Question

a stone of mass 7 kg moving with a velocity of 50m/s collides with another body of mass 5kg moving with a velocity of 10m/s in the opposite direction. if after the impact, two bodies unite and move with common velocity calculate the loss in kinetic energy

Answer 1

ΔK = -2334.66J.  The kinetic energy loss is -2334.66J.

This is an example of elastic collision  which means that the two bodies remain united after the collision, as we know the amount of movement is preserved that is called conservation of momentum. So, we can write the equation as follow:

$P_{i}=P_{f}$ Where $P_{i}$ is the initial momentum and $P_{f}$ is the final momentum.

$P_{i}=m_{1}.v_{1}+m_{2}.v_{2}$

$P_{f}=(m_{1}+m_{2}).v_{f}$

$m_{1}.v_{1}+m_{2}.v_{2}=(m_{1}+m_{2}).v_{f}$

$v_{f}=\frac{m_{1}.v_{1}+m_{2}.v_{2}}{(m_{1}+m_{2})}$

Substituting the values:

$v_{f}=\frac{(7kg)(50\frac{m}{s})+(5kg)(10\frac{m}{s})}{(7kg+5kg)}$

$v_{f}=\frac{350kg\frac{m}{s} +50kg\frac{m}{s} }{12kg}$

$v_{f}=\frac{400kg\frac{m}{s} }{12kg}=33.33\frac{m}{s}$

After the collision the bodies are united and move at the speed of 33.33 m/s.

To calculate the kinetic energy lost in the impact we only have to calculate the energy of each body before the impact and compare it with the energy of the whole after the impact.

ΔK = $K_{f}-K_{i}$

ΔK = $\frac{1}{2}(m_{1}+m_{1})(v_{f})^{2}-[\frac{1}{2}m_{1}(v_{1})^{2}+\frac{1}{2}m_{2}(v_{2})^{2}]$

Substituting the values:

ΔK = $\frac{1}{2}(7kg+5kg)(33.33\frac{m}{s} )^{2}-[\frac{1}{2}(7kg)(50\frac{m}{s} )^{2}+\frac{1}{2}(5kg)(10\frac{m}{s} )^{2}]$

ΔK = $\frac{1}{2}(12kg)(1110.89\frac{m^{2}}{s^{2}})-[\frac{1}{2}(7kg)(2500\frac{m^{2}}{s^{2}} )+\frac{1}{2}(5kg)(100\frac{m^{2}}{s^{2}})]$

ΔK = $\frac{1}{2}(13330.68kg\frac{m^{2}}{s^{2}})-[\frac{1}{2}(17500kg\frac{m^{2}}{s^{2}} )+\frac{1}{2}(500kg\frac{m^{2}}{s^{2}})]$

ΔK = $6665.34kg\frac{m^{2}}{s^{2}}-(8750kg\frac{m^{2}}{s^{2}}+250kg\frac{m^{2}}{s^{2}})$

ΔK = $6665.34kg\frac{m^{2}}{s^{2}}-9000kg\frac{m^{2}}{s^{2}}$

ΔK = $-2334.66J$