A 60.0 kg girl stands up on a stationary floating raft and decides to go into shore. She dives off the 180 kg floating raft with a velocity of 4.0 m/s [W]. Ignore the substantial friction real objects in water experience.

What is the speed of the raft just after the girl dives?

conserve momentum. With everything initially at rest, momentum is zero.

So, the raft moves in the opposite direction when she jumps.

So, the speed s of the raft is such that

180s = 60*4

I note that you asked for speed, so I ignored the direction of the velocity. I expect you can figure that out.

To find the speed of the raft just after the girl dives, we can use the principle of conservation of momentum. According to this principle, the total momentum before the girl dives is equal to the total momentum after she dives.

The momentum of an object is defined as the product of its mass and velocity.

Before the girl dives, the raft and the girl are stationary, so their total momentum is zero. After the girl dives, the momentum of the girl is given by her mass (60.0 kg) multiplied by her velocity (4.0 m/s to the west), which gives a momentum of 240 kg*m/s to the west.

Since the total momentum before the girl dives is zero, the total momentum after she dives must also be zero. Therefore, the momentum of the raft after the girl dives must be equal in magnitude but in the opposite direction. Let's denote the velocity of the raft as v.

The momentum of the raft is given by its mass (180 kg) multiplied by its velocity (v). So, we have:

Momentum of the raft = 180 kg * v

Since the total momentum after the girl dives is zero, we can write:

Momentum of the girl + Momentum of the raft = 0

240 kg*m/s - 180 kg * v = 0

Simplifying the equation, we get:

240 kg*m/s = 180 kg * v

Dividing both sides by 180 kg, we find:

v = 240 kg*m/s / 180 kg

v = 1.33 m/s

So, the speed of the raft just after the girl dives is 1.33 m/s to the east.