Answer:
D. Linear momentum and angular momentum are both conserved.
When the skater jumps onto the board, he/she transfers linear momentum to the board, causing it to start moving with a velocity in the same direction as the skater. However, since the board was at rest initially, its total linear momentum changes from zero to a non-zero value. This conservation of linear momentum can be expressed as:
m(skater) x v(skater) = (m(board) + m(skater)) x v(final)
where m(skater) and v(skater) are the mass and velocity of the skater before the jump, m(board) is the mass of the board, and v(final) is the velocity of the skater and board after the jump.
At the same time, since there is no external torque acting on the board-skater system, the conservation of angular momentum can be expressed as:
I x w(initial) = (I(skater) + I(board)) x w(final)
where I is the moment of inertia of the board-skater system, w(initial) is the initial angular velocity (zero), I(skater) is the moment of inertia of the skater, I(board) is the moment of inertia of the board, and w(final) is the final angular velocity of the system.
Therefore, both linear momentum and angular momentum are conserved in this situation. The conversion of kinetic energy to angular momentum (option B) is not relevant in this case since there is no external torque acting on the system. The conservation of rotational kinetic energy (option C) is also not applicable since the system does not rotate before or after the jump.
Step-by-step explanation: