Block 2 of mass 1.00 kg is at rest on a frictionless surface and touching the end of an unstretched spring of spring constant 200 n/m. the other end of the spring is fixed to a wall. block 1 of mass 2.00 kg, traveling at speed = 4.00 m/s, collides with block 2, and the two blocks stick together. when the blocks momentarily stop, by what distance is the spring compressed?

x = 0.327 m

Explanation:

Block 2 of mass 1.00 kg is at rest

spring constant = 200 N/m            

Block 1 of mass 2 kg moving at 4 m/s

m₁ v₁ = (m₁ + m₂)V                                

2 x 4 = (2 + 1) V                                                      

V = 2.67 m/s                                                        

loss of kinetic energy = gain elastic potential energy

x = 0.327 m

hence, the spring compressed distance is equal to x = 0.327 m

see earlier answer

a cold ice-tea.

pour in room temperature or cold water. cover the pitcher and refrigerate. steep white or green tea for 6 to 8 hours; steep black or oolong tea for 8 to 12 hours.jun 28, 2015