Number of results: 115,703
Bernoulli's equation is an expression of: the conservation of mass the conservation of total energy the conservation of kinetic energy the conservation of momentum the conservation of velocity My ans: the equation states that it should be a constant when there is a streamline ...
Wednesday, December 2, 2009 at 9:29pm by Nikita
Physics! - PLEASE HELP
a) use conservation of energy. b) use conservation of momentum as the source for initial combined velocity, then conservation of energy to transform that KE to height. c. use conservation of momentum.
Friday, August 26, 2011 at 10:36am by bobpursley
it says to use conservation of energy we know conservation of energy when not dealing with heat is u+k=u+k where u is the potential energy and k is you kinetic energy. potential energy is mass x gravity x height. kinetic energy is (1/2)(mass)(velocity)^2. from there you get ...
Monday, November 15, 2010 at 11:57pm by jc
you have to use both conservation of momentum and energy. Start with conservation of momentum, solve for one of the ball's velocity in terms of all the other variables, then substitute that into the conservation of energy. The algebra is substantial, so be prepared.
Tuesday, August 9, 2011 at 11:49pm by bobpursley
I have no idea what the picture is, but you can use conservation of energy (as you have done) to solve it. The speed at C is zero, so you know the energy was lost to friction. Again, conservation of energy.
Thursday, October 25, 2007 at 12:16am by bobpursley
Use conservation of energy. The mass won't matter; it cancels out of the energy conservation equaiton. M g H = (1/2) M V^2 Solve for H, in terms of V.
Sunday, January 2, 2011 at 5:47pm by drwls
A bullet with a mass of 55 grams moving with a speed of 100 m/s slams into and wedges in a 1.9 kilogram block of wood supported by a string initially at rest. Which one, conservation of energy or conservation of momentum, can be applied to this situation? Why? I think it's ...
Thursday, October 14, 2010 at 12:34am by Fred
Use a conservation of energy method to get the velocity of the 1 kg object at the base of the incline. (potential energy loss) = (work done against friction) + (kinetic energy at bottom) Once you have that velocity (V), use conservation of momentum to get the velocity V' of ...
Sunday, December 21, 2008 at 9:16am by drwls
ii) Apply conservation of energy, making sure you count both translational and rotational KE now with this part Im not sure what you mean? i understand the conservation of energy but that wont give me the speed of the wheel
Tuesday, July 20, 2010 at 11:22am by Oisin
The law of conservation of energy states that energy can neither be created nor destroyed. In many instances it may look as though energy is gained or lost, but it is really only changed in form. How is the law of conservation of energy represented when an archer shoots an ...
Wednesday, January 5, 2011 at 6:55pm by Steven
Physics(work and energy)
You have done it well, the principle is conservation of energy: the energy at the top of the wall (PE+KE) is equal the the original energy.
Sunday, May 20, 2012 at 12:50pm by bobpursley
you need to use conservation of momentum and conservation of energy..... v = 2,......
Saturday, January 11, 2014 at 3:36am by Hawk
and what is the question? You cannot use conservation of energy here, only conservation of momentum.
Tuesday, November 17, 2009 at 12:19pm by bobpursley
use the conservation of momentum, conservation of energy. A bit of algebra will be required.
Sunday, November 7, 2010 at 5:06pm by bobpursley
IT is much to complicated to work out here. worke it this way: conservation of momentum in x direction conservation of momentum in y direction conservation of energy. That should give you all you need.
Friday, October 23, 2009 at 11:19pm by bobpursley
Use conservation of momentum and conservation of kinetic energy (since it's an elastic collision.)
Wednesday, January 22, 2014 at 2:50am by Kiwi
Since the collision is elastic everything listed may be calculated from conservation of momentum and conservation of energy.
Saturday, March 5, 2011 at 6:26pm by Damon
You have to use two equations: Conservation of momentum, and conservation of energy. I will be happy to critique your work.
Thursday, March 6, 2008 at 7:56am by bobpursley
No. You cant use that. conservation of energy does not apply when the bullet hits the block. You can use conservation of energy at spring: block, but that is all.
Friday, December 10, 2010 at 2:23pm by bobpursley
Does anyone know what the laws of conservation are when dealing with resistors? Well, I guess you can vaguely state if you have, for example, 2 resistors in series: then, in accordance with the law of conservation of energy, the voltage drop across each resistor (the energy ...
