Question

a 74.9 kg person sits at rest on an icy pond holding a 2.44 kg physics book. he throws the physics book west at 8.25 m/s. what is his recoil velocity? PLEASE HELP ME

Answer 1

His recoil velocity is 0.269 m/s to the East

Step-by-step explanation:

This question is easily solved by using the law of conservation of linear momentum.

The formula for the momentum is

`Mo = mv`

, where m is the mass and v its speed.

The person + the book are at rest which means their momentum is

`Mo_0=0`

After the book is released, they both start to move and their combined momentum is

`Mo_f=m_pv_p+m_bv_b`

Where
`m_p, v_p` are the mass and speed of the person respectively and
`m_b, v_b` are the mass and speed of the book

Knowing that

`m_p=74.9 Kg, m_b=2.44 Kg, v_b=-8.25m/s` (positive speed is assumed to the right or East), and the total momentum is zero:

`m_pv_p+m_bv_b=0` =>

`v_p=-(m_bv_b)/(m_p) =-(2.44 (-8.25))/(74.9)`

`v_p=0.269 m/s`

Since the sign of
`v_p` is positive, it's directed to the East