with a speed of 0.5 m/s. How much kinetic energy is lost in the collision? O 0.50 Joules O 3.75 Joules 2.5 Joules O 1.25 Joules O 5 Joules

Inelastic Collision

An inelastic collision is a type of collision where the kinetic energy is lost due to the internal friction of the particles. For a perfectly inelastic collision, the two colliding bodies stick together and move together as one unit. 

The above problem is an example of perfectly inelastic collision. Let two bodies of masses $m_1$ and $m_2$ be such that the first body is moving with an initial velocity $v_{1i}$ and the second body is at rest. 

The kinetic energy of the system before the collision

$K{E}_i=\frac{1}{2}{m}_1{v}_{1i}^2+\frac{1}{2}{m}_2.0=\frac{1}{2}{m}_1{v}_{1i}^2$

After the collision, the two bodies stick together and the two bodies move as a single unit. If $v_f$ is the velocity of the combined mass then the kinetic energy after the collision is

$K{E}_f=\frac{1}{2}\left(m_1+m_2\right)v_f^2$

The loss of energy in this collision is

$∆K=K{E}_i-K{E}_f=\frac{1}{2}{m}_1{v}_{1i}^2-\frac{1}{2}\left(m_1+m_2\right)v_f^2$

Given in the question, the mass of the first block $m_1=10.0 kg$

the mass of the second block $m_2=10.0 kg$

 the initial velocity of the first block $v_{1i}=1 m/s$.

the velocity of the combined system $v_f=0.5 m/s$

Therefore the loss in kinetic energy

$\begin{array}{rcl}∆KE& =& \frac{1}{2}{m}_1{v}_{1i}^2-\frac{1}{2}\left(m_1+m_2\right)v_f^2=\frac{1}{2}\times 10\times 1^2-\frac{1}{2}\times \left(10+10\right)\times 0.5^2\\ & =& 2.5 J\end{array}$