Conservation of kinetic energy example
Question
A red snooker ball and a blue snooker ball (each with a mass of 160 g) collide. The red ball is travelling from left to right at 0.28 ms-1. The blue ball is travelling from right to left at 0.12 ms-1. This is shown in the diagram below.
After the collision, the blue ball travels from left to right at 0.18 ms-1.
a) Calculate the velocity of the red ball after the collision.
b) Use kinetic energy to determine whether the collision was elastic or not.
a) Total momentum before = total momentum after
\({m_{red}}{u_{red}} + {m_{blue}}{u_{blue}} = {m_{red}}{v_{red}} + {m_{blue}}{v_{blue}}\)
\(0.16 \times 0.28 + 0.16 \times ( - 0.12) = 0.16 \times {v_{red}} + 0.16 \times 0.18\)
Rearranging:
\(0.16 \times {v_{red}} = (0.0448 - 0.0192) - 0.0288\)
\(= 0.0256 - 0.0288\)
\(= \frac{{ - 0.0032}}{{0.16}}\)
\({v_{red}} = - 0.02m{s^{ - 1}}\). This is a negative value, so the direction is from right to left
b) To determine if the collision is elastic or not, you must work out the kinetic energy before and after the collision.
Remember the equation for calculating kinetic energy in general is:
\({E_k} = \frac{1}{2}m{v^2}\)
This example deals with two objects. The kinetic energy before the collision involves their initial speed, so the equation becomes:
\({E_k}=\frac{1}{2}m_{red}({u_{red}})^2+\frac{1}{2}m_{blue}({u_{blue}})^2\)
\(= 0.5 \times 0.16 \times {(0.28)^2} + 0.5 \times 0.16 \times {(0.12)^2}\)
\(= 0.00627 + 0.00115\)
\(= 0.00742J\)
Total kinetic energy after collision:
\(\frac{1}{2}{m_{red}(v_{red})^2+\frac{1}{2}{m_{blue}}(v_{blue})^2}\)
\(=0.5\times0.16\times(0.02)^2+0.5\times0.16\times(0.18)^2\)
\(=0.000032+0.00259\)
\(=0.00262J\)
Total kinetic energy lost:
\(=0.00742-0.00262\)
\(=0.0048\)
48 mJ of energy is lost so the collision is inelastic.
