Two students on roller skates stand face-toface,
then push each other away. One student
has a mass of 95 kg and the second student
60 kg.
Find the ratio of the magnitude of the first
student’s velocity to the magnitude of the
second student’s velocity.
since
m1 v1 = m2 v2,
v1/v2 = m2/m1
To find the ratio of the magnitude of the first student's velocity to the magnitude of the second student's velocity, we need to apply the principle of conservation of momentum.
According to the conservation of momentum, the total momentum before the push is equal to the total momentum after the push.
Before the push:
The momentum is the product of mass and velocity. Let's assume the velocity of the first student is v1 and the velocity of the second student is v2.
Momentum of the first student: p1 = mass1 * v1 = 95 kg * v1
Momentum of the second student: p2 = mass2 * v2 = 60 kg * v2
After the push:
The momenta of the students are given as:
Momentum of the first student: p1' = mass1 * v1'
Momentum of the second student: p2' = mass2 * v2'
According to the conservation of momentum, the total momentum before the push should be equal to the total momentum after the push.
So, we have:
p1 + p2 = p1' + p2'
95 kg * v1 + 60 kg * v2 = 95 kg * v1' + 60 kg * v2'
Now, when pushing off each other, the students exert an equal and opposite force on each other, causing them to move in opposite directions.
Since they are initially standing face-to-face, the velocity of one student (v1 or v2) should have a negative sign to indicate opposite directions.
Let's assume that the first student moves to the right, so their velocity (v1) is positive. The second student moves to the left, so their velocity (v2) is negative.
In this case, after the push, the first student moves to the right (v1' > 0) and the second student moves to the left (v2' < 0).
Now, we can substitute these values into the momentum equation:
95 kg * v1 + 60 kg * v2 = 95 kg * v1' + 60 kg * v2'
Since one student moves to the right (positive velocity) and the other moves to the left (negative velocity), we can assume that the magnitude of v1' is greater than the magnitude of v2':
v1' > v2'
Dividing both sides of the equation by v2:
(95 kg * v1 + 60 kg * v2) / v2 = (95 kg * v1' + 60 kg * v2') / v2
Multiplying both sides by (1/v1):
(95 kg * v1 + 60 kg * v2) / (v2 * v1) = (95 kg * v1' + 60 kg * v2') / (v2 * v1)
Simplifying the equation:
(95 kg / v2) + (60 kg / v1) = 95 kg * (v1' / v2) + 60 kg
Now, since we have assumed v1' > v2':
v1' / v2 > 1
Therefore, the ratio of the magnitude of the first student's velocity (v1) to the magnitude of the second student's velocity (v2) is greater than 1.
Unfortunately, with the given information, we cannot determine the exact ratio value without knowing the specific velocities or the change in velocities (v1' and v2') after the push.
To solve this problem, we can apply the principle of conservation of momentum. According to this principle, the total momentum before the push is equal to the total momentum after the push.
The momentum of an object is given by the product of its mass and velocity: momentum = mass x velocity.
Let's assume that the initial velocity of the first student is v1 and the initial velocity of the second student is v2. After the push, let's assume that their final velocities are v1' and v2', respectively.
According to the conservation of momentum, we have:
(initial momentum of first student) + (initial momentum of second student) = (final momentum of first student) + (final momentum of second student)
(mass1 x velocity1) + (mass2 x velocity2) = (mass1 x velocity1') + (mass2 x velocity2')
Substituting the given values: mass1 = 95 kg, mass2 = 60 kg, we can rewrite the equation as:
(95 kg x v1) + (60 kg x v2) = (95 kg x v1') + (60 kg x v2')
We need to find the ratio of the magnitude of the first student's velocity (v1') to the magnitude of the second student's velocity (v2'). To simplify the equation, we can divide both sides by 5 kg:
19v1 + 12v2 = 19v1' + 12v2'
Now we can find the ratio of v1' to v2'. Rearranging the equation, we get:
19v1' = 19v1 + 12v2 - 12v2'
19v1' + 12v2' = 19v1 + 12v2
v1' = (19v1 + 12v2) / (19)
Therefore, the ratio of the magnitude of the first student's velocity to the magnitude of the second student's velocity is:
v1' / v2' = ((19v1 + 12v2) / (19)) / v2' = (19v1 + 12v2) / (19v2)
Note: The ratio will depend on the specific values of v1 and v2.