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Ivahew [28]
3 years ago
5

The gun, mount, and train car of a railway had a total mass of 1.22 x 10^6 kg. The gun fired a projectile that was 80 cm in diam

eter and weighed 7502 kg. In the firing, the gun has been elevated 20 degrees above the horizontal. If the railway gun at rest before firing and moved to the right at a speed of 4.68 m/s immediately after firing, what was the speed of the projectile as it left the barrel (muzzle velocity)? How far will the projectile travel if air resistance is neglected? Assume that the wheel axles are frictionless.
Physics
1 answer:
kakasveta [241]3 years ago
5 0
1) According to the law of conservation of momentum .. 
<span>Horiz recoil mom of gun (M x v) = horiz. mon acquired by shell (m x Vh) </span>

<span>1.22^6kg x 5.0 m/s = 7502kg x Vh </span>
<span>Vh = 1.22^6 x 5 / 7502 .. .. Vh = 813 m/s </span>

<span>Barrel velocity V .. .. cos20 = Vh / V .. ..V = 813 /cos20 .. .. ►V = 865 m/s </span>

<span>2) Using the standard range equation .. R = u² sin2θ /g </span>

<span>R = 865² x sin40 / 9.80 .. .. ►R = 49077 m .. (49 km)</span>
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Please, see the attached figure for a better understanding of the problem. Notice that the origin of the frame of reference is located at the launching point so that x0 and y0 = 0. In a horizontal launch, initially there is no vertical velocity, then, v0y = 0.

When the cat reaches the ground, the position vector of the cat will be r1 (see figure). The vertical component of r1 is -60 m and the horizontal component will be the horizontal distance traveled by the cat (r1x). Then, using the equation of the y-component of the position vector, we can obtain the time of flight and with that time we can obtain the horizontal distance traveled by the cat:

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