A ball collides with an inclined plane of angle $\alpha$ after falling through a distance h..

Question

If it movies horizontally after the impact, find the coefficient of restitution

Conserving momentum along line of impact $$v_1\cos\alpha=v_2\sin\alpha$$

$$v_2=v_1\cot \alpha$$ Therefore $$e=\frac{v_2\sin\alpha}{v_1\cos\alpha}$$ $$e=1$$ which doesn’t make sense, what am I doing wrong?

This is a physics question, but since Physics SE doesn’t accept such questions, I am posting it on Math SE with the physics tag.

Answer 1

To summarize and organize what has been said in comments, prior to the collision, the ball has velocity components

• $v_1\sin \theta$ parallel to the inclined plane,
• $v_1\cos \theta$ perpendicular to the inclined plane (toward the plane).

After the collision, the ball has velocity

• $v_2\cos \theta$ parallel to the inclined plane,
• $v_2\sin \theta$ perpendicular to the inclined plane (away from the plane).

The coefficient of restitution is the ratio of the two “perpendicular” velocities.

Presumably you are supposed to assume that momentum parallel to the surface is preserved, which implies that the component of velocity is conserved (since the moving mass is the same before and after). This is a simplification, since a real-life ball would have been in contact with the surface for a non-zero amount of time, during which it would experience a frictional force parallel to the surface. But it’s a reasonable simplification.

Instead of setting the parallel velocities equal, however, you decided to write

$$v_1\cos \theta = v_2\sin \theta, $$

which says the two perpendicular velocities are equal. Since $e$ is defined as the ratio of these velocities, in effect you have decided to set $e$ to a ratio of equal quantities, which will always give a ratio of $1.$