Respuesta :
Answer:
A: v1=0, v2=v; B: 0.5v; C: v/3, 4v/3; D: 0.25 V
Explanation:
A)
In this case, the collision can be described by two laws: Momentum conversations law and Energy conservation law. Note, that theoretically, the vehicles will be moving in the unknown directions after the collision, but let's assume that the final velocities are going to have the same direction with the original velocity. Then the two equations are going to be:
[tex]mV=mV1+mV2\\mV^{2}/2=mV1^{2}/2 + mV2^{2}/2[/tex]
After some transformations, the following can be derived:
[tex]\left \{ {{V=V1+V2} \atop {V^{2} =V1^{2}+V2^{2} }} \right.[/tex]
After we square the top equation and substitute the squares of velocities, with the V^2 from the equation below, we are going to get the following:
[tex]2V1V2=0[/tex], from where we can make a logical conclusion: V1=0, as it cannot pass by the second vehicle. In this case, V2=V.
B)
In this case, only Momentum conservation law can be used, as the energy will be lost during the collision. After the collision, objects will move as one object with the same velocity. To find this velocity, we can do the following:
[tex]mV=mV1+mV2\\V1=V2=U\\mV=mU+mU\\U=V/2=0.5V[/tex]
C) This case, similarly to the case A, requires to find two velocities. We can use both conservation laws- Momentum and Energy conservation laws. In this case, the mathematics will look as:
[tex]2mV=2mV1+mV2\\2mV^{2}/2=2mV1^{2}/2 + mV2^{2}/2\\\\2V=2V1+V2\\2V^{2}=2V1^{2}+V2^{2}[/tex]
Substituting V2 in the second equation, we are getting:
[tex]2V^{2}=2V1^{2}+4V^{2}-8V1 V +4V1^{2} \\6V1^{2}-8V1 V+2V^{2}=0[/tex]
From where, by solving a square equation, we are getting two roots:
V1= V and V1=V/3. First answer shows us the initial conditions and the second results- velocity after the collision.
Using this, we can find, that V2=4V/3
So, the answers are: V1=V/3; V2=4V/3
D)
In this case, both objects after the collision will move as a single object with the same velocities. Only the momentum will be conserved in this scenario, so only one equation is required:
[tex]mV=mV1+3mV1\\mV=4mV1\\V1=V/4=V2[/tex]
Based on the law of conservation of momentum, the values of velocities v after collision are given as follows:
- v1 = u - v2 and v2 = u - v1
- v = u/2
- v1 = u - v2/2 and v2 = 2(u - v1)
- v = u/4
What is the law of conservation of momentum?
The law of conservation of momentum states that in an isolated system of colliding bodies, the total momentum is conserved.
From the law of conservation of momentum;
momentum before collision = momentum after collision
momentum = mass × velocity
The masses are the same = m
initial velocity = u
final velocity = v
In elastic collision velocities are different;
momentum before collision = mu
momentum after collision = mv1 + mv2
u = v1 + v2
Thus,
- v1 = u - v2
- v2 = u - v1
If the inelastic collision, the velocity is the same:
momentum before collision = mu
momentum after collision = mv + mv
u = 2v
- v = u/2
If the masses are 2m and m and the collision is elastic:
momentum before collision = 2mu
momentum after collision = 2mv1 + mv2
2u = 2v1 + v2
Thus,
- v1 = u - v2/2
- v2 = 2(u - v1)
If the mass of object 1 is m and the mass of object 2 be 3m and the collision is perfectly inelastic:
momentum before collision = mu
momentum after collision = mv + 3mv
u = 4v
- v = u/4
Therefore, the momentum of the colliding bodies is always conserved.
Learn more about conservation of linear momentum at: https://brainly.com/question/7538238
