A small ball with a mass of 0.6 kg and a velocity of 12 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
1.44 m/s
1.44 m/s
30 m/s
30 m/s
4 m/s
4 m/s
36 m/s
To solve this problem, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision in a closed system.
Let's denote the velocity of the third ball as v3.
Before the collision, the total momentum of the system is given by the sum of the momentum of the first ball (m1 * v1) and the momentum of the second ball (m2 * 0), since the second ball is initially at rest:
Total momentum before collision = (m1 * v1) + (m2 * 0) = m1 * v1
After the collision, the first ball comes to a stop, so its momentum becomes 0. The momentum of the second ball (m2 * v2) gets transferred to the third ball:
Total momentum after collision = (m1 * 0) + (m2 * v2) + (m3 * v3)
Since the system is closed and there are no external forces, the total momentum before the collision is equal to the total momentum after the collision:
m1 * v1 = (m2 * v2) + (m3 * v3)
Substituting the given values:
0.6 kg * 12 m/s = 0.6 kg * v2 + 0.2 kg * v3
Rearranging the equation to solve for v3:
0.2 kg * v3 = 0.6 kg * 12 m/s - 0.6 kg * v2
v3 = (0.6 kg * 12 m/s - 0.6 kg * v2) / 0.2 kg
Since the velocity v2 is not given, we can't determine the exact value of v3 without more information. Therefore, none of the provided responses (1.44 m/s, 30 m/s, 4 m/s, 36 m/s) are correct.