Question

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.

Answer 1

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,

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