Question

A ball is dropped on a floor from a height of 2.0 m. After the collision it rises up to height of 1.5 m. Assume that 40 % of the mechanical energy lost goes as thermal energy into the ball. Calculate the rise in the temperature of the ball in the collision. Heat capacity of the ball is 800 J $K^{-1}$

Answer 1

Let the mass of ball be M kg

$h_1=2m\left(heightbeforehittingthesurface\right)$

$h_2=1.5m\left(heightafterhittingthesurface\right)$

Heat capacity of the ball is 800 J $K^{-1}$

As we know by 3rd equation of motion

$v^2-u^2=2gh;$

$asballisdroppedfromhheight,thereforenoadditionalenergyisimpartedontheball,henceu=0$

$v_1^2-u_1^2=2g{h}_1$

$v_1^2-0=2g{h}_1$

$v_1^2=2g{h}_1=40\frac{m}{s}$

;

$v_1=velocityoftheballjustbeforehittingground$

$aftercollisionwiththesurface,theballlosessomeofitsenergysincetheballonlyreachestoaheight,h=1.5maftercollision,$

$wecancalculatethevelocityoftheballjustafterthecollisionwithwhichitrisesaboveforaheight,h=1.5m,$

$thereforeusing3rdequationofmotion$

$v^2-u^2=2gh$

$0-u_2^2=2(-g)h_2$

$u_2^2=2g{h}_2$

$u_2^2=2g{h}_2=30\frac{m}{s}$

after collision with the surface, the ball loses some of its energy

$\Delta KE=K{E}_f-K{E}_i\Rightarrow \frac{1}{2}M(v_f^2-v_i^2)\Rightarrow \frac{1}{2}M(30-40)$

$\Delta KE=\frac{M}{2}(-10)J$

$\Delta KE=-5MJoule\left(herenegativerepresentlossinenergy\right)$

This loss of energy at a contact point utilized to increase the temperature of the ball

From question,

As 40% of mechanical loss(energy loss) must be equivalent to raise the temperature of the ball, therefore

$Q=MS\Delta T$

$40\%ofheatlossmeans\Rightarrow \frac{40}{100}\times \left(5M\right)=M\left(800\right)\Delta T$

$\Delta T=\frac{1}{400}$

$\Delta T=.0025$

$\Delta T=2.5\times {10}^{-3_0}c$