Final answer:
A gun is monted on a railroad car. The mass of the car, the gun, the shells and the operator is 50m where m is the mass of one shell. If the velocity of het shell with respect to the gun (in its state before firing) is 200 m/s
The recoil speed of the car after the second shot is 200 m/s in the opposite direction.
Step-by-step explanation:
To solve this problem, we can use the principle of conservation of momentum. Initially, the total momentum of the system (car, gun, shells, and operator) is zero since everything is at rest. When the shells are fired, they gain momentum in one direction, causing the car to have an equal magnitude of momentum in the opposite direction to maintain momentum conservation.
Let's first calculate the momentum of the shells before firing.
The mass of the shells is 50m, and their initial velocity relative to the gun is 200 m/s.
The momentum of the shells before firing is given by:
pshells = (50m)(200
) = 10000m
According to the principle of conservation of momentum, the momentum of the car and gun after the second shot is equal in magnitude and opposite in direction to the momentum of the shells.
Therefore, the recoil velocity of the car and gun after the second shot is:
v = - (10000m) / (50m)
= -200m
So, the recoil speed of the car after the second shot is 200 m/s in the opposite direction.