Block 1, with mass m1 and speed 3.6 m/s, slides along an x axis on a frictionles

Question

Block 1, with mass m1 and speed 3.6 m/s, slides along an x axis on a frictionless floor and then undergoes a one-dimensional elastic collision with stationary block 2, with mass m2 = 0.45m1. The two blocks then slide into a region with a coefficient of kinetic friction of 0.50 where they stop. How far into that region do the two blocks slide?

Explanation / Answer

m1*3.6 = m1v1 + 0.4m1v2 =>3.6 = v1 + 0.4v2--(1) =>v2-v1 = u1-u2 =>v2-v1= 3.6-0--(2) solving 1 and 2 we get 3.6 = v2-3.6 +0.4v2 =>7.2=1.4v2 =>v2 = 5.14 =>v1=1.54 we have deceleration = 0.5*9.8 = 4.9m/s^2 =>1.54^2= 2*4.9*s =>s1=0.242 m (a) is 0.242m =>5.14^2=2*4.9s =>s2=2.69 m (b) is 2.69 m

Related Questions

Navigate

• Browse (All)
• Browse B
• Subjects

---