Ball A with a mass of 0.500 kg is moving east at a velocity of 0.800 m/s. It strikes ball B, also of mass 0.500 kg, which is stationary. Ball A glances off B at an angle of 40.0° north of its original path. Ball B is pushed along a path perpendicular to the final path of ball A. What is the momentum (in kg m/s) of ball A after the collision?

well, you know the conservation of momentum applies.

Take the ball B, the North momentum, so BAll A must have that same momentum in the S direction as there was no initial N-S component.

Then, take BAll B East momentum, subtract that from BallA initial East nomentum, so that is the BAll A East momentum.
Final total momentum then will equal initial momentum

6.13

To find the momentum of ball A after the collision, we can use the principle of conservation of linear momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.

The momentum (p) of an object can be calculated by multiplying its mass (m) by its velocity (v). Mathematically, p = m * v.

Given:
Mass of ball A (m₁) = 0.500 kg
Velocity of ball A before the collision (v₁) = 0.800 m/s

After the collision, ball A glances off at an angle of 40.0° north of its original path. This means the final velocity (v₂) of ball A can be resolved into two components: one in the north direction and one perpendicular to it.

To find the velocity of ball A in the north direction, we can use the following formula:
v_north = v₂ * sin(angle)

Given:
Angle (θ) = 40.0°

Using trigonometry, sin(angle) = sin(40.0°) = 0.6428

Therefore, the velocity of ball A in the north direction (v_north) after the collision is:
v_north = v₂ * sin(angle) = v₂ * 0.6428

To find the velocity of ball A perpendicular to the north direction, we can use the following formula:
v_perpendicular = v₂ * cos(angle)

Using trigonometry, cos(angle) = cos(40.0°) = 0.7660

Therefore, the velocity of ball A perpendicular to the north direction (v_perpendicular) after the collision is:
v_perpendicular = v₂ * cos(angle) = v₂ * 0.7660

Now, let's apply the principle of conservation of linear momentum.

Before the collision:
Momentum of ball A before the collision (p₁) = m₁ * v₁

After the collision:
Momentum of ball A after the collision (p₂) = (m₁ * v_north) + (m₁ * v_perpendicular)

Substituting the values, we have:
p₂ = (0.500 kg * (v₂ * 0.6428)) + (0.500 kg * (v₂ * 0.7660))

Simplifying the equation, we get:
p₂ = 0.3214 kg * v₂ + 0.3830 kg * v₂
= 0.7044 kg * v₂

Since the total momentum before the collision is equal to the total momentum after the collision, we can say:
p₁ = p₂

So, we can equate the two equations:
m₁ * v₁ = 0.7044 kg * v₂

Solving for v₂, we get:
v₂ = (m₁ * v₁) / 0.7044 kg

Substituting the given values, we have:
v₂ = (0.500 kg * 0.800 m/s) / 0.7044 kg
= 0.5688 m/s

Now that we have the final velocity of ball A after the collision, we can calculate its momentum.

Momentum of ball A after the collision (p₂) = m₁ * v₂

Substituting the given values, we have:
p₂ = 0.500 kg * 0.5688 m/s
= 0.2844 kg m/s

Therefore, the momentum of ball A after the collision is 0.2844 kg m/s.