A sphere *A, *of mass 3 kg, is moving on a smooth horizontal floor with speed 4 ms–1 in a direction which is at right-angles to a smooth vertical wall. The sphere strikes the wall and rebounds with a speed of 2·8 ms–1.

Find the coefficient of restitution between the sphere and the wall.

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The elasticity of a collision is indicated by the coefficient of restitution, e such that:

Coefficient of restitution=`(Speed after collision)/(Speed before collision)`

When a smooth sphere collides with a smooth flat surface and bounces off it, the velocity of the sphere changes, and therefore the momentum changes. The change in momentum is caused by the impact between the sphere and the surface.

Applying Newton’s law of restitution to the component of velocity perpendicular to the surface,

`vsinbeta=e*usinalpha` (refer to the attached image, where alpha and beta are the angles of incidence and rebound, with respect to the horizontal, respectively).

Here, the collision is occurring at right angles, so `alpha = beta = 90^o`

Plugging in the values,

`2.8*1=e*4.0*1`

`rArr e=2.8/4=0.7`

Therefore, the coefficient of restitution between the ball and the wall is 0.7.

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