Sure. Let's break this problem into several steps to make it easier to comprehend.
Step 1: Understanding the given values.
- Mass of first bullet, M = 5.12 g
- Velocity of the first bullet, V = 229 m/s at an angle of 21.3 degree above horizontal.
- Mass of second bullet, m = 3.05 g
- Velocity of the second bullet, v = 282 m/s at an angle of 15.4 degree above the horizontal.
Step 2: Converting the bullet velocities to their respective x and y components using trigonometry.
- For the first bullet:
V_x = Vcos(21.3 degrees) ≈ 213.36 m/s
V_y = Vsin(21.3 degrees) ≈ 83.18 m/s
- For the second bullet:
v_x = vcos(15.4 degrees) ≈ 271.87 m/s
v_y = vsin(15.4 degrees) ≈ 74.89 m/s
Step 3: Using the principle of conservation of momentum in both the x and y directions to calculate the common velocity's components.
For the x-components (taking right as positive and left as negative as convention):
final_v_x = (M * V_x - m * v_x) / (M + m)
= (5.12 * 213.36 - 3.05 * 271.87) / (5.12 + 3.05)
≈ 32.21 m/s
For the y-components (both are upwards, hence positive):
final_v_y = (M * V_y + m * v_y) / (M + m)
= (5.12 * 83.18 + 3.05 * 74.89) / (5.12 + 3.05)
≈ 80.09 m/s
Step 4: Finding the magnitude of the final common velocity using Pythagoras' theorem.
final_v = sqrt(final_v_x² + final_v_y²)
= sqrt(32.21² + 80.09²)
≈ 86.32 m/s
In conclusion, the magnitude of their common velocity immediately after collision is approximately 86.32 m/s.