A 4.40-kg model rocket is launched, shooting 53.0 g of burned fuel from its exhaust at an average velocity of 650 m/s. What is the velocity of the rocket after the fuel has burned? (Ignore effects of gravity and air resistance.)
To find the velocity of the rocket after the fuel has burned, we can use the law of conservation of momentum. According to this law, the total momentum of an isolated system remains constant before and after an event (in this case, when the fuel is burned and expelled from the rocket).
The equation for conservation of momentum is:
m1v1 + m2v2 = m1v1' + m2v2'
Where,
m1 and v1 are the mass and velocity of the rocket before the fuel is burned,
m2 and v2 are the mass and velocity of the burned fuel,
m1 and v1' are the mass and velocity of the rocket after the fuel is burned, and
m2 and v2' are the mass and velocity of the expelled fuel.
Given:
m1 = 4.40 kg (mass of the rocket)
v1 = unknown (initial velocity of the rocket)
m2 = 53.0 g = 0.053 kg (mass of the burned fuel)
v2 = 650 m/s (velocity of the burned fuel)
Since the rocket was at rest before the fuel is burned, the initial velocity of the rocket (v1) is zero.
Using the conservation of momentum equation, we can solve for v1':
m1v1 + m2v2 = m1v1' + m2v2'
(0 kg)(0 m/s) + (4.40 kg)(0.053 kg)(650 m/s) = (4.40 kg)(v1') + (0.053 kg)(650 m/s)
0 + (0.235 kg⋅m/s) = 4.40 kg(v1') + (0.034 kg⋅m/s)
0.235 kg⋅m/s = 4.40 kg(v1') + 0.034 kg⋅m/s
Rearranging the equation and isolating v1':
4.40 kg(v1') = 0.235 kg⋅m/s - 0.034 kg⋅m/s
4.40 kg(v1') = 0.201 kg⋅m/s
v1' = (0.201 kg⋅m/s) / 4.40 kg
v1' ≈ 0.0457 m/s
Therefore, the velocity of the rocket after the fuel has burned is approximately 0.0457 m/s.