Question

A 60 kg of man and a 50 kg of woman,both on friction less roller skets and facing each other the man pushes the woman and moves away if the speed of the man 1.5 m/s.what is the speed of woman​

Answer 1

Given:

A 60 kg of man and a 50 kg of woman,both on friction less roller skets and facing each other the man pushes the woman and moves away if the speed of the man 1.5 m/s.

To find:

What is the speed of woman?

Calculation:

Since the roller skates are friction-less and there there is no external force acting on the system the net momentum will remain conserved.

Hence , the change in momentum of the men will be equal and opposite to the change in momentum of the woman :

$\therefore \: \Delta P_{man} = - \Delta P_{woman}$

$\implies \:m_{(man)} \times v_{(man)} = - \bigg (m_{woman} \times v_{woman} \bigg)$

$\implies \:60 \times 1.5= - \bigg (50 \times v_{woman} \bigg)$

$\implies \: v_{woman} = - \dfrac{60 \times 1.5}{50}$

$\implies \: v_{woman} = - \dfrac{6 \times 1.5}{5}$

$\implies \: v_{woman} = - 1.8 \: m {s}^{ - 1}$

So, velocity of woman will be 1.8 m/s (negative sign indicates that the velocity of women is opposite to the direction of velocity of man).