Bouncing Rotation

Two blocks of masses 1 kg \SI{1}{kg} and 2 kg \SI{2}{kg} are connected by an ideal spring having spring constant k = 600 N / m k=\SI{600}{N/m} . They are at rest on a smooth horizontal ground. A sphere rotating with initial angular velocity ω = 30 rad / s \omega=\SI{30}{rad/s} , having radius 50 cm \SI{50}{cm} and mass 1 kg \SI{1}{kg} is falling from a height 5 m \SI{5}{m} (from the bottom of the sphere to the top of the block as shown.

The coefficient of restitution and coefficient of friction are given by 1 2 \dfrac{1}{2} and 1 3 \dfrac{1}{3} respectively. Assume The rotation of sphere doesn't cease before it rebounds from the block .

Take g = 10 m / s 2 g=\SI{10}{m/s^2} .

Then

  • Find the velocity of the 1-kg block (that is m m ) just after the collision.

Answer is v m / s \SI{v}{m/s} .

  • Find the maximum elongation in the spring.

Answer is x y \dfrac{x}{y} meters , gcd ( x , y ) = 1 \gcd(x,y)=1 and x , y x,y are positive numbers

Find the value of v + x + y v+x+y .


The answer is 12.

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