Question

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

Answer 1

Given :-

• Initial velocity of first skater = 0 m/s

• Initial velocity of second skater = 0 m/s

• Mass of first skater = 60 kg

• Mass of second skater = 40 kg

• Final velocity of first skater = 2 m/s

To Find :-

• Final velocity of second skater.

Solution :-

According to law of conservation of momentum :-

Final momentum = Initial momentum

$\bf\dag \ \underline{\boxed{\bf m_1v_1+m_2v_2= m_1u_1+m_2u_2}}$

Here :-

• m₁ denotes mass of first skater.

• m₂ denotes mas of second skater.

• v₁ denotes final velocity of first skater.

• v₂ denotes final velocity of second skater.

• u₁ denotes initial velocity of first skater.

• u₂ denotes initial velocity of second skater.

$\longrightarrow \sf ( 60 \times 2 )+ ( 40 \times v_2) = ( 60 \times 0 )+ ( 40 \times 0 ) \\\\\longrightarrow 120+40v_2 = 0+0 \\\\\longrightarrow 120+40v_2 = 0 \\\\\longrightarrow 40v_2 = -120 \\\\\longrightarrow v_2 = \dfrac{-120}{40} \\\\\longrightarrow \bf v_2 = -3 \ m/s$

Velocity of second skater = -3 m/s

The negative sign shows that the two skaters are moving in opposite direction.

Answer 2

Answer:

The velocity of the other skater is – 3 m/s.

Explanation:

Given:

• Initial velocity of first skater (u₁) = 0
• Initial velocity of second skater (u₂) = 0
• Mass of first skater (m₁) = 60 kg
• Mass of second skater (m₂) = 40 kg

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To find:

Velocity of other skater.

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Solution:

We know that according to the principle of conservation of momentum, the final velocity must be equal to the initial velocity.

m₁v₁ + m₂v₂ = m₁u₁ + m₂v₂.

Since the skaters are at rest initial momentum is 0.

m₁v₁ + m₂v₂ = 0

⇒60 × 2 + 40 × v₂ = 0

= – 120/40

= – 30 m/s.

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