Solve 8. CJ5 7.P.032: Puck A & B Final Speed

[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

 

[Andrew Mason]

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

Since there is no net momentum in the direction of the y axis:
m_{Af} v_{Af}sin\alpha + m_{Bf}v_{Bf}sin\beta = 0

Since initial x momentum is conserved:
m_{Af} v_{Af}cos\alpha + m_{Bf}v_{Bf}cos\beta = m_{Ai} v_{Ai}

AM

 

[wais]

I understand that you may be feeling desperate and have limited attempts left on this question, but it is important to remember that cheating or seeking answers from others is not the way to learn and succeed. Instead of focusing on getting the answer, try to understand the concepts and equations involved in solving this problem. This will not only help you in this particular question, but also in future similar problems.

To solve this problem, we can use the conservation of momentum and conservation of kinetic energy equations. The conservation of momentum equation states that the total momentum before a collision is equal to the total momentum after the collision. In this case, we have two pucks colliding, so we can write:

m1v1i + m2v2i = m1v1f + m2v2f

Where m1 and m2 are the masses of pucks A and B respectively, v1i and v2i are their initial velocities, and v1f and v2f are their final velocities.

We also know that the conservation of kinetic energy equation states that the total kinetic energy before a collision is equal to the total kinetic energy after the collision. In this case, we can write:

0.5m1v1i^2 + 0.5m2v2i^2 = 0.5m1v1f^2 + 0.5m2v2f^2

Now, we can solve for the final velocities of both pucks. Plugging in the given values, we get:

0.020 kg (5.5 m/s) + 0.040 kg (0 m/s) = 0.020 kg (v1f) + 0.040 kg (v2f)
0.5(0.020 kg)(5.5 m/s)^2 + 0.5(0.040 kg)(0 m/s)^2 = 0.5(0.020 kg)(v1f)^2 + 0.5(0.040 kg)(v2f)^2

Simplifying, we get:

0.11 kg m/s + 0 = 0.02 kg (v1f) + 0.04 kg (v2f)
0.605 J = 0.01 kg (v1f)^2 + 0.04 kg (v2f)^2

Now, we have two equations with two unknowns