The first car has twice the mass of a second car, but only half as much kinetic energy. When both cars increase their speed by 6.0 m/s , they then have the same kinetic energy.

Question

The first car has twice the mass of a second car, but only half as much kinetic energy. When both cars increase their speed by 6.0 m/sm/s , they then have the same kinetic energy. What were the original speeds of the two cars? Express your answers using two significant figures separated by a comma.

Answer 1

So first, you want to write down the equations you know. We know the relationship for kinetic energy (KE) is as follows:

KE = (1/2)*m*v² , where m is mass and v is velocity. This relationship of KE, mass, and velocity will be VERY important to solving this problem.

Based off reading the first sentence, we know that the first car has 2 times the mass of the second and half the kinetic energy or:

m₁=2m₂ (relationship 1)

KE₁=(1/2)KE₂ (relationship 2)

In the second sentence we find out that if the velocity of both cars increases by 6, their kinetic energies are equal, or:

when V₁ = V₁+6 and V₂=V₂+6:

KE₁=(1/2)*m₁*(V₁+6)²

KE₂=(1/2)*m₂*(V2+6)² and

there kinetic energies are equal at this point or KE₁=KE₂, therefore, substituting for KE₁ and KE₂, we get our next relationship:

(1/2)*m₁*(V₁+6)²=(1/2)*m₂*(V₂+6)² (relationship 3)

Now we have all the relationships we need to answer the question "What were the original speeds of the two cars?"

First we can simplify relationship 2 a bit by substituting KE's for their (1/2)mv² equivalent as follows:

1. KE₁=(1/2)KE₂ (relationship 2) expands to:
2. (1/2)m₁V₁²=(1/2)((1/2)m₂V₂²), then we can use relationship 1 to get rid of the m₁ variable by substituting for m₁ using the relationship m₁=2m₂:
3. (1/2)2m₂V₁²=(1/2)*((1/2)m₂V₂²), then we can cancel out the m₂ term and multiple the numbers out on both sides and this now simplifies to a velocity relationship between car one (V₁) and car two (V₂):
4. V₁²=(1/4)V₂² or 4V₁²=V₂². Take the square root of both sides and we now have a relationship for V₁ and V₂ :
5. 2V₁=V₂

Now let's simplify relationship 3 by substituting for m₁ using the relationship m₁=2m₂:

1. (1/2)*2m₂*(V₁+6)²=(1/2)*m₂*(V₂+6)². Now we can substitute for V₂ using the relationship we found above V₂=2V₁. The equation now simplifies to:
2. m₂*(V₁+6)²=(1/2)*m₂*(2V₁+6)²

The m₂ term on both sides cancels out and now we only have one variable in the equation, V₁, that we can solve for. If we expand out the squared terms, we now get:

1. V₁²+12V₁+36=(1/2)*(4V₁²+24V₁+36), which simplifies to
2. V₁²+12V₁+36=2V₁²+12V₁+18

If we move the variables to one side and the numbers to the other, the V₁ terms cancel out, leaving us with:

1. V₁²=18.

Taking the square root of this this simplifies to the velocity of car 1 which is:

V₁²=3*(2)^(1/2) = 4.24 m/s

Since we know from earlier that 2V₁=V₂, we can find the velocity of car two to be:

V₂= 2*(4.24) = 8.48 m/s

These are the original speeds of the two cars!

Answer 2

Given: m₁ = 2 m₂, 2 K₁ = K₂

Known: K = (1/2) m v²

Implied: K₁ = (1/2) m₁ v₁² and K₂ = (1/2) m₂ v₂²

Replace equations in kinetic energy equality, and replace m₁ with 2 m₂:

2 [(1/2) m₁ v₁² ] = (1/2) m₂ v₂² ==> 2 [(1/2) 2 m₂ v₁² ] = (1/2) m₂ v₂² cancel and reduce,

==> 4 v₁² = v₂² take the square root of both sides: 2 v₁ = v₂ This is the relationship before adding the 6 m/s to each velocity.

Part 2:

Here we are given that K₁ = K₂

so (1/2) m₁ (v₁ + 6)² = (1/2) m₂ (v₂ + 6)² ==> (1/2) 2 m₂ (v₁ + 6)² = (1/2) m₂ (v₂ + 6)²

Cancel and reduce and get 2 (v₁ + 6)² = (v₂ + 6)² In part 1, we found the relationship between the speeds, substitute in for v₂ in terms of v₁, get: 2 (v₁ + 6)² = (2 v₁ + 6)²

In this case, we don't want to take the square root of both sides, because that introduces √2 into the problem, which will complicate things. Multiply out both terms in parenthesis; don't forget cross terms.

2( v₁² + 12 v₁ + 36) = 4 v₁² + 24 v₁ + 36

2 v₁² + 24 v₁ + 72 = 4 v₁² + 24 v₁ + 36

Move all terms with v1 to the right hand side and all constants to the left. You will be left with a constant = 2 v₁² divide through by 2, then take the square root of both sides. This gives you v₁ You can use the relationship between v₁ and v₂ from part 1 to find v₂ ,