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

The recoil velocity is 0.2687 m/s.

Step-by-step 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) -

`p_(i)`=
`p_(f)`

System initially is at rest

∴
`p_(i)`=0

∴ From the above 2 equations

`p_(f)`=0

We know that ,

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

Let
`m_(1)` and
`m_(2)` be the mass of the person and the book respectively and
`v_(1)` and
`v_(2)` be the final velocities of the person and book respectively.

∴
`p_(f)`=
`m_(1)`
`v_(1)`+
`m_(2)`
`v_(2)`=0

From the question ,

`m_(1)` = 74.9 kg

`m_(2)` = 2.44 kg

`v_(2)` = 8.25 m/s

Substituting these values in the above equation we get ,

(74.9 ×
`v_(1)` )+ (2.44×8.25) = 0

∴
`v_(1)` = - 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.