Question

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​

Answer 1

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