A cannon of mass M = 200 kg is positioned a top a narrow wall as shown. It fires a ball of mass m = 2kg horizontally across a plane. Unfortunately the gunners forgot to lock the frictionless wheels of the cannon and it immediately rolls backwards off the wall, landing a distance 8 m from the wall as shown. Neglecting air friction, at what distance x (in meters) from the base of the wall does the ball land ?

Engineering

Physics

Collision

Momentum and Energy Conservation for System of Particles

Question

A cannon of mass M = 200 kg is positioned a top a narrow wall as shown. It fires a ball of mass m = 2kg horizontally across a plane. Unfortunately the gunners forgot to lock the frictionless wheels of the cannon and it immediately rolls backwards off the wall, landing a distance 8 m from the wall as shown. Neglecting air friction, at what distance x (in meters) from the base of the wall does the ball land ?

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By conservation of momentum
       – 200 × v₁ + 2 × v₂ = 0
       100 v₁ = v₂    ... (1)
       v₁ = v₂ / 100

$(\frac{v_{2}}{100}) \cdot t = 8$

v₂ · t = x

$x = v / 2 \cdot \frac{800}{v_{2}}$

x = 800 m