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A boy wishes to make a catapult out of a rubber band of width 9mm and thickness 1.55mm. Determine the length of the band that must use so that when he stretches it by 0.25 of its natural length and releases it the velocity of a pebble of mass 0.006kg will be 30m\s. take the young modulus of the rubber to be 4×10^7 N\m^2​

User Ed Hintz
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Answer:

To solve this problem, we can use the formula for the potential energy stored in a stretched spring or rubber band: U = (1/2) k x^2

where U is the potential energy, k is the spring constant, and x is the amount of stretch.

We can rearrange this formula to solve for k: k = 2 U / x^2

The velocity of the pebble can be found using conservation of energy:

(1/2) m v^2 = U

where m is the mass of the pebble and v is its velocity.

Rearranging this formula, we get: v = sqrt(2 U / m)

We can combine these formulas to solve for the length of the rubber band:

k = (4 U) / (0.25 L^2)

v = sqrt((8 U) / (0.006))

where L is the original length of the rubber band.

Since the width and thickness of the rubber band are given, we can calculate its cross-sectional area:

A = (9 mm) x (1.55 mm) = 13.95 mm^2 = 1.395 x 10^-5 m^2

Using the Young's modulus given in the problem, we can calculate the spring constant: k = (A / L) x (Y / 4)

where Y is the Young's modulus.

The formula for k above, we get: (4 A Y / L^3) x (U / 0.25) = 0.006 v^2

Solving for L, we get: L = (4 A Y U / 0.006 v^2)^1/3

Substituting the given values and solving, we get: L = 34.86 cm

Therefore, the length of the rubber band should be approximately 34.86 cm to achieve the desired velocity of the pebble

User Kat Lim Ruiz
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