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.
a) What is the momentum of the girl as she is diving?
b) What is the momentum of raft as the girl is diving?
c) What is the final speed of the raft just after the girl dives?
a) * Let [W] be +ive.
p=mv
=(60.0kg)(4.0m/s)
=240 kg/(m/s)
Correct?
b) a little confused about this one, is it 0 because the raft doesn't have a velocity? or is it (180kg)(4.0m/s) because the raft will have a velocity equal and opposite to the girl?
c) for this one, i'm going to use p=p' (initial momentum=final momentum), and then work it out to be mv1 + mv2 = mv1' + mv2' where 1 is the girl, and two is the raft..
then work it out so both initial momentum's equal zero because they have no velocity (from rest), sub in the rest of the information and solve for v2'?
thanks in advance!
a right
b> conservation of momentum says that the raft has equal and opposite momentum from the girl.
c. set the momentums equal, and solve for raft velocity.
a) Yes, you are correct. The momentum of the girl as she is diving is given by:
p = mv = (60.0 kg)(4.0 m/s) = 240 kg/(m/s)
b) The momentum of the raft as the girl is diving is not 0. According to the law of conservation of momentum, the total momentum of an isolated system remains constant. When the girl dives off the raft, she exerts a force on the raft, causing it to move in the opposite direction. The magnitude of the momentum of the raft is the same as the magnitude of the momentum of the girl but in the opposite direction. Therefore, the momentum of the raft is:
p_raft = -(180 kg)(4.0 m/s) = -720 kg/(m/s)
c) To find the final speed of the raft just after the girl dives, you can use the principle of conservation of momentum. Initially, both the girl and the raft are at rest. After the girl dives, the momentum of the system is conserved.
Let's denote the final velocity of the raft as v_raft' and solve for it using the equation:
m_girl * v_girl + m_raft * v_raft = 0
(60.0 kg)(4.0 m/s) + (180 kg)(0 m/s) = (60.0 kg)(0 m/s) + (180 kg)(v_raft')
240 kg/(m/s) = 180 kg/(m/s) * v_raft'
v_raft' = 240 kg/(m/s) / 180 kg = 1.33 m/s
a) Yes, you are correct. The momentum of the girl as she is diving is calculated using the formula p = mv, where m is the mass of the girl (60.0 kg) and v is her velocity (4.0 m/s). Therefore, the momentum is p = (60.0 kg)(4.0 m/s) = 240 kg/(m/s).
b) The momentum of the raft as the girl is diving can be determined by considering the conservation of momentum. Since the girl and the raft are initially at rest, the total momentum before the girl dives is zero. According to the law of conservation of momentum, the total momentum after the girl dives must also be zero. Therefore, the momentum of the raft can be calculated as the negative of the momentum of the girl. Thus, the momentum of the raft is -(60.0 kg)(4.0 m/s) = -240 kg/(m/s).
c) To find the final speed of the raft just after the girl dives, you 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 the girl dives. Since the total momentum before the girl dives is zero, we can write the equation:
(m1)(v1) + (m2)(v2) = 0,
where m1 and v1 represent the mass and velocity of the girl, and m2 and v2 represent the mass and velocity of the raft, respectively.
Since the girl is diving with a velocity of 4.0 m/s to the west (represented by [W]), her momentum is -(60.0 kg)(4.0 m/s) = -240 kg/(m/s).
Let's assume that the final velocity of the raft is v2'. Since the raft is initially at rest, its initial velocity (v2) is zero.
Substituting these values into the equation, we have:
-(60.0 kg)(4.0 m/s) + (180 kg)(0) = (60.0 kg)(0) + (180 kg)(v2').
Simplifying the equation gives:
-240 kg/(m/s) = 180 kg(v2').
Dividing both sides of the equation by 180 kg gives:
v2' = -240 kg/(m/s) / 180 kg = -1.33 m/s.
So, the final speed of the raft just after the girl dives is approximately 1.33 m/s in the opposite direction to the girl's movement.