) 3 m/s (d) 6 m/s

## Homework Equations

p = mv

KE = (1/2)mv

^{2}

## The Attempt at a Solution

To start off, one way of solving this problem seems to be knowing that the two masses have equal mass. Since the collision is elastic,

p

_{i}= p

_{f}

m

_{1}v

_{1init}+ m

_{2}v

_{2init}= m

_{1}v

_{1final}+ m

_{2}v

_{2final}

And since v

_{2init}= 0, this simplifies to

m

_{1}v

_{1init}= m

_{1}v

_{1final}+ m

_{2}v

_{2final}

Since the two masses have equal mass, we can assume that the (

**absolute values**of the) two final momentums (m

_{1}v

_{1final}and m

_{2}v

_{2final}) are equal. Therefore, if I just change m

_{2}v

_{2final}to be m

_{1}v

_{1final}, the equation should still yield us a correct answer:

m

_{1}v

_{1init}= m

_{1}v

_{1final}+ m

_{1}v

_{1final}

m

_{1}v

_{1init}=

**2*** m

_{1}v

_{1final}

Now I isolate the equation for v

_{1final}:

v

_{1final}= v

_{1init}/ 2

Substituting variables for their values yields me the correct answer:

v

_{1final}= (6m/s) / 2

v

_{1final}=

**3 m/s**

(Technically, if the final momentums of the two masses are equal and opposite, they should add up to zero! This is also something I am confused with. Is this even correct to do? But that's a later question.)

------------------------

Now, the problem is, I can't reproduce this same answer when solving for the same v

_{1final}through just Conservation of Energy and Conservation of Momentum, with no other assumptions made... Here are the steps I tried:

First, use Conservation of Momentum to find an expression equal to v

_{2final}:

m

_{1}v

_{1init}+ m

_{2}v

_{2init}= m

_{1}v

_{1final}+ m

_{2}v

_{2final}

And again, v

_{2init}= 0, so

m

_{1}v

_{1init}= m

_{1}v

_{1final}+ m

_{2}v

_{2final}

m

_{2}v

_{2final}= m

_{1}v

_{1init}- m

_{1}v

_{1final}

v

_{2final}= (m

_{1}v

_{1init}- m

_{1}v

_{1final}) / m

_{2}

v

_{2final}= (4kg * (6m/s) - 4kg * v

_{1final}) / 4kg

**v**

_{2final}= 6m/s - v_{1final}We can use this later to substitute v

_{2final}into our next equation, from Conservation of Energy:

(1/2)m

_{1}v

_{1init}

^{2}+ (1/2)m

_{2}v

_{2init}

^{2}= (1/2)m

_{1}v

_{1final}

^{2}+ (1/2)m

_{2}v

_{2final}

^{2}

(1/2)m

_{1}v

_{1init}

^{2}+ 0 = (1/2)m

_{1}v

_{1final}

^{2}+ (1/2)m

_{2}v

_{2final}

^{2}

m

_{1}v

_{1init}

^{2}= m

_{1}v

_{1final}

^{2}+ m

_{2}

**v**

_{2final}^{2}

4kg * 6m/s

^{2}= 4kg * v

_{1final}

^{2}+ 4kg *

**(6m/s - v**

_{1final})^{2}

(Omitting units because it gets messy)

4 * 6

^{2}= 4 * v

_{1final}

^{2}+ 4 * (6 - v

_{1final})

^{2}

144 = 4 * v

_{1final}

^{2}+ 4 * (36 - 12v

_{1final}+ v

_{1final}

^{2})

36 = v

_{1final}

^{2}+ (36 - 12v

_{1final}+ v

_{1final}

^{2})

36 = 2v

_{1final}

^{2}- 12v

_{1final}+ 36

0 = 2v

_{1final}

^{2}- 12v

_{1final}

0 = 2v

_{1final}(v

_{1final}- 6)

Therefore:

v

_{1final}= 0 or 6m/s

But the answer found earlier was 3m/s. How come the same number is not produced?

Sorry for the long post, and thank you for your help.

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