Question

A swimmer of mass 64.38 kg is initially standing still at one end of a log of mass 237 kg which is floating at rest in water. He runs toward the other end of the log and dives off with a horizontal speed of 3.472 m/s relative to the water. What is the speed of the log relative to water after the swimmer jumps off

Answer 1

Answer:

0.9432 m/s

Explanation:

We are given;

Mass of swimmer;m_s = 64.38 kg

Mass of log; m_l = 237 kg

Velocity of swimmer; v_s = 3.472 m/s

Now, if we consider the first log and the swimmer as our system, then the force between the swimmer and the log and the log and the swimmer are internal forces. Thus, there are no external forces and therefore momentum must be conserved.

So;

Initial momentum = final momentum

m_l × v_l = m_s × v_s

Where v_l is speed of the log relative to water

Making v_l the subject, we have;

v_l = (m_s × v_s)/m_l

Plugging in the relevant values, we have;

v_l = (64.38 × 3.472)/237

v_l = 0.9432 m/s