Question

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.

Answer 1

x=9.5 cm

Step-by-step explanation:

We have here three parts to analyze.

1. 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.
2. The second one, the cylinder A after the collision, whit a velocity v(fA) and cylinder B after the collision whit velocity v(fB).
3. 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:

`m_(A)v_(iA)=m_(A)v_(fA)+m_(B)v_(fB)` (1)

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

`v_(iA)=v_(fA)+v_(fB)`

`2=v_(fA)+v_(fB)` (2)

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

`e=(v_(fB)-v_(fA))/(v_(iA)-v_(iB))`

`0.77=(v_(fB)-v_(fA))/(v_(iA))`

`0.77*v_(iA)=v_(fB)-v_(fA)`

`1.54=v_(fB)-v_(fA)` (3)

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

`v_(fB)=1.77 m/s`

`v_(fA)=0.23 m/s`

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

`(1)/(2)m_(B)v^(2)=(1)/(2)kx^(2)` (4)

here, v is 1.77 m/s.

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

`x=v\sqrt{(m_(B))/(k)}=9.5 cm`

I hope it helps you!