Thursday, February 1, 2007 at 10:26pm by matt
The sum of the emfs and potential difference around a closed loop equals zero" is a consequence of: a) Newton's third law b) Ohm's law c) Newton's second law d) conservation of energy e) conservation of charge For this question: I had picked d conservation of energy but now I ...
Wednesday, March 28, 2007 at 12:10pm by susane
You can rest assured on this: Conservation of momentum is ALWAYS conserved. You dont have to determine it. Now conservation of Energy is another matter, it is not always conserved. If it is and elastic collions, energy is conserved.
Thursday, August 20, 2009 at 9:59pm by bobpursley
Use the conservation ofmomentum to solve for one veloicty in terms of the other. then put that into the conservation of energy equation.
Sunday, October 25, 2009 at 9:16am by bobpursley
Before applying conservation of energy law, you have to evaluate the speed of the block+wad system by applying conservation of impulse law: m(block+wad)*V = m(wad)*V(wad) So, find V and then use it in conservation of energy law (keep in mind that the the kinetic energy of ...
Friday, November 30, 2012 at 5:51pm by Gene
I am doing my final project on energy conservation.What are some good resources on energy conservation?
Saturday, June 20, 2009 at 10:36pm by sam
(a) Use conservation of energy. The distance it descends is 2.00 sin 30 = 1.00 meter. Mass will cancel out. (b) Use conservation of energy again, but subtract the heat loss due to friction work from the energy available for kinetic energy. The coefficient of kinetic friction ...
Saturday, March 26, 2011 at 9:19pm by drwls
Conservation of energy
Can someone explain to me what happens when you take a slap shot in hockey according to conservation of energy? I think that when you wind up you have potential energy, then when you swing the stick towards the puck you have kinetic energy, then when the stick flexes you have ...
Monday, June 13, 2011 at 8:11pm by Kyle
Use conservation of momentum to get the final velocity of the stuck-together blocks. Then compute the new total kinetic energy using that velocity. For the final step, use conservation of energy. Initial KE of the stuck blocks equals the potential energy gain at their highest ...
Tuesday, November 17, 2009 at 9:49pm by drwls
physics please help
inelastic equation: only conservation of momentum can be used, I dont think you will solve it. elastic equation; you can use both conservation of momentum and conservation of energy, you can solve it. A little algebra is required.
Sunday, November 7, 2010 at 2:45pm by bobpursley
Use the conservation of momentum, and conservation of energy equations to find the velocities of the ball and person after collision. I will be happy to critique your thinking.
Sunday, November 7, 2010 at 11:58am by bobpursley
Physics help please
remember, conservation of momentum is a vector, so do it in two directions, x, and y. Conservation of energy is just one equation. You can get three equations out of these.
Sunday, November 7, 2010 at 7:10pm by bobpursley
physics!-concervation of energy equations
wait...if this is conservation of energy, better to use: convert the potential energy into kinetic energy! (sorry). potential energy is mgh, and kinetic enery is (1/2)m*v^2
Thursday, December 4, 2008 at 9:33pm by DanH
The use of conservation of energy approach will tell you right away that V = sqrt(2 g H) loss of potential energy = M g H = gain in kinetic energy = (1/2) M V^2 Therefore 2gH = V^2
Monday, November 24, 2008 at 11:50am by drwls
Use conservation of energy and subtract 319 from the initial potential energy of MgH = 413 J. The difference is the kinetic energy at the bottom of the slide. Get the speed from that.
Thursday, March 8, 2012 at 9:09pm by drwls
What is your question? It this a ballistic pendulum situation? If so, apply conservation of momentum to the impact and conservation of energy to how far the pendulum swings.
Sunday, November 21, 2010 at 10:11pm by drwls
problem 1 is conservation of momentum m (v) + M(0) = (M+m)V problem 2 is conservation of energy Ke after collision = (1/2)(m+M)V^2 which = (M+m)g h
Friday, May 1, 2009 at 12:28am by Damon
1. figure with conservation of momentum the initial velocity of gayle/sled 2. Now, with conservation of energy, the speed at bottome of hill. KEbottom=KEattop+PEtop 1/2(Mg+ms)vbottom^2=1/2(Mg+ms)Vtop^2 + (Mg+Ms)g*height figure vbottom. Now, with conservation of momentum, ...
Sunday, October 17, 2010 at 5:30pm by bobpursley
Since the masses stick together, there is conservation of momentum but no conservation of energy. Let x be the new velocity. Hence, 0.5*0.4 + 0.8*(-0.20) = (0.5+0.8)*x X = (0.2-0.16)/1.3 = 0.0307 m/s KE = 0.5*1.3*0.0307^2 = 0.000615 J.
