A 74.1 kg high jumper leaves the ground with
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Answer:
The velocity of the launcher after the projectile is launched is -5.011 m/s
Explanation:
Here we have the mass of the cannon and cart, m₁ = 4.65 kg
Velocity of cannon and cart, v₁ = 2.00 m/s
Mass of projectile, m₂ = 50.0 g = 0.05 kg
Velocity of projectile, v₂ = 647 m/s
Velocity of the launcher, v₃ = Required
Mass of cannon and cart, launcher after launching projectile m₃ = 4.65-0.05
= 4.6 kg
Therefore, from the principle of the conservation of linear momentum, we have
Total initial momentum = Total final momentum
m₁ × v₁ = m₂ × v₂ + m₃ × v₃
Substituting gives
4.65 kg × 2.00 m/s = 0.05 kg × 647 m/s + 4.6 kg × v₃
4.65 kg × 2.00 m/s - 0.05 kg × 647 m/s = 4.6 kg × v₃
-23.05 kg·m/s = 4.6 kg × v₃
v₃ = -5.011 m/s.