Question

You (50 kg) are standing on a floating log (200 kg). Both are floating down a river at 1 m/s. The log points in the direction along the river. You walk in a direction that is down river for 5 seconds and your speed is 1.5 m/s as measured by a ground observer at the end of the 5 seconds.

a. What is the speed of the log at the end of the 5 seconds?
b. What is the average force between you and the log during those 5 seconds?

Answer 1

0.875 m/s

5 N

Step-by-step explanation:

`m_1` = Mass of person = 50 kg

`m_2` = Mass of log = 200 kg

`v_1` = Velocity of person = 1.5 m/s

`v_2` = Velocity of log

v = Velocity of log with respect to shore = 1 m/s

t = Time taken = 5 seconds

As the momentum of system is conserved we have

`(m_1+m_2)v=m_1v_1+m_2v_2\\\Rightarrow v_2=((m_1+m_2)v-m_1v_1)/(m_2)\\\Rightarrow v_2=((50+200)1-50* 1.5)/(200)\\\Rightarrow v_2=0.875\ m/s`

Velocity of the log at the end of the 5 seconds is 0.875 m/s

Force is given by

`F=(m(v-u))/(t)\\\Rightarrow F=(50(1.5-1))/(5)\\\Rightarrow F=5\ N`

The average force between you and the log during those 5 seconds is 5 N