Conservation of Momentum problem-don't understand the solution

[skier07]

[SOLVED] Conservation of Momentum problem-don't understand the solution

Homework Statement

The drawing shows a collision between two pucks on an air-hockey table. Puck A has a mass of 0.025kg 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 0.050kg and is initially at rest. The collision is not head-on. After the collision, the two pucks fly apart with the angles shown in the drawing. Find the final speed of (a) puck A and (b) puck B.

Homework Equations

The Attempt at a Solution

X direction:


Y direction:






Plug into (2)

 

[MasterTinker]

More Information?

Homework Statement

After the collision, the two pucks fly apart with the angles shown in the drawing.

...

Plug into (2)

No one has posted yet to help you but I think that you should include the drawing and demonstrate what equation (2) is exactly. The lack of information makes your problem difficult to solve...

 

[Moetasim]

to get V2f = 3.7m/s, which is the final speed of puck A. For puck B, we can use conservation of momentum to solve for its final velocity in the x-direction:


V2f = 2.4m/s

The solution provided is correct. It seems that you have correctly applied the conservation of momentum principle to solve for the final velocity of puck A. To solve for the final velocity of puck B, you can use the equation V2f = (m1 * V1i + m2 * V2i)/(m1 + m2), where m1 and m2 are the masses of the two pucks and V1i and V2i are their initial velocities. This equation takes into account the conservation of momentum in both the x and y directions. Plugging in the values given in the problem, we get V2f = (0.025kg * 5.5m/s * cos(45°) + 0.050kg * 0m/s * cos(0°))/(0.025kg + 0.050kg) = 2.4m/s. This is the final velocity of puck B in the x-direction. In the y-direction, the final velocity can be found using the equation V2f = (m1 * V1i + m2 * V2i)/(m1 + m2) * sin(θ), where θ is the angle shown in the drawing. Plugging in the values, we get V2f = (0.025kg * 5.5m/s * sin(45°) + 0.050kg * 0m/s * sin(0°))/(0.025kg + 0.050kg) = 1.85m/s. Therefore, the final velocity of puck B is 2.4m/s in the x-direction and 1.85m/s in the y-direction. I hope this helps clarify the solution for you.