From a boat weighing 200 kg, moving at a speed of 1 m / s, a load of 100 kg fell out. What is the speed of the boat.

ml = 200 kg.

Vl1 = 1 m / s.

mg = 100 kg.

Vl2 -?

Momentum conservation law
For a closed system, which consists of a boat and cargo, the total impulse remains unchanged.

The law of conservation of momentum in vector form will have the form: pl1 + pg1 = pl2 + pg2, where

rl1 – initial impulse of the boat;
pr1 is the initial impulse of the load;
rl2 is the final impulse of the boat;
pr2 is the final impulse of the load.
Body impulse
The momentum of a body p is a vector physical quantity equal to the product of the body’s velocity V by its mass m.

The momentum of the body p is determined by the formula: p = m * V.

The initial and final impulses for the boat and cargo will look like:

pl1 = ml * Vl1;
pr1 = mg * Vg1;
pl2 = ml * Vl2;
pr2 = mg * Vg2.
Let’s write the law of conservation of momentum:

ml * Vl1 + mg * Vg1 = ml * Vl2 + mg * Vg2.

Since after the drop of the load from the boat, the load will have a speed equal to 0, Vg2 = 0, then the law of conservation of momentum will have the form:

ml * Vl1 + mg * Vg1 = ml * Vl2.

Let us express the speed of the boat Vl2 after the drop of the load.

Vl2 = (ml * Vl1 + mg * Vg1) / ml.

Since the initial speeds of the boat and the load are the same Vl1 = Vg1, the load was in the boat, the formula for determining the speed of the boat Vl2 after ejection of the load will be determined by the formula:

Vl2 = (ml * Vl1 + mg * Vl1) / ml = Vl1 * (ml + mg) / ml.

Substitute the numerical values ​​from the problem statement into the formula.

Vl2 = 1 m / s * (200 kg + 100 kg) / 200 kg = 1.5 m / s.

Answer: after ejection of the cargo, the speed of the boat will increase and become Vl2 = 1.5 m / s.