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A block of mass [tex]m_1 = 2.0kg[/tex] slides along a frictionless table with a speed of 10m/s. Directly in front of it, and moving in the same direction, is a block of mass [tex]m_2 = 5.0kg[/tex] moving at 3.0m/s. A massless spring with a spring constant k=1120N/m is attached to the backside of [tex]m_2[/tex]. when the block collide, what is the maximum compression of the spring? Assume that the spring does not bend and always obeys Hooke's law.

Ans: 0.25 m

Let K= spring constant

e= extension

Here's how i tried to do,

i calculate that relative speed of [tex]m_1[/tex] to [tex]m_2[/tex] = 7m/s

then using this value calculate the KE and equate it to E= [tex]\frac{1}{2} k e^2[/tex].

but the ans is wrong.

Then,

i used the conservation of inelastic collision formula,

[tex] m_1 u_1 + m_2 u_2 = m_(1+2) V[/tex]

used the velocity V and calculate the KE and Equate is into E= [tex]\frac{1}{2} k e^2[/tex] but didn't work.

how should i solve this prob?

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