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poizon [28]
3 years ago
7

A system of inertia 0.47 kg consists of a spring gun attached to a cart and a projectile. The system is at rest on a horizontal

low-friction track. A 0.050-kg projectile is loaded into the gun, then launched at an angle of 40∘ with respect to the horizontal plane.
Physics
1 answer:
Vlad [161]3 years ago
4 0

Answer:

v'=0.83m/a and v=10.2m/s

Explanation:

The information that we have is:

m_1=0.050kg\\m_2=0.47kg\\\theta = 40\\h=2.2m

The maximum height of the projectile is given by the equation

h=\frac{v^2sin^2\theta}{2g}

So, rearrange for the velocity,

v=\sqrt{\frac{2gh}{sin^2\theta}}\\v=\sqrt{\frac{2*9.8*2.2}{sin^2(40)}}\\v=10.2m/s

Apply the conservation of momentum,

mvcos(\theta)=m'v'

Then rearrange the recoil speed,

v'=\frac{mvcos\theta}{m_2}\\v'=\frac{0.05*10.2*cos40}{0.47}\\v'=0.83m/s\\

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