# If a block attached to a spring fixed at one end is given a velocity perpendicular to the spring, will the spring eventually stretch?

Consider an ideal spring which is fixed to the ground (frictionless) at one end, and has a block (of mass $$m$$, say) attached to it at the other end. When the spring is in its natural length, the block is given a velocity perpendicular to the spring. After this, what would happen?

(This is supposed to be top-down view. The squiggle connecting the block and circle is supposed to be a spring, say with stiffness constant $$K$$, and natural length $$l$$).

1. Let's assume that the spring stretches. This means that the system would gain potential energy with magnitude $$\cfrac{1}{2}Kx^2$$ (where $$x$$ is the stretch in the spring). Since energy of system is conserved, it must lose its kinetic energy. However, the only force acting on the block is perpendicular to its velocity, so it shouldn't affect its speed and thus its kinetic energy should remain unchanged. This means the spring shouldn't stretch.

2. Let's assume the spring doesn't stretch. This would imply that the block undergoes uniform circular motion, which requires a force of magnitude $$\cfrac{mv^2}{l}$$ to act continously towards the centre. But an unstreched spring can't exert any force on the block. This means that the spring should stretch.

So, will the spring stretch or not? The 2nd scenario seems outright impossible to me, but I don't know what is the flaw in my reasoning about the 1st scenario either.

• Initially, the block moves vertically upwards. Then you should see that by Pythagorean theorem, the spring stretches. The spring then pulls backward diagonally, and diagonal forces have vertical components, i.e. the block will slow down. Jan 23 at 16:52

The spring stretches. Your reasoning for case #1, suggesting that the force of the spring only acts perpendicular to motion, is only true at one point in time, $$t=0$$. The instant the block begins to move, the spring applies no force at all and the block moves directly north. Just after $$t=0$$, the spring has a component pointed in the vertical direction, aligned with the velocity of the block.