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4 years ago

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A cart of mass 6.0 kg moves with a speed of 3.0 m/s towards a second stationary cart with a mass of 3.0 kg. The carts move on a

frictionless surface and when the carts collide they stick together. What is the ratio of the initial kinetic energy of the two-cart system to the final kinetic energy of the two-cart system?

1 answer:

IgorLugansk [536]4 years ago

5 0

Answer:1.5

Explanation:

Given

mass of first cart $m_1=6 kg$

initial Velocity $u_1=3 m/s$

mass of second cart $m_2=3 kg$

$u_2=0 m/s$

In the absence of External Force we can conserve momentum

$m_1u_1+m_2u_2=(m_1+m_2)v$

$v=\frac{m_1u_1+m_2u_2}{m_1+m_2}$

$v=\frac{6\times 3+3\times 0}{6+3}$

$v=2 m/s$

Final kinetic Energy of two masses

$K.E._2=\frac{1}{2}(m_1+m_2)v^2$

$K.E._2=\frac{1}{2}\cdot (3+6)\cdot (2)^2$

$K.E._2=18 J$

Initial Kinetic Energy

$K.E._1=\frac{1}{2}m_1u_1^2+\frac{1}{2}m_2u_2^2$

$K.E._1=\frac{1}{2}6\times 3^2+0$

$K.E._1=27 J$

$ratio =\frac{K.E._1}{K.E._2}=\frac{27}{18}=1.5$

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