# Answer to Question #347462 in Mechanics | Relativity for carl

Question #347462

A 1.72 kg ball rolls on a frictionless horizontal surface at 4.19 m/s and collides with a stationary block of mass 1.23 kg which is attached to horizontal spring whose force constant is 1695 N/m. Right after impact, the ball bounces on the opposite direction at 1.59 m/s. To what distance will the spring be compressed?

1
2022-06-02T10:25:39-0400

Given:

"m_1=1.72\\:{\\rm kg},\\; m_2=1.23\\:{\\rm kg}"

"v_{1x}=4.19\\:{\\rm m\/s},\\; v_{2x}=0"

"v_{1x}'=-1.59\\:{\\rm m\/s},\\; k=1695\\:\\rm N\/m"

The law of conservation of momentum says

"m_1v_{1x}+m_2v_{2x}=m_1v_{1x}'+m_2v_{2x}'"

Hence, the velocity of the block after collision with ball

"v_{2x}'=\\frac{m_1v_{1x}+m_2v_{2x}-m_1v_{1x}'}{m_2}"

"v_{2x}'=\\frac{1.72*1.23-1.72*(-1.59)}{4.19}=1.16\\:\\rm m\/s"

The law of conservation of energy says

"\\frac{m_2v_{2x}'^2}{2}=\\frac{k(\\Delta x)^2}{2}"

Thus, the compression of the spring

"\\Delta x=\\sqrt{\\frac{m_2}{k}}v_{2x}=\\sqrt{\\frac{4.19}{1695}}*1.16=5.76*10^{-2}\\:\\rm m"

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