Damon is question 2,:
5*10^4 x 2.5= 125000kgm/s
then 4*10^4x 0= 0
so momentum before = 125000
so v=p/m
so 125000/5*10^4 divided by 4*10^4 which is 1.38recurring ms-1
so is the initial velocity of the coupled carriages after the impact 1.38recurring ms-1??? sorry i cant do a recurring sign
Sorry i meant this is what i think of question 3 for question 2 b i got so stuck please help
i am also completely baffled byquestion 4 too
To determine the initial velocity of the coupled carriages after the impact, you need to consider the principle of conservation of momentum.
In the given problem, you already calculated the momentum before the impact, which is 125,000 kg m/s. Now, you want to find the initial velocity of the coupled carriages.
To do this, you can use the formula v = p/m, where v is the velocity, p is the momentum, and m is the mass.
The mass of the system does not change, so you can use the same mass for both carriages. Let's assume the mass of each carriage is m kg.
From the given information, the momentum before the impact is 125,000 kg m/s. Therefore, the momentum after the impact should also be 125,000 kg m/s because momentum is conserved.
Let's consider the momentum of the system after the impact. The momentum of each carriage can be calculated using the formula p = mv.
Since both carriages have the same mass, the momentum of each carriage can be written as 125,000/2.
Now, we set up equations for momentum after the impact:
(125,000/2) + (125,000/2) = (125,000/2)m1 + (125,000/2)m2,
where m1 is the mass of one carriage and m2 is the mass of the other carriage.
Since the masses are the same (m1 = m2 = m), we can simplify the equation:
125,000 = (125,000/2)m + (125,000/2)m.
Now we solve for m:
125,000 = 125,000m
Dividing both sides of the equation by 125,000, we find:
1 = m
So the mass of each carriage is 1 kg.
Now, we can find the initial velocity using the formula v = p/m:
v = (125,000/2) / 1
Simplifying, we get:
v = 62,500 m/s
Therefore, the initial velocity of the coupled carriages after the impact is 62,500 m/s.