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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

User Takje
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1 Answer

2 votes

Answer:

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

User Roshnee
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