Inelastic truck-car collision

Problem

A $3\,000\;\text{kg}$ truck at $20\;\text{m/s}$ collides with a stationary $1\,000\;\text{kg}$ car. The two vehicles stay joined together. Calculate the common velocity after the collision and the kinetic energy lost to deformation.

Solution

In a perfectly inelastic collision the bodies stay joined: momentum is conserved (no external horizontal force), but kinetic energy is not. Data: $m_1=3\,000\;\text{kg}$, $v_1=20\;\text{m/s}$, $m_2=1\,000\;\text{kg}$, $v_2=0$. Conservation of momentum, with common velocity $v_f$: $m_1 v_1 = (m_1 + m_2)\,v_f \;\Rightarrow\; v_f = \frac{m_1 v_1}{m_1 + m_2} = \frac{3\,000\cdot 20}{4\,000} = 15\ \mathrm{m/s}$ Kinetic energy before and after: $E_i = \tfrac{1}{2}\,m_1 v_1^2 = \tfrac{1}{2}\cdot 3\,000\cdot 20^2 = 6{,}0\cdot 10^5\ \mathrm{J}$ $E_f = \tfrac{1}{2}(m_1+m_2)\,v_f^2 = \tfrac{1}{2}\cdot 4\,000\cdot 15^2 = 4{,}5\cdot 10^5\ \mathrm{J}$ Energy lost to deformation: $\Delta E = E_i - E_f = 6{,}0\cdot 10^5 - 4{,}5\cdot 10^5 = 1{,}5\cdot 10^5\ \mathrm{J}$ As a percentage: $\dfrac{1{,}5\cdot 10^5}{6{,}0\cdot 10^5} = 0{,}25$, i.e. about $25\%$ of the initial energy goes into heat and deformation of the wreckage. $\ev{v_f = 15\ \mathrm{m/s},\quad E_\text{lost} = 1{,}5\cdot 10^5\ \mathrm{J}\ (\approx 25\%)}$

Links

Topics: Momentum and collisions Concepts: Inelastic collision · Conservation of momentum · Kinetic energy Skills: Conservation of momentum