A ball of mass 1kg is dropped from a height of 3.2 m on smooth inclined plane. The coefficient of restitution for the collision is $e = \frac{1}{2}$ . The ball's velocity becomes horizontal after the collision.
Take g = 10 m/s².

Question

A ball of mass 1kg is dropped from a height of 3.2 m on smooth inclined plane. The coefficient of restitution for the collision is $e = \frac{1}{2}$ . The ball's velocity becomes horizontal after the collision.

Ball dropped vertically from 3.2 m onto smooth inclined plane at theta

Take g = 10 m/s².

TestHub · Accepted Answer

Before collision, the speed after falling 3.2 m is

$u = 2 g h = 8 m/s$

Resolve the velocity parallel and perpendicular to the smooth plane.

$v_{x} = u (1 + e) sin θ cos θ$

$v_{y} = u (e cos^{2} θ - sin^{2} θ)$

The rebound is horizontal, so its vertical component is zero.

$tan^{2} θ = e = \frac{1}{2}$

$v_{x} = u tan θ = \frac{8}{2} = 42 m/s$

The kinetic energy loss is 16 J, not 8 J. The return to the sloped plane is not vertically downward. A and B are correct.

Physics - Center of Mass Question with Solution | TestHub

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