A sled of mass m is coasting at a constant velocity on the ice covered surface of a lake. Three birds, with a combined mass 0.5m, gently land at the same time on the sled. The sled and birds continue sliding along the original direction of motion. How does the kinetic energy of the sled and birds compare with the initial kinetic energy of the sled before the birds landed

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A sled of mass m is coasting at a constant velocity on the ice covered surface of a lake. Three birds, with a combined mass 0.5m, gently land at the same time on the sled. The sled and birds continue sliding along the original direction of motion. How does the kinetic energy of the sled and birds compare with the initial kinetic energy of the sled before the birds landed

Complete Question A sled of mass m is coasting at a constant velocity on the ice covered surface of a lake. Three birds, with a combined mass 0.5m, gently land at the same time on the sled. The sled and birds continue sliding along the original direction of motion. How does the kinetic energy of the sled and birds compare with the initial kinetic energy of the sled before the birds landed?
a) The final kinetic energy is one half of the initial kinetic energy.
b) The final kinetic energy is one third of the initial kinetic energy.
c) The final kinetic energy is one quarter of the initial kinetic energy.
d) The final kinetic energy is one ninth of the initial kinetic energy.
e) The final kinetic energy is equal to the initial kinetic energyAnswer:The correct option is bExplanation:From the question we are told that The mass of the sled is m The mass of the three birds is [tex]m_b = 0.5 m[/tex]Generally from the law of momentum conservation, The initial momentum of the boat = final momentum of the boat Hence [tex]m * v = (m_b+ m) * v_2[/tex]=> [tex]m v = (0.5m+ m) * v_2[/tex]=> [tex]m v = 1.5m* v_2[/tex]Here v is the velocity of the boat before the birds landed and [tex]v_2[/tex] is the velocity after the birds landed So [tex]v_2 = \frac{2}{3} v[/tex]Generally the initial velocity of the boat before the birds landed is mathematically represented as [tex]K_i = \frac{1}{2} * m * v ^2[/tex]Generally the final velocity of the boat when the birds have landed is mathematically represented as [tex]K_f = \frac{1}{2} * (m_b + m ) * v_2^2[/tex]=> [tex]K_f = \frac{1}{2} * (0.5m + m ) * (\frac{2}{3} v)^2[/tex]=> [tex]K_f = \frac{1}{2} * 1.5m * (\frac{2}{3} v)^2[/tex]=> [tex]K_f = \frac{1}{2} * 1.5m * \frac{4}{9} v^2[/tex]=> [tex]K_f = \frac{1}{2}m v^2 * \frac{1}{3}[/tex]Comparing this equation with that of the initial kinetic energy => [tex]K_f = \frac{1}{3} K_i[/tex]

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