Question

A block oscillating up and down on a spring comes to the top of its path, where it is momentarily at rest. At the instant it is at rest, what is the direction of the rate of change of its momentum?

Answer 1

Opposite to the direction of the velocity which led it to its current position.

Explanation:

The direction of momentum when a vertically oscillating block comes to the rest momentarily will be opposite to the direction of the velocity that it has just followed to reach reach its current position.

The direction of change in momentum at the bottom will be upwards and at the top will be downwards.

The change in momentum is mathematically defined as:

$\Delta P=m.v_f-m.v_i$

where:

$m=$ mass of the block

$v_f=$ final velocity of the block

$v_i=$ initial velocity of the block

When the block comes to rest it is due to the result of continuously decreasing velocity.