A steel ball of mass 0.50 kg, moving with a velocity of
2.0 m/s [E], strikes a second ball of mass 0.30 kg, initially
at rest. The collision is a glancing one, causing the
moving ball to have a velocity of 1.5 m/s [30° N of E] after
the collision. Determine the velocity of the second ball
after the collision.
First, we can calculate the momentum of the system before and after the collision.
Before the collision:
Initial momentum = (0.5 kg)(2.0 m/s) = 1.0 kg*m/s [E]
Initial momentum of second ball = 0
After the collision:
Final momentum = (0.5 kg)(1.5 m/s) = 0.75 kg*m/s [30° N of E]
Now, we can break down the final momentum into x and y-components:
Final momentum (x) = 0.75 kg*m/s * cos(30°) = 0.65 kg*m/s [E]
Final momentum (y) = 0.75 kg*m/s * sin(30°) = 0.375 kg*m/s [N]
Since the momentum of the system is conserved, the momentum of the second ball after the collision must be equal to the final momentum.
Let the velocity of the second ball after the collision be V. Therefore, the momentum of the second ball after the collision is:
0.3 kg * V = 0.65 kg*m/s [E] + 0.375 kg*m/s [N]
Since we have two equations:
0.5 kg * 2.0 m/s = 0.3 kg * V + 0
0.3 kg * V = 0.65 kg*m/s [E] + 0.375 kg*m/s [N]
By solving these two equations simultaneously, we get V = 1.83 m/s [69.3° N of E].
Therefore, the velocity of the second ball after the collision is 1.83 m/s [69.3° N of E].