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In an elastic collision between two objects do both objects have the same kinetic energy after the collision as before?

Let's do the problem and see

The problem below will prove that in an elastic collision between two objects the objects do not have the same kinetic energy after the collision as before?

But the KE before collision = KE after collision

A 2 kg object with velocity of 8 m/s E hits a 5 kg object with a velocity of 5m/s W. What is the velocity of both objects after colision?

East is positive, West is negative

Draw a picture showing the 2 objects before and after collisions.

…..8 m/s……………....-5 m/s………….V2…………….V5

...2kg object..-->….<--5 kg object…..2kg object…….5kg object W……………………………………………………………….E

Mom = +16………..Mom = -25………..2V2……………5V5

Momentum = mass * velocity

Since the 5 kg object has more momentum, both objects will continue to move West. They should have negative velocities!!

Momentum = mass * velocity

Total momentum before collision = (2 * 8) + (5 * -5) =

Total momentum before collision = 16 + -25 = -9

Total momentum before collision = -9

Total momentum after collision = (2 * V2) + (5 * V5)

Momentum is always conserved.

Total momentum after collision = Total momentum before collision

2V2 + 5V5 = -9 = (add - 2V2 and +9 to both sides)

2V2 = -9 - 5V5 , (divide by -2)

V2 = -4.5 - 2.5V5

Eq. momentum = V2 = -4.5 - 2.5V5

Since the collision is elastic, kinetic energy is conserved

KE before collision = KE after collision

KE ½ mass * velocity ^2

KE before collision = ½ * 2* 8^2 + ½ * 5 * 5^2 =

KE before collision = 64 + 62.5 =

KE before collision = 126.5

KE after collision = ½ * 2 * V2^2 + ½ *5 *V5^2

KE before collision = KE after collision

126.5 = ½ * 2 * V2^2 + ½ *5 *V5^2

126.5 = V2^2 + 2.5 *V5^2

Eq. energy = V2^2 + 2.5 *V5^2 -126.5 =0

Now we have 2 equations in 2 unknowns

Substitute the value of V2 from Eq.m into Eq.e

V2 = -4.5 - 2.5V5

(-4.5 - 2.5V5)^2 + 2.5V5^2 -126.5 = 0

(20.25 + 22.5V5 + 6.25V5^2) + 2.5V5^2 -126.5 =0

Add (6.25V5^2 + 2.5V5^2) = 8.75V5^2

Subtract (20.25 - 126.5) = -106.25

8.75V5^2 + 22.5V5 -106.5 =0

Use quadratic equation to solve for V5

V5 =[ -22.5 ± [22.5^2-(4 * 8.75 * -106.25)]^0.5] ÷ (2 * 8.75)

V5 = [-22.5 ± [506.25 + 3718.75]^0.5] ÷ 17.5

V5 = [-22.5 ± [4225]^0.5] ÷ 17.5

The square root of 4225 = ± 65, I will try +65 first and try add +65 first.

V5 = [-22.5 + 65] ÷ 17.5

V5 = +42.5 ÷ 17.5

V5 = +2.42857

This means the 5 kg object is going East. I made this statement at the beginning. "Since the 5 kg object has more momentum, both objects will continue to move West." V5 = +2.4286 is wrong

I will try subtract +65.

V5 = [-22.5 - 65] ÷ 17.5

V5 = -87.5 ÷ 17.5

V5 = -5

That can not be true, because that was the velocity of the 5 kg in the beginning.

I will try using - 65 for the square root of 4225, and add.

V5 = [-22.5 + -65] ÷ 17.5

V5 = -87.5 ÷ 17.5

V5 = -5

That can not be true, because that was the velocity of the 5 kg in the beginning.

Last but not least, I will using - 65 for the square root of 4225, and subtract.

V5 = [-22.5 - -65] ÷ 17.5

V5 = +42.5 ÷ 17.5

V5 = +2.42857

I know the answer has to be V5 = -2.42857.

If the -22 was +22.5, I would get the correct answer..

If you find the mistake, email me at morrison60957@yahoo.com

V5 = [+22.5 - 65] ÷ 17.5

V5 = -42.5 ÷ 17.5

V5 = -2.42857

I will copy, paste the area where I believe my mistake is at the bottom of this work. If you find my mistake let me know!!

V5 = -2.42857

Eq. momentum = V2 = -4.5 - 2.5V5

V2 = -4.5 - (2.5* -2.42857)

V2 = -10.5714

Let's see if the momentum = -9

Total momentum after collision = (2 * -10.5714) + (5 * -2.42857)

Total momentum after collision = -21.142853 + 12.14285 = -9.000003

Now let's see if the Kinetic energy is conserved

KE before collision = 126.5

KE after collision = ½ * 2 * V2^2 + ½ *5 *V5^2

126.5 = ½ * 2 * (-10.5714)^2 + ½ *5 *(-2.42857)^2

126.5 = 111.754498 + 14.75588061

126.5 ≈ 126.49993786 OK

KE before collision = 64 + 62.5

KE of 2 kg object = 64 J

KE of 5 kg object = 62.5 J

Sum of KE = 126.5

KE after collision = 64 + 62.5

KE of 2 kg object after collision = 111.754498 J

KE of 5 kg object after collision = 14.75588061 J

Sum of KE after collision = 126.49993786

I have proved that in an elastic collision between two objects the objects do not have the same kinetic energy after the collision as before?

But the KE before collision = KE after collision

Below is the work where I suspect I have made a mistake, If you find the mistake, email me at morrison60957@yahoo.com

Momentum = mass * velocity

Total momentum before collision = (2 * 8) + (5 * -5) =

Total momentum before collision = 16 + -25 = -9

Total momentum before collision = -9

Total momentum after collision = (2 * V2) + (5 * V5)

Momentum is always conserved.

Total momentum after collision = Total momentum before collision

2V2 + 5V5 = -9 = (add - 2V2 and +9 to both sides)

2V2 = -9 - 5V5 , (divide by -2)

V2 = -4.5 - 2.5V5

Eq. momentum = V2 = -4.5 - 2.5V5

V2 = -4.5 - 2.5V5

(-4.5 - 2.5V5)^2 + 2.5V5^2 -126.5 = 0

(20.25 + 22.5V5 + 6.25V5^2) + 2.5V5^2 -126.5 =0

Add (6.25V5^2 + 2.5V5^2) = 8.75V5^2

Subtract (20.25 - 126.5) = -106.25

8.75V5^2 + 22.5V5 -106.5 =0

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Q: In an elastic collision between two objects do both objects have the same kinetic energy after the collision as before?
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