Question #593cb

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Question #593cb

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Answer 1 · 2015-10-12T21:02:14

Force $(\to F)$ $(vecF)$ is a push or a pull and is measured in Newtons (N)

Impulse $(\to I)$ $(vec I)$ gives a measure of how long the force acts on the objects and is equal to the change in momentum brought about.
Its units is hence either #N.s or kg.m//s#

From Newton's 2nd Law of Motion, the resultant force acting on an objects is equal tot he rate of change of momentum brought about, ie
$\sum \to F = \frac{d \to p}{d t} = \frac{d}{d t} (m \to v)$ $sum vecF=(dvecp)/(dt)=d/dt(mvecv)$
Hence $\to F d t = \to I = m d \to v$ $vecFdt = vecI = m d vecv$

Answer 2 · 2015-10-12T21:05:55

The impulse gives you the average over time of a rapidly varying force.

Explanation:

Consider a force $F (t)$ $F(t)$ not constant but function of time such as the one experienced by an object (tennis ball) during an impact:
The force varies rapidly between two instants $t_{1}$ $t_1$ and $t_{2}$ $t_2$ so that:
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where $F_{m}$ $F_m$ is the mean value of the force so that the grey square area is equal to the red total area under the original curve.

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Question #593cb

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