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The Blocks of Masses 10 Kg and 20 Kg Moving at Speeds of 10 M S−1 and 20 M S−1respectively in Opposite Directions, Approach Each Other and Collide.

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प्रश्न

The blocks of masses 10 kg and 20 kg moving at speeds of 10 m s−1 and 20 m s−1respectively in opposite directions, approach each other and collide. If the collision is completely inelastic, find the thermal energy developed in the process.

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उत्तर

Given:-

Mass of the first block, m1 = 10 kg

Mass of the second block, m2 = 20 kg

Initial velocity of the first block, u1 = 10 m/s

Initial velocity of the second block, u2 = 20 m/s

Let the velocity of the blocks after collision be v.

Applying conservation of momentum, we get

m2u2 − m1u1 = (m1 + m2)v

⇒ 20 × 20 − 10 × 10 = (10 + 20)v

⇒ 400 − 100 = 30 v

⇒ 300 = 30 v

⇒ v = 10 m/s

Initial kinetic energy is given by

`K_i=1/2m_1u_1^2+1/2m_2u_2^2`

`K_i=1/2xx10xx(10)^2+1/2xx20xx(20)^2`

`K_i=500+4000=4500`

Final kinetic energy is given by

`K_f=1/2(m_1+m_2)"v"^2`

`K_f=1/2(10+20)(10)^2`

`K_f=(30/2)xx100=1500`

∴ Total change in KE = 4500 J − 1500 J = 3000 J

Thermal energy developed in the process = 3000 J

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पाठ 25: Calorimetry - Exercises [पृष्ठ ४७]

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एचसी वर्मा Concepts of Physics Volume 1 and 2 [English]
पाठ 25 Calorimetry
Exercises | Q 15 | पृष्ठ ४७

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