A fisherman throws the anchor off her boat at an angle of 3.0° above the horizontal.If the anchor weighs 2.0 times as much as she does and the boat moves with speed 2.7 m/s during the launch,how fast did she throw the anchor? (Solve for her frame of reference.Assume the mass of the boat is negligible. )

Question

Quizplus · Accepted Answer

Since we are solving for the fisherman's frame of reference, we can use conservation of momentum in the vertical direction to relate the velocity of the anchor to the velocity of the boat.

Let's call the mass of the fisherman M and the mass of the anchor m. The initial momentum in the vertical direction is zero since the system is at rest. After the fisherman throws the anchor, the momentum in the vertical direction is given by:

p = (M + m) * v_anchor - M * v_boat * sin(3.0°)

where v_anchor is the velocity of the anchor, v_boat is the velocity of the boat, and sin(3.0°) is the vertical component of the angle at which the anchor is thrown. We can rearrange this equation to solve for v_anchor:

v_anchor = [M * v_boat * sin(3°)] / (M + m)

Substituting the given values, we get:

v_anchor = [M * 2.7 m/s * sin(3°)] / (M + 2M) = 0.70 m/s

Therefore, the speed at which the fisherman threw the anchor is approximately 0.70 m/s or 1.4 m/s in the opposite direction to the boat's velocity.

The correct answer is B.

A Fisherman Throws the Anchor Off Her Boat at an Angle