Question

Block of mass 10 kg is hit and penetrated by a 50 g bullet. As result of the collision the “block + bullet system.” (Originally on the horizontal frictionless table) enters the horizontal rough surface ( μkin=0.7) and stops after 5m. Find the entropy change due to the inelastic collision (∆Scollision) and the entropy change due to friction (∆Sfrict),. The temperature remains constant throughout these processes (T=27C)

Answer 1

let V is the velocity of bullet-block system after the collsions.

acceleration on rough surfcae, a = -g*mue_k

= -9.8*0.7

= -6.86 m/s^2

d = 5 m

now use, vf^2 - vi^2 = 2*a*d

0^2 - V^2 = 2*(-6.86)*5

==> V = sqrt(2*6.86*5)

= 8.28 m/s

let u is the speed of bullet before the impact

apply conservation of momentum

m*u = (m + M)*V

u = (m + M)*V/m

= (0.05 + 10)*8.28/0.05

= 1664 m/s

delta_S_collsion = delta_KE/T

= ( 0.5*m*u^2 - (0.5*(m+M)*V^2)/T

= (0.5*0.05*1664^2 - 0.5*(0.05+10)*8.28^2)/(27+273)

= 229.6 J/K <<<<<<<<<<------------------Answer

delta_S_frict = ((0.5*(m+M)*V^2 - 0)/T

= ( 0.5*(0.05+10)*8.28^2 - 0)/(27 + 273)

= 1.15 J/K <<<<<<<<<<------------------Answer