A WWII weapon known as the PIAT used a spring to throw an armor piercing project

Question

A WWII weapon known as the PIAT used a spring to throw an armor piercing projectile into the side of an enemy tank, without leaving a tell-tale smoke trail behind it. When the weapon was elevated 30 Degree above the horizontal, the spring could propel a 1.1 kg projectile through a Range of 143 meters. What was the spring constant of this spring, assuming it was compressed by 1 meter to arm the weapon? Do you believe one soldier could reload this weapon (recompress the spring)? Why, or why not? Use numerical values to justify your answer.

Explanation / Answer

Range of projectile, R = v^2 sin2@ / g

143 = v^2 sin(2 x 30) / 9.8

v = 40.23 m/s

Using energy conservation,

m v^2 /2 = k x^2 /2

1.1 (40.23^2) = k (1^2)

k = 1780 N/m .......Ans

to compress 1m force needed ,

= kx = 1780(1) = 1780 N
It is quite difficult to exert that kind of force.

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