Question

A cannon is mounted on a cart which sits on the ground, supported by frictionless wheels. The mass of the cannon and cart is 4.65 kg, and it is initially traveling at 2.00 m/s. A 50.0 g projectile is then launched with a horizontal velocity of 647 m/s from the cannon. What is the velocity of the launcher after the projectile is launched

Answer 1

Answer:

v = -4.96 m/s

Explanation:

The mass of the cart and the cannon, M₁ = 4.65 kg

The initial velocity of the cart, v₁ = 2 m/s

The mass of the projectile, M₂ = 50 g = 0.05 kg

The initial velocity of the projectile, v₂ = 647 m/s

Velocity of the launcher after the projectile is launched, v = ?

Using the principle of momentum conservation:

Momentum of the launcher before the launch = Momentum of the projectile + momentum of the launcher after the launch

M₁v₁  = M₂v₂ + M₁v

(4.65*2) = (0.05*647) + (4.65v)

9.3 = 32.35 + 4.65v

9.3 - 32.35 = 4.65v

-23.05 = 4.65v

v = -23.05/4.65

v = -4.96 m/s

Answer 2

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

$v_3 = \frac{-23.05 \, kg\cdot m/s}{4.6 \, kg} = \frac{-461}{92} m/s$

v₃ = -5.011 m/s.