A steel rod is initially at \( 20^\circ\mathrm{C} \) and has a length of \( 2\,\mathrm{m} \). It has a cross-sectional area of \( 0.001\,\mathrm{m^2} \) If it is heated to \( 120^\circ\mathrm{C} \), how large a force must be applied to its end to restore the original length? Use \( 12\times10^{-6}\mathrm{(C^\circ)^{- 1}} \).​

Answermap (2) two

Part a:   using momentum conservation we have:   direction x:   (15) (7.0) + 0 = (15) (4.2) cos (-20 °) + (10) v₂ (f) cosθ  (10) v₂ (f) cosθ = (15) (7.0) - (15) (4.2) cos (-20 °) (eq 1)  direction y:   0 + 0 = (15) (4.2) sin (-20 °) + (10) v₂ (f) sinθ  (10) v₂ (f) sinθ = - (15) (4.2) sin (-20 °) (eq 2)  we divide both equations (1) and (2):   left side:   (10) v₂ (f) sinθ / (10) v₂ (f) cosθ = tanθ  right side:   - (15) (4.2) sin (-20 °) / (15) (7.0) - (15) (4.2) cos (-20 °)  clearing the angle we have:   θ = tan⁻¹ [- (15) (4.2) sin (-20 °) / (15) (7.0) - (15) (4.2) cos (-20 °)]  θ = 25 °  we use any of the equations to find the speed.  from (2) we have:   (10) v₂ (f) sinθ = - (15) (4.2) sin (-20 °)  v₂ = - (15) (4.2) sin (-20 °) / (10) sin25 °  answer:   v₂ = 5.1m / s  note:     θ = 25 ° (direction is in the first quadrant, north of east)  part b:   initial kinetic energy:   k (i) = 0.5 (15) (7.0) ²  = 370j (rounded)  final kinetic energy:   ke (f) = 0.5 (15kg) (4.2m / s) ² + 0.5 (10kg) (5.1m / s) ²  = 260j (rounded)  answer:     a difference of 110j, so it was an inelastic collision.