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

Answer:

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)  -

$p_{i}$=$p_{f}$

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.