Conservation of Momentum Problem

• skier07
In summary, the two air hockey pucks, A and B, with masses of 0.025 kg and 0.05 kg respectively, collide on an x-axis table. Puck A, initially moving at 5.5 m/s, collides with Puck B, initially at rest. After the collision, Puck A and Puck B fly apart at angles of 65 degrees and 37 degrees respectively. Using the equation for conservation of momentum, it can be determined that the final speeds of Puck A and Puck B are 5.46 m/s and 5.10 m/s, respectively.
skier07

Homework Statement

(understood: two air hockey pucks make contact--the x-axis is the table)
Puck A has a mass of .025 kg and is moving along the x-axis with a velocity of +5.5m/s. It makes a collision with puck B, which has a mass of .05kg and is initially at rest. The collision is NOT head on. After the collision, the two pucks fly apart with the following angles.
Find the final speed of a) puck A and b) puck B

A:
Ma=.025kg
Voa=5.5m/s
Vfa= ?

B:
Mb=.05 kg
Vob = 0m/s
Vfb= ?
......A
...../
....../ 65 deg
-----A--->--B-------------
......\ 37 deg
.....\
......B

Homework Equations

(Ma)(Vfa) + (Mb)(Vfb) = (Ma)(Voa) + (Mb)(Vob)

Also, this equation can be written in terms of X and Y

X: (Ma)(Vfax) + (Mb)(Vfbx) = (Ma)(Voax) + (Mb)(Vobx)
Y: (Ma)(Vfay) + (Mb)(Vfby) = (Ma)(Voay) + (Mb)(Voby)

The Attempt at a Solution

Since it is an air hockey table, friction is negligible.
therefore, Momentum is conserved, and Fnet=0

Initial momentum in Y = 0
Initial momentum in x = (Ma)(Voax) + 0 = .1375 kgm/sec

Last edited:
Momentum in Y after collision = (Ma)(Vfay) + (Mb)(Vfby) = 0 Momentum in x after collision = (Ma)(Vfax) + (Mb)(Vfbx) = .1375 kgm/secTherefore,X: (Ma)(Vfax) + (Mb)(Vfbx) = (Ma)(Voax) + 0Y: (Ma)(Vfay) + (Mb)(Vfby) = 0 Using trigonometry, Vfay = Voa * sin(65) = 3.53 m/sVfox = Voa * cos(65) = 4.19 m/s Vfby = Voa * sin(37) = 2.43 m/sVfbx = Voa * cos(37) = 4.98 m/sTherefore, Vfa = √(Vfay^2 + Vfox^2) = 5.46 m/sVfb = √(Vfby^2 + Vfbx^2) = 5.10 m/s

Final momentum in Y = (Ma)(Vfay) + (Mb)(Vfby) = 0
Final momentum in X = (Ma)(Vfax) + (Mb)(Vfbx) = .1375 kgm/sec

Using the given angles, we can split the final velocities into their X and Y components using trigonometry.

For puck A:
Vfax = Vfa*cos(65) = 2.39 m/s
Vfay = Vfa*sin(65) = 4.99 m/s

For puck B:
Vfbx = Vfb*cos(37) = 4.02 m/s
Vfby = Vfb*sin(37) = 2.99 m/s

Plugging these values into the conservation of momentum equation, we get:

(Ma)(Vfax) + (Mb)(Vfbx) = (Ma)(Voa) + (Mb)(Vob)
(.025 kg)(2.39 m/s) + (.05 kg)(4.02 m/s) = (.025 kg)(5.5 m/s) + (.05 kg)(0 m/s)
.06 kgm/s + .2 kgm/s = .1375 kgm/s + 0
.26 kgm/s = .1375 kgm/s

Therefore, the final speed of puck A is Vfa = .26 kgm/s and the final speed of puck B is Vfb = .1375 kgm/s.

1. What is conservation of momentum?

Conservation of momentum is a fundamental law of physics that states that the total momentum of a closed system remains constant, unless acted upon by an external force.

2. How is momentum defined?

Momentum is defined as the product of an object's mass and velocity, and is a vector quantity with both magnitude and direction.

3. What is an example of a conservation of momentum problem?

An example of a conservation of momentum problem is a collision between two objects, where the total momentum before the collision is equal to the total momentum after the collision.

4. What is the equation for conservation of momentum?

The equation for conservation of momentum is pi = pf, where p represents momentum and the subscripts i and f represent the initial and final states, respectively.

5. How does conservation of momentum apply to real-world situations?

Conservation of momentum applies to real-world situations such as rocket propulsion, car crashes, and sports like billiards and ice skating. It is also used in engineering and design to ensure the safety and efficiency of various systems.

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