Two students on roller skates stand face-toface,

then push each other away. One student
has a mass of 96 kg and the second student
56 kg.
Find the ratio of the magnitude of the first
student’s velocity to the magnitude of the
second student’s velocity.

V1/V2 = M2/M1 = 56kg/96kg = 0.583

V1/V2 = - M2/M1

= 56/96

To find the ratio of the magnitude of the first student's velocity to the magnitude of the second student's velocity, we'll use the principle of conservation of momentum.

According to the principle of conservation of momentum, the total momentum before the push is equal to the total momentum after the push. The formula for momentum is:

momentum = mass × velocity

Let's assume the initial velocities of the two students are v1 and v2 (with v1 being the velocity of the first student and v2 being the velocity of the second student).

Initially, the total momentum of the system is:

momentum_before = (mass1 × velocity1) + (mass2 × velocity2)

After the push, the first student moves in one direction, and the second student moves in the opposite direction. Let's assume the final velocities of the two students are v1' and v2' (with v1' being the final velocity of the first student and v2' being the final velocity of the second student).

The total momentum after the push is:

momentum_after = (mass1 × velocity1') + (mass2 × velocity2')

Since momentum is conserved, we can equate the initial and final momenta:

(mass1 × velocity1) + (mass2 × velocity2) = (mass1 × velocity1') + (mass2 × velocity2')

Given the masses and initial velocities of the two students, we have:

(96 kg × v1) + (56 kg × v2) = (96 kg × v1') + (56 kg × v2')

Now, let's solve for the ratio of v1' (final velocity of the first student) to v2' (final velocity of the second student):

(96 kg × v1) + (56 kg × v2) = (96 kg × v1') + (56 kg × v2')

Rearranging the equation:

96 kg × (v1 - v1') = 56 kg × (v2' - v2)

Now, let's divide both sides of the equation by 56 kg × (v2' - v2):

(96 kg × (v1 - v1')) / (56 kg × (v2' - v2)) = 1

Finally, let's simplify the equation:

(96 kg / 56 kg) × [(v1 - v1') / (v2' - v2)] = 1

Simplifying further:

(12/7) × [(v1 - v1') / (v2' - 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 12/7.

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 conservation of momentum.

The conservation of momentum states that the total momentum before an interaction is equal to the total momentum after the interaction, assuming no external forces act on the system.

Let's assume the first student's initial velocity is v1 and the second student's initial velocity is v2. After they push each other, the first student's final velocity is v1' and the second student's final velocity is v2'.

Using the conservation of momentum, we have:

(mass1 * v1) + (mass2 * v2) = (mass1 * v1') + (mass2 * v2')

We know the masses of the students:
mass1 = 96 kg
mass2 = 56 kg

Now, let's consider the direction of the velocities. Since they are facing each other, the first student's initial velocity is positive, and the second student's initial velocity is negative.

Therefore, the equation becomes:

(96 kg * v1) + (-56 kg * v2) = (96 kg * v1') + (-56 kg * v2')

Simplifying the equation:

96v1 - 56v2 = 96v1' - 56v2'

Now, let's use the fact that the first student pushes the second student away, resulting in opposite velocities. We can say that v1' = -v1 and v2' = -v2.

Substituting these values:

96v1 - 56v2 = 96(-v1) - 56(-v2)

Simplifying further:

96v1 - 56v2 = -96v1 + 56v2

Rearranging the equation:

96v1 + 96v1 = 56v2 + 56v2

192v1 = 112v2

Divide both sides by 112v2:

v1/v2 = 112/192

Simplifying the ratio:

v1/v2 = 7/12

Therefore, the ratio of the magnitude of the first student's velocity to the magnitude of the second student's velocity is 7/12.