2 Questions dealing with energy and momentum.

[cruisx]

Homework Statement

Hi guys, i have the following questions that i am having trouble with and need some help to solve.

http://img692.imageshack.us/img692/7827/capturey.jpg

Homework Equations

+ The attempt at a solution


Well for the first Question, i tried to uses

W = 1/2 mV₂² - 1/2 mV²

and for mass i subbed in the mass of the bullet and subbed in the two different velocities and the Work was 81 888.125J
I then subbed that into (over here i used the mass of the object. )

h = 81 888.125J/ ( 0.245 * 9.8m/s² )

and found h to be 34 105m but that seems wrong so i don't know how to solve this one.

Problem 2---------------------------------------------------------------------------------------------------------

Now for the second problem, i used

Et1 = Et2
Et1= mg(1.5m)
Et2 = 1/2 mV₂² + 1/2 Kx₂²
mg(1.5m) = 1/2 mV₂² + 1/2 (588N/M)(0.50m)²
147J = 1/2 10KgV₂² + 147J


And V2 was 3.16m/s

is this correct? Help would be appreciated. Thankyou. Our teacher said that we should try to solve these questions for tomorrows lessons so we can take them up.

 

[anathema]

I can see that you have put in a lot of effort into solving these questions, but there are a few mistakes in your attempts. Let's go through them one by one.

For the first question, you are correct in using the work-energy theorem to solve for the height, but you have made a mistake in calculating the work done. The correct formula for work is W = Fd, where F is the force applied and d is the displacement. In this case, the force applied is the weight of the bullet, which is mg, and the displacement is the height h. So the correct equation would be W = mgh. Substituting the values, we get W = 0.245kg * 9.8m/s^2 * 34,105m = 81,888.125J. Therefore, your answer for the work done is correct, but the height should be 34,105m.

For the second question, you have made a small error in the equation. The correct equation for the conservation of energy would be Et1 = Et2, where Et1 is the initial total energy and Et2 is the final total energy. In your attempt, you have equated Et1 to mg(1.5m), which is the potential energy at the starting point. However, the initial total energy should include both potential and kinetic energy, so the correct equation would be Et1 = mg(1.5m) + 1/2 mv^2. Substituting the values, we get 147J = 1/2 * 10kg * v^2 + 147J. Solving for v, we get v = 3.16m/s, which is correct.

I hope this helps you to understand and solve these questions more accurately. Keep up the good work and don't hesitate to ask for help when needed. Best of luck for your lesson tomorrow!

 

[kerimek]

Hi there, it's great to see you working on these problems and seeking help when needed. Here are my thoughts on your solutions:

Question 1:

Your attempt at using the work-energy theorem is a good start. However, I believe you may have made a mistake in your calculation. The work done by the bullet can be found using the equation W = Fd, where F is the force and d is the distance. In this case, the force is equal to the weight of the object, which is given by mg. So the work done is W = mgd. Substituting in the values given, we get W = (0.245 kg)(9.8 m/s^2)(35 m) = 84.665 J. This is the work done by the bullet on the object, and it is equal to the change in kinetic energy of the object. So we can use the equation W = 1/2 mv^2 to find the final velocity of the object, which is 3.53 m/s.

To find the height, we can use the conservation of energy principle, where the initial potential energy is equal to the final kinetic energy. So we have mgh = 1/2 mv^2. Solving for h, we get h = v^2/2g = (3.53 m/s)^2 / (2)(9.8 m/s^2) = 0.65 m. This is the correct answer for the height.

Question 2:

Your approach to this problem is correct. The total energy on both sides should be equal, so we can set them equal to each other. However, I believe you may have made a mistake in your calculation. The potential energy in this case is not equal to mg(1.5 m), but rather mg(0.5 m) since the object has only risen 0.5 m.

So we have mg(0.5 m) = 1/2 mv^2 + 1/2 kx^2. Substituting in the values given, we get (10 kg)(9.8 m/s^2)(0.5 m) = 1/2 (10 kg)(3.16 m/s)^2 + 1/2 (588 N/m)(0.5 m)^2. This gives us a final velocity of 3.16 m/s, which is the correct answer