Respuesta :
Given:
The initial velocity of one of the objects is v.
The mass of that same object is 1/3 of the mass of the other object.
To find:
The fraction of the original kinetic energy that the stuck pair carry.
Explanation:
Let us assume that the mass of the object that was initially stationary is m. Thus the mass of the other object is m/3
From the law of conservation of momentum,
[tex]\frac{m}{3}v=(\frac{m}{3}+m)u[/tex]Where u is the velocity of the objects after the collision.
On simplifying the above equation,
[tex]\begin{gathered} \frac{m}{3}v=\frac{4m}{3}u \\ \implies v=4u \\ \implies u=\frac{v}{4} \end{gathered}[/tex]The total initial kinetic energy of the system is,
[tex]\begin{gathered} KE_i=\frac{1}{2}\times\frac{m}{3}v^2 \\ =\frac{mv^2}{6} \end{gathered}[/tex]The kinetic energy of the stuck pair is,
[tex]\begin{gathered} KE_f=\frac{1}{2}(\frac{m}{3}+m)u^2 \\ =\frac{1}{2}\times\frac{4m}{3}\times\frac{v^2}{4^{^2}} \\ =\frac{mv^2}{24} \end{gathered}[/tex]On taking the ratio of the final to initial kinetic energy of the system,
[tex]\begin{gathered} \frac{KE_f}{KE_i}=\frac{\frac{mv^2}{24}}{\frac{mv^2}{6}} \\ =\frac{6}{24} \\ =\frac{1}{4} \end{gathered}[/tex]Final answer:
The stuck pair carry 1/4 th of the original kinetic energy.