Tuesday, October 30, 2012 at 1:15am by Robert
Do this in three steps. I would need to see your figure to provide numerical values. I assume m1 initially slides down some ramp a certain distance. (1) Use conservation of energy to get the speed of m1 just before impact (2) Apply cnservation of energy and momentum to the ...
Friday, July 16, 2010 at 1:44am by drwls
percent efficiency: (final energy)/initial energy * 100 so I see about 1059/1130 * 100=93.7 percent Now, on a: demonstrate the law of conservation of energy: Does initial energy= final energy? not exactly, some mechanical energy was lost, probably to friction.
Sunday, January 31, 2010 at 11:28am by bobpursley
this case the energy of sistem is conservation of energy: (1/2)kx² = (1/2)mv² then you can find the energy of springs, this energy would be the energy of total sistem because no exist friccion. the maximum velocity of the mass v= square root((kx²)/ m) the speed when it has ...
Saturday, June 2, 2012 at 12:07am by Delbarre
Which of the following statements is a consequence of equation E=mc^2? Energy is released when matter is destroyed Mass and energy are equivalent The law of conservation of energy must be modified to state that mass and energy are conserved in any process all of the above ...
Wednesday, April 11, 2012 at 9:12pm by kevin
Because of conservation of energy, kinetic energy K is equal to the potential energy of the spring: K = U. So (mv^2)/2 = (kA^2)/2. Solving for k, we have k = (mv^2)/A^2, which is d).
Sunday, March 25, 2007 at 2:17pm by Pable
The correct answer is 21.276 Consider the law of conservation of energy in solving this problem. i.e. gravitational potential energy = elastic potential energy. So, m g (h-3) = 1/2 k x^2.
Tuesday, July 23, 2013 at 12:52pm by Ozone
Here's two or three clues. For the velocity of A just before impact (Va), use conservation of energy. g*h = (1/2)*Va^2 For the velocity of and b stuck together after impact (Vfinal) , use conservation of momentum. (It is not an elastic collision) Ma*Va = Vfinal*(Ma + Mb) = 15*...
Saturday, March 17, 2012 at 4:55pm by drwls
The easy way is to use conservation of energy. At the vertical position, the potential energy decrease, M g L/2, equals the kinetic energy increase, which is (1/2) I w^2 = (1/6)ML^2*w^2 gL/2 = L^2 w^2/6 w^2 = 3 g/L w = sqrt(3g/L) = 5.42 rad/s (since L = 1 meter)
Tuesday, October 25, 2011 at 12:44pm by drwls
No, not for a solid sphere. You can find it from the experiment. Conservation of energy: final energy= energy at top of incline 1/2mv^2 + 1/2 I w^2= mgh= 1.6*9.8*2.3sin28 but w=v*r, so 1/2 mv^2+1/2 I v^2 r^2= 1.6*9.8*2.3sin28 Solve for I.
Wednesday, October 24, 2007 at 1:31pm by bobpursley
Use energy: gravitational potential energy: GPE = mgh kinetic energy: KE = (1/2)m*v^2 By conservation of energy, GPEi+KEi = GPEf + KEf. (Final energy = initial energy). At the bottom of the hill, GPEf = 0 (h=0). At the top of the hill, KEi = 0(v=0). Find the final kinetic ...
Sunday, January 2, 2011 at 5:47pm by Marth
The kinetic energy when it reaches its final height must be 0. By conservation of energy, the energy it has at its final height is equal to the energy when it is launched.
Sunday, January 10, 2010 at 5:51pm by Marth
Assume conservation of mechanical energy (which means neglecting friction). The elevation change of the roller coaster is 50 sin 45 = 35.36 m The Potential Energy loss equals the Kinetic Energy gain.
Monday, July 12, 2010 at 9:56pm by drwls
How do I use Kirchhoff's First Law -- the law of conservation of energy and the law of conservation of charge to prove the equivalent resistance formula. Rt = R1 + R2 + R3 ......
Sunday, May 27, 2012 at 7:24pm by Art
Use conservation of energy. For the correct answer, you need to include the rotational kinetic energy of the ball.
