Asystem 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.

with what speed does the cart recoil if the projectile rises 2.3 m at its maximum height?

0.95 m/s

Explanation:

Find the vertical component of the projectile's initial velocity.

Given:

Δy = 2.3 m

v = 0 m/s

a = -9.8 m/s²

Find: v₀

v² = v₀² + 2aΔy

(0 m/s)² = v₀² + 2 (-9.8 m/s²) (2.3 m)

v₀ = 6.71 m/s

Find the horizontal component of the projectile's initial velocity using trigonometry.

tan 40° = 6.71 m/s / v

v = 8.00 m/s

Momentum is conserved:

(0.47 kg) (0 m/s) = (0.050 kg) (8.00 m/s) + (0.47 kg − 0.050 kg) v

v = -0.95 m/s

The cart recoils at 0.95 m/s.

answer: good job it's all correct !

explanation: