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Physic I

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Gulcan4 | eNotes Newbie

Posted October 20, 2013 at 10:21 PM via web

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Physic I

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valentin68 | College Teacher | (Level 3) Associate Educator

Posted October 21, 2013 at 6:08 AM (Answer #1)

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a) The linear momentum of an mass m that is moving at speed v is a vector quantity that is equal to the product between the mass and the speed vector.

P= m*v   (written as vectors).

Thus the absolute value of the initial moment is

`P_i =m*v_i`

and the absolute value of the final moment is

`P_f =0`

since its speed is zero.

The magnitude of the momentum change is simply

`Delta(P) =P_f-P_i =0-mv_i =-m*v_i `

b)

Since there is no time involved into the definition of momentum, the total momentum change is the same regardless you catch the egg more slowly or not.

c)

I would prefer to catch the egg more slowly. This is because when catching more slowly my hand would apply a smaller deceleration to the egg, and thus (as second principle says) the egg will be subjected to a smaller force.

d)

From the theorem of momentum one can write for the expression of applied external force

`F = (Delta(P))/(Delta(t))`   (1)

Therefore the average force one exerts is inversely proportional to the time `Delta(t)` during which it is applied. For `Delta(t)` small the force is big and the egg might break. The above expression of force (1) is equivalent to what has been said at point c):

`F = -(mv_i)/(Delta(t)) = m*(0-v_i)/(Delta(t)) = m*a`

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