Question

Block 2 with mass m₂=5.0 kg is at rest on a frictionless surface and connected to a spring constant k=64.0 N/m. The other end of the spring is connected to a wall, and the spring is initially at its equilibrium (unstretched) position. Block 1 with mass m₁=10.0 is initially traveling with speed v₁=4.0 m/s and collides with block 2. The collision is instantaneous, and the blocks stick together after the collision. Find the speed of the blocks immediately after the collision AND how far the spring is compressed from equilibrium (d) before the blocks momentarily come to a stop. Select one answer for speed (v) and one answer for distance (d).

Answer 1

Mass of block 1 = m₁ = 10 kg

Mass of block 2= m₂ = 5 kg

Initial velocity of block 1= V₁ = 4 m/s

Initial velocity of block 2 = V₂ = 0 m/s (At rest)

Velocity of both the blocks after the collision = V₃

By conservation of linear momentum,

V₃ = 2.67 m/s

Force constant of the spring connected to block 2 = k = 64 N/m

Distance the spring compresses when the blocks momentarily come to rest = d

By conservation of energy the kinetic energy of the blocks immediately after the collision is converted into the potential energy of the spring as it compresses.

d = 1.29 m

a) Speed of the blocks immediately after the collision = 2.67 m/s

b) Distance the spring compresses when the blocks momentarily come to a stop = 1.29 m