3. With a mass of 114 kg, Baby Bird is the smallest monoplane ever flown. Suppose the Baby Bird and pilot are coasting along the runway when the pilot jumps horizontally to the runway behind the plane. The pilot’s velocity upon leaving the plane is 5.32 m/s backward. After the pilot jumps from the plane, the plane coasts forward with a speed of 3.40 m/s. If the pilot's mass equals 60.0 kg, what is the velocity of the plane and pilot before the pilot jumps?

Calculate the total momentum before the pilot jumps:
Calculate the total momentum after the pilot jumps:
- Momentum of plane after the jump = (114 kg) * 3.40 m/s = 387.6 kg*m/s
- Momentum of pilot after the jump = (60 kg) * (-5.32 m/s) = -319.2 kg*m/s (negative because the pilot jumps backward)
- Total momentum after the jump = 387.6 kg*m/s - 319.2 kg*m/s = 68.4 kg*m/s
Set the total momentum before the jump equal to the total momentum after the jump:
- (174 kg) * v = 68.4 kg*m/s
- v = 68.4 kg*m/s / 174 kg
- v = 0.393 m/s

Therefore, the velocity of the plane and pilot before the pilot jumps is 0.393 m/s.

Qwtea Qwtea · Answer 2 · 2024-08-03T02:52:59-04:00

Final answer:

The velocity of the plane and the pilot before the pilot jumps is -0.48 m/s. The negative sign indicates the direction is backwards relative to the plane's motion.

Explanation:

To solve this problem, we apply the principle of conservation of momentum. Before the pilot jumps, the momentum of the plane and the pilot together is their total mass times their velocity. This momentum must be conserved after the pilot jumps.

The formula for momentum is mass times velocity, so we get:

114 kg x v + 60 kg x v = 114 kg x 3.40 m/s + 60 kg x (-5.32 m/s)

Where v is the speed of the plane and the pilot before the pilot jumps.

Solving this equation gives us the velocity v = -0.48 m/s, which is backwards relative to the direction of the plane's motion.

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