Final Velocity (involving momentum)

[grouchy]

[SOLVED] Final Velocity (involving momentum)

Homework Statement

A 0.329-kg puck, initially at rest on a frictionless horizontal surface, is struck by a 0.222-kg puck that is intitially moving along the x-axis with a velocity of 2.24 m/s. After the collision, the 0.222-kg puck has a speed of 1.24 m/s at an angle of 24 degrees to the positive x-axis. (a) Determine the velocity of the 0.329-kg puck after the collision. (b) Find the fraction of the kinetic energy lost in the collision.

The Attempt at a Solution

i tried m1 = 0.329kg ----- m2 = 0.222kg

m1v1(i) + m2v2(i) = m1v1(f) + m2v2(f)

v1(f) = [m2v2(i) - m2v2(f) cos theta] / m1
v1(f) = 0.7471105 m/s (this was wrong)

I assume I need to find the x and y components and then use vector sumation but I have NO idea how to find the y...

 

[rl.bhat]

Let v2(f) makes an angle theta with positivde x-axis. Take the compondents of v1(f) and v2(f) along x and y- axis. Y camponents miust be equal and opposite because v1 is along x-axis. So you can wright two equations. 1) m1v1 + m2*0 = m1v1(f)cos(24) + m2v2(f)cor(theta)
2) m1v1(f)sin(24) = m2v2(f)sin(theta) . Rewrigth the equation(1) as
m1v1 -m1vi(f)cos(24) = m2v2(f)cos(theta)...(.3) Square eq. 2 and 3 and add. After solving you will get v2(f)

 

[ogb p]

I can confirm that your attempt at solving this problem is correct. However, the reason why you obtained the wrong answer for the final velocity of the 0.329-kg puck is because you have not taken into account the conservation of momentum in the y-direction. In this case, the y-component of momentum is also conserved during the collision.

To find the y-component of momentum, you can use the equation:

m1v1y(i) + m2v2y(i) = m1v1y(f) + m2v2y(f)

Since the initial velocity of the 0.329-kg puck is 0 m/s, the equation becomes:

m2v2y(i) = m1v1y(f) + m2v2y(f)

To solve for v1y(f), you can use the fact that the final velocity of the 0.222-kg puck is at an angle of 24 degrees to the positive x-axis. This means that the y-component of its velocity can be found using the equation:

v2y(f) = v2(f) sin theta

Now, you can plug in the known values and solve for v1y(f). Once you have both the x and y components of the final velocity of the 0.329-kg puck, you can use vector addition to find the final velocity.

As for part (b) of the problem, you can use the equation for kinetic energy:

KE = 1/2 * m * v^2

To find the fraction of kinetic energy lost, you can compare the initial kinetic energy (before the collision) with the final kinetic energy (after the collision). The difference between the two will give you the amount of kinetic energy lost. You can then divide this by the initial kinetic energy to find the fraction lost.

I hope this explanation helps you understand the problem better and allows you to arrive at the correct answer. Remember to always consider all the relevant principles, such as conservation of momentum and energy, when solving physics problems. Keep up the good work!