Question

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 ????1 = 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?

Answer 1

Answer:

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

$\dfrac{1}{2}mv^2 = \dfrac{1}{2}kx^2$                  

$\dfrac{1}{2}\times (3)\times 2.67^2 = \dfrac{1}{2}\times 200 \times x^2$

x = 0.327 m

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