a ball moving with velocity of 6m/s strikes an identical stationary ball. after collision each ball moves at an angle of 30° with the original line of motion. what are the speeds of the balls after the collision?

To solve this problem, we need to apply the principles of conservation of momentum and the geometry of the situation. When the moving ball strikes the stationary ball, both balls will move at an angle of 30° to the original line of motion after the collision. Let's break this down step by step.

Understanding the Collision

In this scenario, we have two identical balls. One ball is initially moving with a velocity of 6 m/s, while the other is at rest. After the collision, both balls move at an angle of 30° to the original direction of the moving ball. We can denote the mass of each ball as m, which will cancel out in our calculations since they are identical.

Applying Conservation of Momentum

The principle of conservation of momentum states that the total momentum before the collision must equal the total momentum after the collision. We can analyze this in two dimensions: the x-direction (the original line of motion) and the y-direction (perpendicular to the original line).

Initial Momentum

Before the collision, the total momentum in the x-direction is:

• Momentum of the moving ball: \( p_{initial} = m \cdot 6 \, \text{m/s} \)
• Momentum of the stationary ball: \( 0 \, \text{m/s} \)

Thus, the total initial momentum in the x-direction is \( 6m \, \text{kg m/s} \), and in the y-direction, it is \( 0 \, \text{kg m/s} \).

Final Momentum

After the collision, let’s denote the speeds of both balls as \( v_1 \) and \( v_2 \). Since both balls move at an angle of 30°, we can break their velocities into components:

• For the first ball (moving at angle 30°):
  • Horizontal component: \( v_1 \cos(30°) \)
  • Vertical component: \( v_1 \sin(30°) \)
• For the second ball (also moving at angle 30°):
  • Horizontal component: \( v_2 \cos(30°) \)
  • Vertical component: \( v_2 \sin(30°) \)

Setting Up the Equations

Now, we can set up our equations based on the conservation of momentum:

X-Direction

The total momentum in the x-direction after the collision must equal the total momentum before the collision:

\( 6m = m(v_1 \cos(30°) + v_2 \cos(30°)) \)

Dividing through by m, we get:

\( 6 = v_1 \cos(30°) + v_2 \cos(30°) \)

Y-Direction

For the y-direction, since the initial momentum is zero, the total momentum after the collision must also equal zero:

\( 0 = m(v_1 \sin(30°) - v_2 \sin(30°)) \)

Again, dividing through by m gives us:

\( 0 = v_1 \sin(30°) - v_2 \sin(30°) \Rightarrow v_1 \sin(30°) = v_2 \sin(30°) \Rightarrow v_1 = v_2 \)

Solving the Equations

Now that we know \( v_1 = v_2 \), we can substitute this into our x-direction equation:

\( 6 = v_1 \cos(30°) + v_1 \cos(30°) \Rightarrow 6 = 2v_1 \cos(30°) \Rightarrow v_1 = \frac{6}{2 \cos(30°)} \Rightarrow v_1 = \frac{6}{2 \cdot \frac{\sqrt{3}}{2}} \Rightarrow v_1 = \frac{6}{\sqrt{3}} \Rightarrow v_1 = 2\sqrt{3} \, \text{m/s} \approx 3.46 \, \text{m/s} \

Final Speeds of the Balls

Thus, after the collision, both balls move with a speed of approximately 3.46 m/s at an angle of 30° to the original line of motion. This result illustrates how momentum is conserved in collisions, even when the objects involved are moving at angles post-collision.