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2 years ago

6

A 8.0 n force acts on a 0.70-kg object for 0.50 seconds. by how much does the object's momentum change (in kg-m/s)? (never inclu

de units in the answer to a numerical question.)

1 answer:

inysia [295]2 years ago

4 0

For Newton's second law, the force applied to an object is equal to the product between the mass of the object and its acceleration:
$F=ma$
Rewriting the acceleration as the increment of velocity $\Delta v$ in a time $\Delta t$ : $a= \frac{\Delta v}{\Delta t}$ , F becomes
$F=m \frac{\Delta v}{\Delta t}$
But given the definition of momentum: $p=mv$ , then $m \Delta v$ represents the momentum change. So we can rewrite F as
$F= \frac{\Delta p}{\Delta t}$
And re-arranging the formula we can calculate the value of the change in momentum:
$\Delta p = F \Delta t=(8.0 N)(0.50 s)=4 kg m/s$

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