Question

A ball strike a horizontal floor at an angle 45° the coefficient of restitution between the ball and the floor is 1÷2 the fraction of the kinetic energy lost in collision is

Answer 1

Place the values in this equation and you'll get your answer.

Answer 2

Answer: Fraction of kinetic energy $=\frac { 1+e^{ 2 } }{ 2 }$

Given data is the coefficient of restitution as $\frac { 1 }{ 2 }$ when ball strikes at an angle of 45 degrees.

Therefore the fraction of kinetic energy is the final kinetic energy/initial kinetic energy.

Thereby fraction of kinetic energy lost $=\quad \frac { \frac { 1 }{ 2 } m{ v }^{ 2 } }{ \frac { 1 }{ 2 } m{ u }^{ 2 } }$

Where $v\quad =\quad \sqrt { { \left( \frac { u }{ \sqrt { 2 } } \right) }^{ 2 }+{ \left( \frac { eu }{ \sqrt { 2 } } \right) }^{ 2 } } \quad and\quad v^2\quad is\quad simply\quad u^2$

Therefore, the fraction of kinetic energy $=\quad \frac { \left[ \sqrt { { \left( \frac { u }{ \sqrt { 2 } } \right) }^{ 2 }+{ \left( \frac { eu }{ \sqrt { 2 } } \right) }^{ 2 } } \right] }{ { u }^{ 2 } }$

Therefore, Fraction of kinetic energy $=\frac { 1+e^{ 2 } }{ 2 }$