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# Answer to Question #116933 in Mechanics | Relativity for sridhar

Question #116933
From a point on a smooth horizontal plane a body is projected with a velocity 20m/s at an angle of 45° to the horizontal. If the coefficient of restitution is 0.5 the total horizontal distance it travels before coming to rest after several rebounds is (Acceleration due to gravity =10ms^(-2) )
1)40m
2)60m
3)80m
4)120m
1
2020-05-19T10:59:50-0400

Let us determine the distance from the point of the beginning of the motion to the point of first impact to the ground. This distance can be calculated (see https://courses.lumenlearning.com/boundless-physics/chapter/projectile-motion/) as

"D_1 = \\dfrac{v_0^2\\sin2\\theta}{g} = \\dfrac{20^2\\sin90^\\circ}{g} = 40\\,\\mathrm{m}."

Coefficient of restitution (see https://en.wikipedia.org/wiki/Coefficient_of_restitution) is defined as the ratio of the velocity after impact and initial velocity. The distance is proportional to "v_0^2" , so "D_1 = 0.5^2\\cdot D_1 = 10\\,\\mathrm{m}" . , "D_3 = 0.5^2\\cdot D_2 = 2.5\\,\\mathrm{m}," and so on. The sum will be "\\approx 53.3\\,\\mathrm{m}."

But if we define the coefficient of restitution as the ratio of kinetic energy, we get the distances to decrease as "0.5^1" , so "D_2 = 0.5D_1 = 20\\,\\mathrm{m}," "D_3 = 0.5D_2 = 10\\,\\mathrm{m}, D_4 = 0.5D_3 = 5\\,\\mathrm{m}," and so on. The sum will be "40+20+10+5+2.5+1.25 + \\ldots \\approx 80\\,\\mathrm{m}."

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