Let <em>m</em> be the mass of the second car, so the first car's mass is 2<em>m</em>.
Let <em>K</em> be the kinetic energy of the second car, so the first car's kinetic energy would be <em>K</em>/2.
Let <em>u</em> and <em>v</em> be the speeds of the first car and the second car, respectively. At the start,
• the first car has kinetic energy
<em>K</em>/2 = 1/2 (2<em>m</em>) <em>u</em> ² = <em>mu</em> ² ==> <em>K</em> = 2<em>mu</em> ²
• the second car starts with kinetic energy
<em>K</em> = 1/2 <em>mv </em>²
It follows that
2<em>mu</em> ² = 1/2 <em>mv</em> ²
==> 4<em>u</em> ² = <em>v</em> ²
When their speeds are both increased by 2.76 m/s,
• the first car now has kinetic energy
1/2 (2<em>m</em>) (<em>u</em> + 2.76 m/s)² = <em>m</em> (<em>u</em> + 2.76 m/s)²
• the second car now has kinetic energy
1/2 <em>m</em> (<em>v</em> + 2.76 m/s)²
These two kinetic energies are equal, so
<em>m</em> (<em>u</em> + 2.76 m/s)² = 1/2 <em>m</em> (<em>v</em> + 2.76 m/s)²
==> 2 (<em>u</em> + 2.76 m/s)² = (<em>v</em> + 2.76 m/s)²
Solving the equations in bold gives <em>u</em> ≈ 1.95 m/s and <em>v</em> ≈ 3.90 m/s.
