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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
and the absolute value of the final moment is
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 `
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.
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.
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`
Posted by valentin68 on October 21, 2013 at 6:08 AM (Answer #1)
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