Respuesta :
Answer:
D. The momentum of car B is four times as great in magnitude as that of car A.
Explanation:
Hi there!
Since the collision is elastic the momentum and the kinetic energy of the system are conserved, i.e, they remain constant (initial momentum = final momentum and initial kinetic energy = final kinetic energy). The momentum of the system is given by the sum of the momenta (p) of each car:
pA + pB = pA´ + pB´
Where:
pA = momentum of car A before the collision.
pB = momentum of car B before the collision.
pA´ = momentum of the car A after the collision.
pB´ = momentum of the car B after the collision.
The momentum is calculated as follows:
p = m · v
Where:
m = mass of the car.
v = velocity.
Then, the momentum of the system before the collision will be:
initial momentum of the system = m · v + mB · vB
Where:
m = mass of car A
v = velocity of car A
mB = mass of car B
vB = velocity of car B
We know that initially, the velocity of car B is zero and mB = 2m. Then, the initial momentum will be:
initial momentum = m · v
Let´s express the final momentum of the system:
final momentum of the system = m · vA´ + 2m · vB´
Where vA´ and vB´ are the final velocities of car A and B respectively.
Since initial momentum of the system = final momentum of the system, then:
m · v = m · vA´ + 2m · vB´
dividing both sides of the equation by m:
v = vA´ + 2 vB´
Solving for vA´:
v - 2 vB´ = vA´
Now, using the coservation of kinetic energy (KE):
intial kinetic energy = final kinetic energy
1/2 · m · (v)² + 1/2 · 2 · m · (vB)² = 1/2 · m · (vA´)² + 1/2 · 2 · m · (vB´)²
vB = 0, then:
1/2 · m · (v)² = 1/2 · m · (vA´)² + 1/2 · 2 · m · (vB´)²
dividing by 1/2 m both sides of the equation:
(v)² = (vA´)² + 2 · (vB´)²
replacing vA´ = v - 2 vB´
(v)² = (v - 2 vB´)² + 2 (vB´)²
(v)² = (v)² - 4 vB´ v + 4 (vB´)² + 2 (vB´)²
4 vB´ v = 6 (vB´)²
divide both sides by vB´
4 v = 6 vB´
2/3 v = vB´
Now, we can calculate the final velocity of car A:
vA´ = v - 2 vB´
vA´ = v - 2 · 2/3 v
vA´ = v - 4/3 v
vA´ = -1/3 v
Now let´s calculate the momentum of each car after the collision:
Car A:
pA = m · vA´
pA = -1/3 mv
Car B:
pB = mB · vB´
pB = 2 m · 2/3 v
pB = 4/3 mv
Then:
4|pA| = |pB|
The momentum of car B is four times as great in magnitude as that of car A.
Option D will be right that is the momentum of car B is four times as great in magnitude as that of car A.
What will be the momentum of car A with respect to cae B?
It is given that the collision of the car is elastic so the momentum of the car will remain conserved. That is the sum of the momentum of the car before the collision will be equal to the sum of the momentum of the car after a collision.
[tex]P_{A} +P_{B} =P'_{A} +P'_{B}[/tex]
Here
[tex]P_{A}[/tex] = momentum of car A before the collision.
[tex]P_{B}[/tex] = momentum of car B before the collision.
[tex]P'_{A}[/tex] = momentum of car A after the collision.
[tex]P'_{A}[/tex] = momentum of car B after the collision.
We know that the momentum is given as
[tex]P=m\times V[/tex]
Here:
m = mass of the car.
v = velocity.
Now the momentum before collision will be
The initial momentum of the system = [tex]m_{A} \times V_{A} +m_{B} \times V_{B}[/tex]
Here
[tex]m_{A}[/tex]= mass of car A
[tex]V_{A}[/tex]= velocity of car A
[tex]m_{B}[/tex] = mass of car B
[tex]V_{B}[/tex] = velocity of car B
Since the initial velocity of the car [tex]V_{A}=0[/tex]
Also [tex]m_{B}=2m_{A}[/tex]
So the initial momentum will become = [tex]m_{A}\times V_{A}[/tex]
And final momentum of the system =[tex]m_{A}\times V'_{A}+2m_{A}\times V'_{B}[/tex]
Where [tex]V'_{A}[/tex] and [tex]V'_{A}[/tex] are the final velocities of car A and B respectively.
Since initial momentum of the system = final momentum of the system, then:
[tex]m_{A}\times V_{A} = m_{A}\times V'_{A}+2m_{A}\times V'_{B}[/tex]
Now by solving both we get
[tex]V_{A}=V'_{A}+2V'_{B}[/tex]
[tex]V'_{A}'=V_{A}-2V'_{B}[/tex]
Now, using the conservation of kinetic energy (KE):
initial kinetic energy = final kinetic energy
[tex]\dfrac{1}{2} m_{A}V_{A}^{2} +\dfrac{1}{2} 2m_{A}V_{B}^{2} =\dfrac{1}{2} m_{A}V'_{A}^{2} +\dfrac{1}{2} 2m_{A}V'_{B}^{2}[/tex]
[tex]V_{B} =0[/tex] So
[tex]\dfrac{1}{2} m_{A}V_{A}^{2} =\dfrac{1}{2} m_{A}V'_{A}^{2} +\dfrac{1}{2} 2m_{A}V'_{B}^{2}[/tex]
[tex]V_{A}^{2} = V'_{A}^{2} + 2 V'_{B}^{2}[/tex]
now put
[tex]V'_{A}'=V_{A}-2V'_{B}[/tex]
[tex]V_{A}^{2} = (V_{A} - 2V'_{B} )^{2} + 2 V'_{B}^{2}[/tex]
[tex]V_{A}^{2} = V_{A} ^{2} -4V_{A} V'_{B} +4V'_{B} ^{2} +2V'_{B} ^{2}[/tex]
[tex]4V'_{B} V_{A}=6 V'_{B}^{2}[/tex]
[tex]\dfrac{2}{3} V_{A} = V'_{B}[/tex]
Now, For calculating the final velocity of car A:
[tex]V'_{A}'=V_{A}-2V'_{B}[/tex]
[tex]V'_{A}'=V_{A}-2\times \dfrac{2}{3} V_{A}[/tex]
[tex]V'_{A}'=\dfrac{-1}{3} V_{A}[/tex]
Now By calculating the momentum for each car
For car A
[tex]P'_{A} =m_{A} \times V'_{A}[/tex]
[tex]P'_{A} =\dfrac{-1}{3} m_{A} V_{A}[/tex]
For Car B
[tex]P'_{B}=m_{B} \times V'_{B}[/tex]
[tex]P'_{B} = 2\times \dfrac{2}{3} V_{B}\times m_{A}= \dfrac{4}{3} \times V_{A}\times m_{A[/tex]
[tex]P'_{B} =\dfrac{4}{3} m_{A} V_{A}[/tex]
SO
[tex]P'_{B} =4P_{A}[/tex]
Thus Option D will be right that is the momentum of car B is four times as great in magnitude as that of car A.
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