Question

Two objects each moving with speed v travel in opposite directions along a straight line passing through both their centers. The objects stick together when they collide and move together with speed v/4 after the collision.
(a) What is the ratio of the final kinetic energy to the initial kinetic energy? (answer: Kf/Ki = 1/16)
(b) What is the ratio of the mass of the more massive object to the mass of the less massive object? (answer: mmore massive/mless massive = 5/3)

Answer 1

Answer:

$\dfrac{1}{16}$

$\dfrac{5}{3}$

Explanation:

$m_1$ = Mass of first object

$m_2$ = Mass of second object

v = Speed of both objects

$\dfrac{v}{4}$ = Combined velocity

The ratio of final kinetic energy to initial kinetic energy will be

$\dfrac{K_f}{K_i}=\dfrac{\dfrac{1}{2}(m_1+m_2)(\dfrac{v}{4})^2}{\dfrac{1}{2}(m_1v^2+m_2v^2)} \\\Rightarrow \dfrac{K_f}{K_i}=\dfrac{(m_1+m_2)\dfrac{v^2}{16}}{m_1v^2+m_2v^2}\\\Rightarrow \dfrac{K_f}{K_i}=\dfrac{(m_1+m_2)\dfrac{1}{16}}{m_1+m_2}\\\Rightarrow \dfrac{K_f}{K_i}=\dfrac{1}{16}$

The ratio is $\dfrac{1}{16}$

As the linear momentum is conserved

$m_1v-m_2v=(m_1+m_2)\dfrac{v}{4}\\\Rightarrow m_1-m_2=(m_1+m_2)\dfrac{1}{4}$

Divide by $m_2$ on both sides

$\dfrac{m_1}{m_2}-1=\dfrac{m_1}{4m_2}+\dfrac{1}{4}\\\Rightarrow \dfrac{m_1}{m_2}-\dfrac{m_1}{4m_2}=\dfrac{1}{4}+1\\\Rightarrow \dfrac{3m_1}{4m_2}=\dfrac{5}{4}\\\Rightarrow \dfrac{m_1}{m_2}=\dfrac{5\times 4}{3\times 4}\\\Rightarrow \dfrac{m_1}{m_2}=\dfrac{5}{3}$

The ratio of mass is $\dfrac{5}{3}$