Solve Elastic Collision: 2D Homework

• Valenti
In summary, two rubber pucks with masses of 40 grams each collide in an elastic collision on a frictionless air table. One is initially at rest while the other is moving at 4 m/s in the x direction. After the collision, the first puck travels in the +30 degree direction while the second puck travels in the -55 degree direction. Using conservation of energy and momentum, the final velocities of the pucks were found to be 4.6 m/s and 2.5 m/s, respectively. However, due to rounding errors, it is recommended to keep more significant digits during calculations for a more accurate result.
Valenti

Homework Statement

A red and a blue rubber puck are free to slide along a frictionless air table. Each has a mass of 40 grams. They collide in an elastic collision. Initially the red one is at rest and the blue one is traveling in the x direction with a speed of 4 m/s. After the collision the blue one is traveling in the direction +30 degrees, with the red one traveling in the direction ‐55 degrees. Using conservation of energy and momentum find the speed of each puck after the collision.

Homework Equations

m1v1i+m2v2i = m1v1f + m2v2f

The Attempt at a Solution

Solve for the momentum in each direction
m1v1i+m2v2i = m1v1f + m2v2f
X Direction
m1v1i+m2v2i = m1v1f + m2v2f
0.04kg(4m/s) + 0.04kg(0m/s) = 0.04 (v1f cos30) + 0.04(v2f cos-55)
0.16kgm/s = 0.03v1f + 0.02v2f

Y Direction
m1v1i+m2v2i = m1v1f + m2v2f
0.04kg(0) + 0.04kg(0m/s) = 0.04kg(v1f sin30) + 0.04kg (v2f sin-55)
0 = 0.02v1f - 0.03 v2f

Solve for one of the Velocities
0 = 0.02v1f - 0.03 v2f
0.03v2f = 0.02v1f
v2f = 2/3 v1f

Replace with new velocity
0.16kgm/s = 0.03v1f + 0.02v2f
0.16kgm/s = 0.03v1f + 0.02(2/3)v1f
0.16kgm/s = 0.03v1f + 1/75 v1f
0.16kgm/s = 13/300 v1f
3.69m/s = v1f

Solve for v2f
v2f = 2/3 v1f
v2f = 2/3 (3.69 m/s)
v2f = 2.46 m/s

Rounding to only 1 significant digit at intermediate steps might be making your final answer very inaccurate.
Best is to avoid plugging in numbers until the final step. Work entirely symbolically, even providing symbolic variables to represent data given as numbers. It has numerous advantages.
When you do plug in the numbers, keep 3 or 4 significant digits in order to quote an answer of 2 or 3 digits. Even then, watch out if at some point you find you take the difference between two numbers that are close in value.
I reconstructed the initial velocity from your answers and the x momentum equation and got 4.6 m/s.

haruspex said:
Rounding to only 1 significant digit at intermediate steps might be making your final answer very inaccurate.
Best is to avoid plugging in numbers until the final step. Work entirely symbolically, even providing symbolic variables to represent data given as numbers. It has numerous advantages.
When you do plug in the numbers, keep 3 or 4 significant digits in order to quote an answer of 2 or 3 digits. Even then, watch out if at some point you find you take the difference between two numbers that are close in value.
I reconstructed the initial velocity from your answers and the x momentum equation and got 4.6 m/s.
Ah alright thanks I'll be sure to do that for upcoming questions. As for this question not counting my errors in significant digits was it wrong, or did i have the right procedure?

Valenti said:
Ah alright thanks I'll be sure to do that for upcoming questions. As for this question not counting my errors in significant digits was it wrong, or did i have the right procedure?
The method was fine.

You round off too much during the calculation. If you keep one significant digit only, the result will be less accurate than one digit. So you can write 4m/s = v1f and v2f = 2m/s. The data were given with two digits. You need keep at last 3 digits during the calculation to reach the result with accuracy of one digit.
By the way, the problem gives both angles, so you can solve for the velocities using conservation of momentum alone. According to the results, the collision is not elastic, energy is not conserved. Your result gives more energy after collision it was before, so it is impossible. Calculating with enough accuracy, the final energy is a bit less than the initial one, a Physically realistic result.

1. What is an elastic collision in 2D?

An elastic collision in 2D is a type of collision where the total kinetic energy of the system is conserved. This means that the objects involved in the collision do not lose any kinetic energy during the collision and the total momentum of the system is also conserved.

2. How do you solve for elastic collisions in 2D?

To solve for elastic collisions in 2D, you need to use the principles of conservation of momentum and conservation of kinetic energy. You also need to use the equations for calculating momentum and kinetic energy, as well as the equations for calculating the velocities of the objects before and after the collision.

3. What are the key variables involved in solving elastic collisions in 2D?

The key variables involved in solving elastic collisions in 2D are the masses, velocities, and angles of the objects before and after the collision. Other important variables may include the coefficient of restitution, which is a measure of the elasticity of the collision, and any external forces acting on the objects during the collision.

4. Can elastic collisions occur in real-life scenarios?

Yes, elastic collisions can occur in real-life scenarios, although they may be difficult to observe. Examples of elastic collisions in real-life include collisions between gas molecules in a container, collisions between billiard balls, and collisions between particles in particle accelerators.

5. What are some common misconceptions about elastic collisions in 2D?

One common misconception is that elastic collisions only occur between perfectly spherical objects. In reality, elastic collisions can occur between any objects, regardless of their shape. Another misconception is that the total kinetic energy of the system must be conserved in all directions, when in fact it is only conserved in the direction of the collision.

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