Answer to Question #116933 in Mechanics | Relativity for sridhar

Question

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

Answer 1

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 $40+20+10+5+2.5+1.25 + \ldots \approx 80\,\mathrm{m}.$