Question

Two ice skaters stand together as illustrated in the attached figure. They "push off" and travel directly away from each other, the boy with a velocity of v = +0.50 m/s. If the boy weighs 710 N and the girl 480 N, what is the girl's velocity in m/s after they push off? (Consider the ice to be frictionless.)

Answer 1

When the boy and the girl push off and travel directly away from each other on frictionless ice, the girl's velocity after they push off is approximately -0.744 m/s.

Step-by-step explanation:

When the boy and the girl push off and travel directly away from each other, their total momentum remains constant. The initial momentum of the system is zero because they were initially at rest. The boy's mass is 710 N/9.8 m/s^2 = 72.45 kg, and the girl's mass is 480 N/9.8 m/s^2 = 48.98 kg.

The girl's velocity can be calculated using the principle of conservation of momentum:

Initial momentum = Final momentum

Since the initial momentum is zero, we have:

0 = (Mass of boy x Velocity of boy) + (Mass of girl x Velocity of girl)

Substituting the known values:

0 = (72.45 kg x 0.50 m/s) + (48.98 kg x Velocity of girl)

Solving for Velocity of girl:

Velocity of girl = (-72.45 kg x 0.50 m/s) / 48.98 kg

Velocity of girl = -0.744 m/s

Therefore, the girl's velocity after they push off is approximately -0.744 m/s.

Answer 2

By definition, the momentum is given by:
p = m * v
Where,
m = mass
v = speed.
On the other hand,
F = m * a
Where,
m = mass
a = acceleration:
For the boy we have:
p1 = m * v
p1 = (F / a) * v
p1 = ((710) / (9.81)) * (0.50)
p1 = 36.19 Kg * (m / s)
For the girl we have:
p2 = m * v
p2 = (F / a) * v
p2 = ((480) / (9.81)) * (v)
p2 = 48.93 * v Kg * (m / s)
Then, we have:
p1 + p2 = 0
36.19 + 48.93 * v = 0
Clearing v:
v = - (36.19) / (48.93)
v = -0.74 m / s (negative because the velocity is in the opposite direction of the boy's)
Answer:
the girl's velocity in m / s after they push off is -0.74 m / s