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 GIRLS SPEED IS?
The boys change of momentum is equal and opposite to the girls change of momentum.
GirlMomentum=-Boy momentum
30kg*V=-25kg*1.0m/s
0.83
To find the girl's speed, 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. Let's use the following variables:
- Girl's mass (m1) = 30 kg
- Boy's mass (m2) = 25 kg
- Girl's initial velocity (u1) = unknown
- Boy's initial velocity (u2) = 0 m/s (since he is initially at rest)
According to conservation of momentum:
(m1 * u1) + (m2 * u2) = (m1 * v1) + (m2 * v2)
Given:
m1 = 30 kg
m2 = 25 kg
u2 = 0 m/s (initially at rest)
v2 = 1.0 m/s (boy's velocity after the push)
Applying the principle of conservation of momentum:
(30 kg * u1) + (25 kg * 0 m/s) = (30 kg * v1) + (25 kg * 1.0 m/s)
30 kg * u1 = 30 kg * v1 + 25 kg * 1.0 m/s
Since the friction between the roller blades is negligible, the total momentum does not change. Therefore, we can simplify the equation as:
30 kg * u1 = 30 kg * v1 + 25 kg * 1.0 m/s
Now, let's solve for the girl's speed (v1):
30 kg * u1 = 30 kg * v1 + 25 kg * 1.0 m/s
Divide by 30 kg:
u1 = v1 + (25 kg * 1.0 m/s) / 30 kg
Since we are looking for the girl's speed after the push (v1), we rearrange the equation:
v1 = u1 - (25 kg * 1.0 m/s) / 30 kg
Now, we need to know the value of u1 (the girl's initial velocity) to calculate v1. If the initial velocity of the girl is given, substitute that value into the formula to find the girl's final speed (v1).