A74.9 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?me

The recoil velocity is 0.2687 m/s.

Explanation:

∵ The person is sitting on an icy surface , we can assume that the surface is frictionless.

∴ There is no force acting acting on the person and book as a system in horizontal direction.

Hence , momentum is conserved for this system in horizontal direction of motion.

If 'i' and 'f' be the initial and final states of this system , then by principle of conservation of momentum(p)  -

=

System initially is at rest

∴=0

∴ From the above 2 equations

=0

We know that ,

Momentum(p)=Mass of the body(m)×velocity of the body(v)

Let and be the mass of the person and the book respectively and and be the final velocities of the person and book respectively.

∴=+=0

From the question ,

= 74.9 kg

= 2.44 kg

= 8.25 m/s

Substituting these values in the above equation we get ,

(74.9 × )+ (2.44×8.25) = 0

∴  = - 0.2687 m/s (Negative sign suggests that the motion of  the person is opposite to that of the book)

∴ The recoil velocity is 0.2687 m/s.

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