A long spring with a spring constant of 75.0 N/m is sitting vertically with one end attached to the ground. A 10.0 kg block (block 1) sits at rest on top of the spring. A 4.00 kg block (block 2) is held at rest, 1.20 m above block 1 as it sits on top of the spring. Block 2 is released and lands on top of block one, sticking to it. a) What is the compression of the spring with block 1 sitting on it?

b) How fast are the two blocks moving immediately after their collision? (hint: this is a collision… what is conserved in every collision?)

(a) The 10.0 kg block is in equilibrium and the restoring force s of the spring has the same magnitude as the block's weight mg :

∑ F = s - mg = 0   ==>   s = mg = (10.0 kg) (9.80 m/s²) = 98.0 N

With spring constant k = 75.0 N/m, it follows by Hooke's law that the spring is compressed a distance x such that

s = kx   ==>   98.0 N = (75.0 N/m) x   ==>   x ≈ 1.31 m

(b) When the 4.00 kg block is 1.20 m above the first block, it has potential energy

P = (4.00 kg) (9.80 m/s²) (1.20 m) = 47.04 J

which gets converted to kinetic energy as it falls. Ignoring air resistance, the block's kinetic energy K at the moment it reaches the first block is the same as its potential energy P before beginning to fall (i.e. energy is conserved), so that

47.04 J = 1/2 (4.00 kg) v ²

Solve for the block's speed v :

==>   v ≈ 2.42 m/s

Immediately prior to the blocks' collision, the first block has zero momentum (the quantity that the hint is referring to) because it's at rest, while the second block has momentum

(4.00 kg) (2.42 m/s) ≈ 9.70 kg•m/s

The collision is inelastic, and since momentum is conserved, the sum of the blocks' individual momenta is equal to the momentum of the two blocks combined:

0 + 9.70 kg•m/s = (10.0 kg + 4.00 kg) v

where v is now the speed of the two blocks stuck together. Solving for v gives

==>   v ≈ 0.693 m/s

the independent variable, or the one that is intentionally manipulated is the speed.