Question

A steel ball of mass m₁ = 1 kg moving with velocity 50 m/sec collides with another steel ball of mass m₂ = 200 gm lying on the ground and both come to rest. During the collision their internal energies changes equally and T₁ and T₂ are the rise in temperature of masses m₁ and m₂ respectively. If s_steel = 0.105 cal/gm ŗC and J = 4.18, then -

• A. T₁ = 7.1⁰C, T₂ = 1.47⁰C
• B. T₁ = 1.42⁰C, T₂ = 7.1⁰C
• C. T₁ = 3.4 K, T₂ = 17.0 K
• D. None of these

Answer 1

Answer: (B)

T₁ = 1.42⁰C, T₂ = 7.1⁰C

Answer 2

Half of KE is shared by each ball $\frac{1}{2}$KE = m₁s₁T₁ = m₂s₂T₂