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Luda [366]
3 years ago
13

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)
Physics
1 answer:
Tpy6a [65]3 years ago
6 0

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}

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