Question

A block of mass, m traveling on a frictionless surface at speed, v impacts an ideal spring and is brought to rest when the spring is compressed by a displacement, z. If the initial speed of the block is doubled, what happens to the displacement of the spring?

1) The displacement of the spring remains the same
2) The displacement of the spring is halved
3) The displacement of the spring is doubled
4) The displacement of the spring is quadrupled

Answer 1

If the initial speed of the block is doubled, the displacement of the spring is quadrupled due to the energy conservation principle.

Step-by-step explanation:

The question asks about the relationship between the initial speed of a block and the displacement of a spring that brings the block to rest. According to the conservation of energy principle, the initial kinetic energy of the block is converted to the potential energy stored in the spring when it is compressed. The kinetic energy of the block is given by KE = 1/2 m v^2, where m is the mass and v is the velocity of the block. The potential energy stored in the spring is given by PE = 1/2 k z^2, where k is the spring constant and z is the displacement of the spring. Setting these equal to each other, we get 1/2 m v^2 = 1/2 k z^2. Now, if the initial speed v is doubled, v becomes 2v, and the new equation would be 1/2 m (2v)^2 = 1/2 k z'^2. Simplifying, we get 2m v^2 = k z'^2, meaning that the new displacement z' equals to the initial displacement z times 2, which means the displacement of the spring is quadrupled. Therefore, if the initial speed of the block is doubled, the displacement of the spring is quadrupled.