Cylinder A is moving to the right with speed v = 2.0 m/s when it impacts the initially stationary cylinder B. Both cylinders have mass m = 2.2 kg, and the coefficient of restitution for the collision is e = 0.77. Determine the maximum deflection δ of the spring of modulus k = 760 N/m. Neglect friction.

The first one related to the cylinder A moving to the right with speed v(iA) = 2.0 m/s with the cylinder B at rest, v(iB) = 0.
The second one, the cylinder A after the collision, whit a velocity v(fA) and cylinder B after the collision whit velocity v(fB).
The third one, the cylinder B compresses the spring. The velocity of the cylinder B at this moment will be zero.

Let's use conservation of momentum in the first part, to find the one equation:

[tex]m_{A}v_{iA}=m_{A}v_{fA}+m_{B}v_{fB}[/tex] (1)

Where m(A) = m(B) = 2.2 kg, so the masses canceled out from equation (1):

[tex]v_{iA}=v_{fA}+v_{fB}[/tex]

[tex]2=v_{fA}+v_{fB}[/tex] (2)

We can use the coefficient of restitution to find the second equation:

[tex]e=\frac{v_{fB}-v_{fA}}{v_{iA}-v_{iB}}[/tex]

[tex]0.77=\frac{v_{fB}-v_{fA}}{v_{iA}}[/tex]

[tex]0.77*v_{iA}=v_{fB}-v_{fA}[/tex]

[tex]1.54=v_{fB}-v_{fA}[/tex] (3)

We can find v(fA) and v(fB) combining (3) and (4) and solving the system of equations:

[tex]v_{fB}=1.77 m/s[/tex]

[tex]v_{fA}=0.23 m/s[/tex]

Now, using the conservation of energy, related to the cylinder B, we have:

[tex]\frac{1}{2}m_{B}v^{2}=\frac{1}{2}kx^{2}[/tex] (4)

here, v is 1.77 m/s.

Solving the equation (4) for x, we have:

[tex]x=v\sqrt{\frac{m_{B}}{k}}=9.5 cm[/tex]

I hope it helps you!