a 30 kg girl and a 25 kg boy face each other on friction-free roller blades.the girl pushes the boy,who moves away at a speed of 1.0 m/s.the girl,s speed is _ m/s.

The law of conservation of linear momentum

0 = m1•v1 - m2•v2,
v2 =30•1/25 = 1.2 m/s².

Well, it seems like the girl really gave the boy a push! Let me calculate her speed for you.

To solve this problem, we can use the principle of conservation of momentum, which states that the total momentum before and after an interaction remains the same.

The momentum formula is given by:

momentum = mass × velocity

Let's assume that the girl's speed after pushing the boy is v (in m/s). Initially, the total momentum is zero since they are both at rest. So, the total momentum after the girl pushes the boy is also zero to satisfy conservation of momentum.

Therefore, we can write the equation:

(30 kg × v) + (25 kg × 1.0 m/s) = 0

Simplifying the equation:

30 kg × v = -25 kg × 1.0 m/s

v = (-25 kg × 1.0 m/s) / 30 kg

v ≈ -0.833 m/s

So, the girl's speed is approximately -0.833 m/s. Now, I must admit, she seems to be going in the opposite direction with quite the velocity. She might want to work on her balance on those roller blades!

To solve this problem, we can use the principle of conservation of momentum. According to this principle, the total momentum before the girl pushes the boy should be equal to the total momentum after the push. The formula for momentum is given by:

Momentum = mass × velocity

Let's denote the girl's mass as m1 (30 kg) and the boy's mass as m2 (25 kg). The initial velocity of the girl is denoted as v1 (unknown), and the boy's initial velocity is v2 (0 m/s since he is at rest). After the girl pushes the boy, the boy's final velocity is v2' (1.0 m/s), and the girl's final velocity becomes v1' (unknown).

Using the conservation of momentum, we can set up the equation:

(mass of girl × initial velocity of girl) + (mass of boy × initial velocity of boy) = (mass of girl × final velocity of girl) + (mass of boy × final velocity of boy)

(30 kg × v1) + (25 kg × 0 m/s) = (30 kg × v1') + (25 kg × 1.0 m/s)

30 kg × v1 = 30 kg × v1' + 25 kg

Since we have two variables (v1 and v1'), we cannot solve this equation for a unique solution. However, we can determine the relationship between the girl's initial and final velocities.

Let's rearrange the equation:

30 kg × v1 = 30 kg × v1' + 25 kg

30 kg × v1 - 25 kg = 30 kg × v1'

Divide both sides of the equation by 30 kg:

v1 - (25 kg/30 kg) = v1'

Simplifying further:

v1 - 5/6 = v1'

Therefore, the girl's final velocity, v1', is equal to her initial velocity, v1, minus 5/6.

To find the girl's speed after pushing the boy, we can use the principle of conservation of momentum. According to this principle, the total momentum before and after the push should remain the same.

The momentum of an object is defined as the product of its mass and velocity (p = m * v). In this case, we have a girl with a mass of 30 kg and a boy with a mass of 25 kg facing each other. The girl pushes the boy, causing him to move away at a speed of 1.0 m/s.

Let's assume the girl's initial speed before the push is v_girl, and the boy's initial speed is v_boy (both speeds are in the opposite direction). After the push, the girl's speed is v_girl' and the boy's speed is v_boy'.

According to the conservation of momentum:

Initial momentum = Final momentum

(mass girl * initial speed girl) + (mass boy * initial speed boy) = (mass girl * final speed girl) + (mass boy * final speed boy)

(30 kg * v_girl) + (25 kg * (-v_boy)) = (30 kg * v_girl') + (25 kg * v_boy')

Since the roller blades are friction-free, there are no external forces involved, and the momentum is conserved. Therefore, we can assume that the initial momentum is zero because the girl and boy are initially at rest.

0 = (30 kg * v_girl') + (25 kg * 1 m/s)

Simplifying the equation:

0 = 30 kg * v_girl' + 25 kg

Rearranging the equation to solve for the final speed of the girl (v_girl'):

30 kg * v_girl' = -25 kg

v_girl' = -25 kg / 30 kg

v_girl' = -0.83 m/s

Therefore, the girl's speed after pushing the boy is approximately -0.83 m/s. Note that the negative sign indicates that the girl is moving in the opposite direction to her initial position.