Question

A lumberjack (mass = 110 kg) is standing at rest on one end of a floating log (mass = 206 kg) that is also at rest. The lumberjack runs to the other end of the log, attaining a velocity of +3.09 m/s relative to the shore, and then hops onto an identical floating log that is initially at rest. Neglect any friction and resistance between the logs and the water. (a) What is the velocity of the first log (again relative to the shore) just before the lumberjack jumps off? (b) Determine the velocity of the second log (again relative to the shore) if the lumberjack comes to rest relative to the second log.

Answer 1

a). The velocity of the first log is -1.65 m/s.

(b). The velocity of the second log is 1.07 m/s.

Step-by-step explanation:

Given that,

Mass of lumberjack M= 110 kg

Mass of log m= 206 kg

Final velocity = 3.09 m/s

(a). We need to calculate the velocity of the first log just before the lumberjack jumps off

Using conservation of momentum

`Mu_(1)+mu_(2)=Mv_(1)+mv`

Put the value into the formula

`0=110*3.09+206v`

`v=-(110*3.09)/(206)`

`v=-1.65\ m/s`

The velocity of the first log is -1.65 m/s.

(b). If the lumberjack comes to rest relative to the second log

We need to calculate the velocity of the second log

`(M+m)v=Mv_(1)`

`v=(Mv_(1))/(M+m)`

Put the value into the formula

`v=(110*3.09)/(110+206)`

`v=1.07\ m/s`

The velocity of the second log is 1.07 m/s.

Hence, (a). The velocity of the first log is -1.65 m/s.

(b). The velocity of the second log is 1.07 m/s.