Problem 8 is for the Mass 3m find the time that is required to reach the floor (from the top of the ramp to the floor).
I use three formulas v0sin theta =V0y, y = v0yt =1/2 at^2, and the quadratic formula and you solve for t.
Problem 9 is Calculate the velocity of the mass 3m at the point of...
thanks tiny tim
problem 1 is the same as before
for problem two and three we use the formula PE = (mV1^2)/2 + (3mV2^2)/2 to calculate conservation of energy and mv1 =3mv2 to calculate conservation of momentum. this will give you 8.39 m/s for problem 2 and 2.8 m/s for problem three.
problem 4...
I'm running out of time can't you just tell me what I need to do? I can do the rest and I have shown an attempt at the problem. Also I know that 1,4,5,and 7 are correct. and their were actually 11 problems and I can solve those If I can find the answer to 2 and 3.
OK I'm not quite sure how to do that, But this is my attempt. Because of conservation of momentum I know PE = (.5V1^2)/2 + (1.5V2^2)/2 but that leaves me with two variables, so what do I do? M =.5 and 3M = 1.5, which means I am using the formula KE = (mv^2)/2
final answers:
for problem 1: PE =KE, where PE = potential energy and KE = kinetic energy formula two equals .5K delta X^2. this means PE = (750*.25^2)/2 = 23.44 J
for problem 2 I believe you use V = square root (2KE)/m = square root 2(23.44)/.5 = 9.68m/s
I believe problem 3 is solved the same...
The way I see it I found the x direction velocity in problem 3 so I need to find V that accommodates two dimensions (the x as well as the y). so i am using the formula v0cos 20 = vx so vx/cos20 = v0 which gives me 5.55m/s. So I'm wondering if my method is correct?
I'm confused how does that help me find the speed of mass 3m when it leaves the ramp? Your answer seems to be the perspective I need to look at this problem, is their something wrong with how I have been solving the problems? If so where did I mess up and if not using work/energy formulas how do...
I believe I made a mistake on # 5 so what I think I was suppose to do was take the work of the spring - the work of friction = (mv^2)/2 now multiply the new work by 2 divide by m and square root both sides and you get velocity. so the answer should be 9.568 round to 9.57.
Homework Statement
Two masses m and 3m are joined together by a compressed massless spring of constant k = 750 N/m. When spring is released it pushes masses apart and they slide on a horizontal tabletop as shown below. Mass 3m slides to the right and its track is frictionless. Mass m slides to...
my final answers
part a: 5.93 * 10^6 m/s
part b: 1.686 * 10^7 s
part c: 8.78 * 10^11 m/s^2
part d .0125 m
part e: 6.078 * 10^6 m/s
part f: 45.71 degrees
final questions
can I uses 5.93 * 10^6 m/s from part a as V0 in part e?
can I uses 6.078 * 10^6 m/s from part e as velocity y in part f
for part d I was suppose to find vertical displacement where gravity is negligible so I used the formula x = (at^2)/2 so my answer was .02496
for part e. i need to find the magnitude of velocity so I use the formula V = V0 + at I used part a's velocity for V0 and I used vertical...
so because it is an electron q = 1.6 *10^-19 C and M = 9.11* 10 ^-31. and the Volts in the parallel plates is 1 and the change in vertical distance is .1 M so it should be (q(1/.1))/M. if this is correct then the answer should be a = 1.756 * 10^12. did i do this right if not please tell me where...
I was already their my question was if how I found F was correct i applied f = qE, E = k(q/r^2), and V= k(q/r) and derived F = V^2/K. to get the answer you use a= F/m but my question was how to get F.
can anyone that just helped mocambo help me my problem is at https://www.physicsforums.com/showthread.php?p=1627736#post1627736
I really need help, anything will be greatly appreciated.
Ok i now am confused? I don't know how to find a I just try two different ways but I think they are both wrong.
first way: ended up using the Pythagorean theorem to find distance traveled and I plugged it into x = vt +.5at^2 where I use my horizontal velocity as the initial velocity. I think...
I do not think so because the parallel plates voltage is +1 which would change the force which would also means it changes the velocity. unless the x component of velocity is constant which I'm not certain of. also check out that pic I think my reasoning will be clearer if you see it.