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A steel ball of mass m 1 = 1 kg moving with velocity 50 m s - 1 collides with another ball of mass m 2 = 200 g lying on the ground. Due to collision, the KE is lost and their internal energies change equally and T 1 and T 2 are the temperature changes of masses m 1 and m 2 , respectively. If the specific heat of steel is unity and J = 4 . 18 J cal - 1 , then

Options

  1. AT 1   =   7 . 1   ° C and T 2   =   1 . 47   ° C
  2. BT 1   =   1 . 47   ° C and T 2   =   7 . 1   ° C
  3. CT 1   =   3 . 4   ° C and T 2   =   17 . 0   ° C
  4. DT 1   =   17 . 0   ° C and T 2   =   3 . 4   ° C

Correct answer

C. T 1   =   3 . 4   ° C and T 2   =   17 . 0   ° C

Step-by-step solution

Half of KE is attained as heat by each ball 1 2 KE = m 1 s 1 T 2 ⇒ 1 2 × 1 × 5 0 2 = 1 × 0.105 × 4 1 8 × 1 0 3 × 1 0 3 × T 1 T 1 = 5 0 × 5 0 2 × 0.105 × 4.18 × 1 0 3 = 2 5 2.1 × 4.18 = 2 5 8.778 ≅ 3.4 K As m 2 = m 1 5 , so T 2   =   5 T 1   =   17   K

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