a 40kg girl and a 30kg boy face each other on friction free roller skates. the boy pushes the girl,
and she moves away from him at 3/4 his speed.
Wright
Here is a step-by-step explanation of what happens when the 30kg boy pushes the 40kg girl:
1. The boy applies a force on the girl in the forward direction. Let's denote this force as Fboy→girl.
2. According to Newton's third law of motion, for every action, there is an equal and opposite reaction. So, the girl exerts a force on the boy in the opposite direction. Let's denote this force as Fgirl→boy.
3. Since the roller skates are friction-free, there is no opposing force present from the ground. Therefore, the only horizontal forces acting on the system are Fboy→girl and Fgirl→boy.
4. According to Newton's second law of motion, the net force acting on an object is equal to the product of its mass and acceleration (F = ma). In this case, the net force acting on the girl is Fboy→girl, and the net force acting on the boy is Fgirl→boy.
5. Since the girl has a mass of 40kg, her acceleration can be calculated using Fboy→girl = mgirl × agirl, where agirl is the acceleration of the girl.
6. Similarly, since the boy has a mass of 30kg, his acceleration can be calculated using Fgirl→boy = mboy × aboy, where aboy is the acceleration of the boy.
7. As the boy pushes the girl, she will experience an acceleration in the forward direction, while the boy will experience an acceleration in the backward direction.
8. The magnitude of the acceleration of the girl can be calculated using Fboy→girl = mgirl × agirl. Rearranging the equation, agirl = Fboy→girl / mgirl.
9. Similarly, the magnitude of the acceleration of the boy can be calculated using Fgirl→boy = mboy × aboy. Rearranging the equation, aboy = Fgirl→boy / mboy.
10. As the boy applies a force on the girl, they both experience accelerations, but the girl's acceleration will be smaller than the boy's due to her larger mass.
11. The exact value of the accelerations and resulting motions can be determined by knowing the specific magnitudes of the applied forces Fboy→girl and Fgirl→boy.
Note: If you provide the magnitude of the applied forces, I can help you calculate the accelerations and resulting motions more accurately.
To determine the resulting motion of the girl and the boy, we need to consider 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, assuming no external forces act on the system.
Momentum (p) is calculated by multiplying mass (m) by velocity (v), so we have:
Initial total momentum = Final total momentum
Before the push:
Total momentum = (mass of girl x velocity of girl) + (mass of boy x velocity of boy)
After the push:
Total momentum = (mass of girl x new velocity of girl) + (mass of boy x new velocity of boy)
Since the roller skates have no friction, we can assume there is no external force acting on the system, so the total momentum is conserved.
Let's assume the girl's velocity is v1 before the push, and the boy pushes her with a force that gives her a new velocity v2.
Therefore, we can set up an equation:
(mass of girl x v1) + (mass of boy x 0) = (mass of girl x v2) + (mass of boy x v2)
Substituting the given values:
(40kg x v1) + (30kg x 0) = (40kg x v2) + (30kg x v2)
Simplifying:
40kgv1 = 70kgv2
Now, solve for the ratio of velocities:
v1/v2 = 70/40
By simplifying:
v1/v2 = 7/4
This ratio indicates that the velocity of the girl is 7 times that of the boy after the push on friction-free roller skates.