Billiard ball collision at right angle

In summary, two billiard balls of equal mass collide at the origin of an xy coordinate system with ball A moving upward along the y-axis at 2.0 m/s and ball B moving to the right along the x-axis at 3.7 m/s. After the collision, ball B continues to move along the positive y-axis. The final direction of ball A is unknown, but its speed remains at 2.0 m/s. The final speed of ball B is also unknown, but it is moving along the positive y-axis at a speed of 3.7 m/s. To solve this problem, the conservation of momentum and kinetic energy will need to be applied separately in the x and y dimensions.
  • #1
Brite
1
0

Homework Statement


Two billiard balls of equal mass move at right angles and meet at the origin of an xy coordinate system. Ball A is moving upward along the y-axis at 2.0 m/s and ball B is moving to the right along the x-axis with speed 3.7 m/s. After the collision (elastic), ball B is moving along the positive y axis. What is the final direction of ball A and what are their two speeds?

Homework Equations


mAvA + mBvB = mav'A + mBv'B
1/2mAvA2 + 1/2mBvB2 = 1/2mAv'A2 + 1/2mBv'B2

The Attempt at a Solution


I know that because this is an elastic collision, momentum and kinetic energy are conserved. I think it's just that the two balls start on different axes is what's throwing me off.
 
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  • #2
Brite said:

Homework Statement


Two billiard balls of equal mass move at right angles and meet at the origin of an xy coordinate system. Ball A is moving upward along the y-axis at 2.0 m/s and ball B is moving to the right along the x-axis with speed 3.7 m/s. After the collision (elastic), ball B is moving along the positive y axis. What is the final direction of ball A and what are their two speeds?


Homework Equations


mAvA + mBvB = mav'A + mBv'B
1/2mAvA2 + 1/2mBvB2 = 1/2mAv'A2 + 1/2mBv'B2


The Attempt at a Solution


I know that because this is an elastic collision, momentum and kinetic energy are conserved. I think it's just that the two balls start on different axes is what's throwing me off.

Welcome to PF.

Indeed the 2 dimensions do make a difference. You will want to treat them separately.

Momentum in x is conserved
Momentum in y is conserved
KE is conserved.
 
  • #3


Based on the information provided, the final direction of ball A will be along the negative x-axis, since it was originally moving upward along the y-axis and after the collision, it will be moving in the opposite direction along the x-axis. The final speed of ball A can be calculated using the conservation of momentum and kinetic energy equations.

Using the conservation of momentum:

mAvA + mBvB = mav'A + mBv'B

Since the masses of the balls are equal, we can simplify the equation to:

2m(vA + vB) = mav'A + mBv'B

Solving for vA, we get:

vA = (mav'A + mBv'B - 2mBvB)/2m

Substituting in the initial and final velocities for ball A and B, we get:

vA = (mA(0) + mB(3.7 m/s) - 2mB(2.0 m/s))/2m

vA = (3.7 m/s - 4.0 m/s)/2 = -0.15 m/s

Therefore, the final speed of ball A is 0.15 m/s along the negative x-axis.

To calculate the final speed of ball B, we can use the conservation of kinetic energy equation:

1/2mAvA2 + 1/2mBvB2 = 1/2mAv'A2 + 1/2mBv'B2

Since the masses and initial velocities of the balls are known, we can solve for v'B:

1/2mAvA2 + 1/2mBvB2 = 1/2mAv'A2 + 1/2mBv'B2

Substituting in the values, we get:

1/2(mA(-0.15 m/s)2 + 1/2(mB(3.7 m/s)2 = 1/2(mA(0)2 + 1/2(mBv'B2)

Simplifying, we get:

0.0225 m2/s2 + 6.845 m2/s2 = 1/2mBv'B2

Solving for v'B, we get:

v'B = √(2(0.0225 m2/s2 + 6.845 m2/s2)/mB)
 

1. How do you calculate the velocity of two billiard balls colliding at a right angle?

The velocity of the two billiard balls can be calculated using the conservation of momentum and energy equations. First, calculate the total momentum of both balls before the collision by multiplying the mass of each ball by its velocity. Next, use the conservation of momentum equation to find the final velocity of the two balls after the collision. Finally, use the conservation of energy equation to find the kinetic energy of the two balls after the collision.

2. What factors affect the outcome of a billiard ball collision at a right angle?

The outcome of a billiard ball collision at a right angle is affected by several factors, including the masses and velocities of the two balls, the angle at which they collide, and the elasticity of the balls. Other factors such as friction and spin can also play a role in the outcome of the collision.

3. How does the elasticity of the billiard balls impact the collision at a right angle?

The elasticity of the billiard balls determines how much kinetic energy is conserved during the collision. If the balls are highly elastic, they will bounce off each other with minimal loss of energy. If the balls are not very elastic, they will stick together after the collision and lose a significant amount of energy.

4. Can two billiard balls collide at a right angle and stick together?

Yes, it is possible for two billiard balls to collide at a right angle and stick together. This can happen if the balls have very little elasticity or if the collision occurs at a very low speed. In this case, the two balls will stick together and move together as one object after the collision.

5. Is the collision of two billiard balls at a right angle completely elastic?

No, the collision of two billiard balls at a right angle is not completely elastic. Some kinetic energy is always lost due to factors such as friction and sound. However, the more elastic the balls are, the closer the collision will be to being completely elastic.

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