A helium balloon and its gondola, of mass M, are in the air and stationary with respect to the ground. A passenger, of mass m, then climbs out and slides down a rope with speed v, measured with respect to the balloon. With what speed (relative to Earth) does the balloon then move? (Express your answer in terms of the variables M, m, and v.)
To find the speed of the balloon after the passenger slides down the rope, we need to consider the principle of conservation of momentum.
Before the passenger slides down, the total momentum of the system (balloon + gondola + passenger) is zero because the system is stationary. Momentum is defined as the product of mass and velocity, so we can write:
Total momentum before = 0
The momentum of the passenger is given by m * v (mass of the passenger times the speed with respect to the balloon).
After the passenger slides down, we can consider the system as two parts: the balloon + gondola and the passenger. Now, the total momentum is no longer zero because the passenger has momentum in one direction.
Total momentum after = Momentum of balloon + gondola + Momentum of passenger
Since no external forces act on the system, the total momentum is conserved. Therefore, we can equate the initial and final momenta:
0 = Momentum of balloon + gondola + Momentum of passenger
0 = (M + m) * Vfinal + m * Vrelative
Vfinal represents the speed of the balloon with respect to the Earth after the passenger has slid down, and Vrelative represents the relative speed of the passenger with respect to the balloon.
Now, we can rearrange the equation to solve for Vfinal:
-(M + m) * Vfinal = m * Vrelative
Dividing both sides by -(M + m), we get:
Vfinal = - (m / (M + m)) * Vrelative
Thus, the speed of the balloon (Vfinal) relative to the Earth is given by -(m / (M + m)) * Vrelative.