A small ball with a mass of 0.7 kg and a velocity of 13 m/s hits another ball with the same mass. The first ball stops moving, and the second ball then moves forward and hits a third ball with a mass of 0.2 kg. If the system is closed, what is the velocity of the third ball?(1 point) Responses 45.5 m/s 45.5 m/s 36 m/s 36 m/s 9.3 m/s 9.3 m/s 1.82 m/s
The principle of conservation of momentum states that the total momentum before an event is equal to the total momentum after the event in a closed system.
Before the collision, the momentum of the first ball is given by:
momentum = mass * velocity = 0.7 kg * 13 m/s = 9.1 kg*m/s
Since the masses of the first and second ball are equal, the second ball will also have a momentum of 9.1 kg*m/s after the collision.
Now, the momentum of the second ball is given by:
momentum = mass * velocity
We can rearrange this equation to find the velocity of the second ball:
velocity = momentum / mass = 9.1 kg*m/s / 0.7 kg = 13 m/s
Therefore, the velocity of the second ball is 13 m/s.
Now, the second ball collides with the third ball. Since the second and third balls have different masses, we need to calculate the momentum of the second ball before and after the collision separately.
The momentum of the second ball before the collision is given by:
momentum = mass * velocity = 0.7 kg * 13 m/s = 9.1 kg*m/s
Using the principle of conservation of momentum, the momentum of the second and third ball after the collision will also be 9.1 kg*m/s.
Now, we can find the velocity of the third ball after the collision:
velocity = momentum / mass = 9.1 kg*m/s / 0.2 kg = 45.5 m/s
Therefore, the velocity of the third ball is 45.5 m/s.
So the correct answer is 45.5 m/s.