# instantaneous velocity

**1. The problem statement, all variables and given/known data**

A stone of mass 5 g is projected with a rubber catapult. If the catapult is streched through a distance of 7 cm by an average force of 70 N, calculate the instantaneous velocity of the stone when released.

**2. Relevant equations**

work done in elastic material w = 1/2 fe = 1/2 ke^{2} = 1/2 mv^{2} – the kinetic energy of the moving stone.

(a) 1/2 fe = 1/2 mv^{2}

Or

(b) 1/2 ke^{2} = 1/2 mv^{2}

**3. The attempt at a solution**

mass m = 5g = 0.005kg; extension e = 7cm = 0.07m;

force f = 70 N; velocity = ?;

using:

1/2 fe = 1/2 mv^{2}

v = square root ( fe/m)

v = (70 * 0.07/0.005)

= 31. 304 m/s

Using either of the equations gives the same result. Am not sure wether I have arrived because the book am using gave a different solution for the problem. So you watching at this thread, what do you think?

http://ift.tt/1iCP7uc

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