Physics, asked by answer6310, 2 days ago

Two ice skaters initially at rest, push each other. If one skater whose mass is 60kg has a velocity of 2 m/s, what is the velocity of other skater, whose mass is 40kg​

Answers

Answered by IlMYSTERIOUSIl
3

Required Answer :-

Given :-

Case 1 ( for first ice skater ) -

  • Initial velocity (u) = \sf{ {u}_{1}} = 0m/s
  • Mass (m) = \sf{ {m}_{1}} = 60 kg
  • Final Velocity (v) = \sf{ {v}_{1}} = 2m/s

Case 2 ( for second ice skater )

  • Initial velocity (u) = \sf{ {u}_{2}} = 0m/s
  • Mass (m) = \sf{ {m}_{2}} = 40 kg
  • Final Velocity (v) = \sf{ {v}_{2}} = ?

To Find :-

  • Final velocity of second skater

Solution :-

We know that, From conservation of momentum

→ Final momentum = initial momentum

The change in momentum

\sf :\implies{m}_{1} {v}_{1} + {m}_{2} {v}_{2}  =  {m}_{1} {u}_{1} + {m}_{2} {u}_{2}

\sf :\implies60 \times 2 + 40 {v}_{2}  =  0 + 0

\sf :\implies120 + 40 {v}_{2}  =  0

\sf :\implies {v}_{2}  =   \dfrac{ \cancel{ - 120}}{ \cancel{40} }

\sf :\implies {v}_{2}  =   - 3ms ^{ - 1}

velocity of second skater is −3ms

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