Answer Momentum Question: Find Puck A & B Final Speeds

[Katerina Villanova]

I am desparate and only have 1 submission left on this question. Can someone help me with the answer please?
8. [CJ5 7.P.032.] The drawing shows a collision between two pucks on an air-hockey table. Puck A has a mass of 0.020 kg and is moving along the x-axis with a velocity of +5.5 m/s. It makes a collision with puck B, which has a mass of 0.040 kg and is initially at rest. After the collision, the two pucks fly apart with the angles shown in the drawing.

Puck a goes into quadrant 1 at an angle of 65.
Puck b goes into quadrant 4 at an angle of 37.


Find the final speed of
(a) puck A and
m/s
(b) puck B.
m/s

 

[rhia]

Hi Katerina,
Use the vector equation of conservation of linear momentum.The motion in your problem is restricted to a plane so you need to deal with only x and y components of momentum.Steps involved:
1) Choose your system such that total external forces along x and y directions is zernly then momentum will be conserved in these directions.The obvious choice is (A+B)
2)Write the total momentum of your system before collision in x direction and equate it with the total momentum of the same system after collision in same direction.Repeat for y direction.
Remember to take x and y components of final momenta as they are not strictly in x and y directions.

 

[wais]

To find the final speeds of puck A and puck B, we can use the conservation of momentum equation, which states that the total momentum before a collision is equal to the total momentum after the collision.

Before the collision, the total momentum is given by:
P = m1v1 + m2v2
Where m1 and m2 are the masses of puck A and B, and v1 and v2 are their respective velocities.

Plugging in the values given in the problem, we get:
P = (0.020 kg)(5.5 m/s) + (0.040 kg)(0 m/s)
P = 0.11 kg*m/s

After the collision, the total momentum will still be 0.11 kg*m/s, but it will be divided between the two pucks. We can use this to find their individual final speeds.

(a) To find the final speed of puck A, we can use the momentum equation again, but this time with the mass and velocity of puck B (since puck A will be at rest after the collision).

P = m1v1 + m2v2
0.11 kg*m/s = (0.040 kg)(v1) + (0.020 kg)(0 m/s)
Solving for v1, we get:
v1 = 2.75 m/s

Therefore, the final speed of puck A is 2.75 m/s.

(b) To find the final speed of puck B, we can use the same equation, but this time with the mass and velocity of puck A (since puck B will be at rest after the collision).

P = m1v1 + m2v2
0.11 kg*m/s = (0.020 kg)(0 m/s) + (0.040 kg)(v2)
Solving for v2, we get:
v2 = 2.75 m/s

Therefore, the final speed of puck B is also 2.75 m/s.

In conclusion, the final speeds of puck A and B are both 2.75 m/s. It is important to note that these values are only approximate, since we are assuming an ideal elastic collision where no energy is lost. In a real-life scenario, some energy would be lost as heat and sound, resulting in slightly different final speeds.