Q: A ball of mass 1 kg is dropped on a floor from a height of 2.0 m. after the collision it rises up to a height of 1.5 m. assume that 40% of the mechanical energy lost goes as thermal energy into the ball. The rise in the temperature of the ball is : (Heat capacity of the ball is 800 J/K)

(a) 10^{-3} °C

(b) 1.5 × 10^{-3} °C

(c) 2 × 10^{-3} °C

(d) 2.5 × 10^{-3} °C

Ans: (d)

Sol: 40% of Loss in Potential Energy = Thermal Energy

$\large \frac{40}{100} \times m g \Delta h = m’ s \Delta T$

$\large 0.4 \times 1 \times 10 \times (2-1.5) = 800 \Delta T$

ΔT = 2.5 × 10^{-3} °C