Tuesday, November 22, 2011 at 10:12am by drwls
physics help (elena)
Friday, October 18, 2013 at 2:24pm by bobpursley
I was trying in this way - one equation is conservation of momentum, m2*v2 - m1*v1, and second conservation of energy, which is writen by Daoine. Two equations with two unknowns, but it gave me wrong answer
Saturday, January 11, 2014 at 3:36am by Anonymous
Use conservation of energy and be sure to include the rotational kinetic energy, (1/2) I w^2 For a cylinder, the moment of inertia is I = (1/2) M R^2 w is the angular velocity, which equals V/R. M and R will cancel out. Total kinetic energy = (1/2) M V^2 + (1/2) I w^2 = (1/2)M...
Thursday, June 2, 2011 at 9:40pm by drwls
I will be glad to critique your thoughts. Surely you know the difference between "conserving" (not wasting) energy and the Law of Conservation of Energy"
Thursday, August 26, 2010 at 2:03pm by drwls
Since it is frictionless, use conservation of energy. The work done equals the sum of kinetic and potential energy gains.
Wednesday, October 5, 2011 at 2:09pm by drwls
Friday, October 18, 2013 at 6:22am by Elena
There is also a shortcut way to do this using "energy conservation" that will give the same answer. At the maximum height H, the potential energy increase M g H will equal the initial kinetic energy, (1/2) M Vo^2) Therefore H = Vo^2/(2g)
Saturday, February 7, 2009 at 1:57pm by drwls
One method to solve this is conservation of energy. initial kinetic energy = final potential energy let m = mass of stone let g = acceleration of gravity let v = initial velocity let h = maximum height (1/2)mv^2 = mgh So, v^2 = 2gh h = (v^2)/g
Friday, September 25, 2009 at 12:35am by Quidditch
A. Apply conservation of momentum in the horizontal direction. B. Apply conservation of energy during comopression (1/2)kXmax^2 = (1/2)MV^2 V is the velocity from part a. Solve for Xmax c) Fmax = k Xmax
Sunday, January 29, 2012 at 9:48pm by drwls
(a) Use conservation of momentum. The mechanical energy converted by the girl equals the increase in the combined kinetic energy of both
Tuesday, February 23, 2010 at 7:24pm by drwls
Try using conservation of energy. Don't forget the frictional work term. There will be a change in the potential energy of the compressed spring.
Tuesday, March 13, 2012 at 2:03am by drwls
A quick way to do this is to use conservation of energy. The potential energy of the moving charge in the E-field of the fixed charge is: P.E. = k Q q/R where k = 8.99*10^9 N m^2/C^2 The kinetic energy of the particle is (1/2) m V^2 The initial energy, which is all potential, ...
Monday, January 28, 2008 at 10:23pm by drwls
You write that: "(A)fter the collision, both ball(s) move at 30 degree and 15 degree from horizontal. " Which is it? Why use the word "both" if they leave at different angles? This is an exercise in momentum and kinetic energy conservation. Since there are only two unknowns, ...
Wednesday, April 6, 2011 at 6:06am by drwls
Alcohol flows smoothly through a horizontal pipe that tapers in cross-sectional area from A1 = 41.9 cm2 to A2= A1/3. The pressure difference Äp between the wide and the narrow sections of the pipe is 11.1 kPa. What is the volume flow rate ÄV/Ät of the alcohol? The density of ...
Wednesday, December 2, 2009 at 8:02pm by Rebecca
Science 8R - help
How does the Law of conservation of energy compared to the conservation of matter? please help
Tuesday, November 27, 2012 at 8:22pm by Laruen
Use energy conservation. Potential energy change = maximum kinetic energy M g L (1 - cos 45) = (1/2) M Vmax^2 Mass M cancels out. L is the string length Vmax^2 = g*L*(0.5858) Vmax = 0.7654 sqrt(g L)
Friday, May 27, 2011 at 7:38am by drwls
Physics: Collison/Momentum Problem
conservation of energy. energy before= energy after 15*5^2+10*1^2=15*3^2+10*v^2 V^2= 1.5*25+1-1.5*9 v= 5m/s check the math.
Wednesday, January 25, 2012 at 8:03pm by bobpursley
Base on the law of conservation of energy, energy cannot be created nor destroyed. Knowing this, potential energy=kinetic energy. Find the potential energy. PE=mgh m=mass, g=gravity, h=height
Thursday, December 9, 2010 at 8:16pm by TutorCat
What is it you are failing to understand. Your subject listed is broad. angular motion angular momentum moments of inertia translation parallel axis theorem conservation of momentum conservation of energy coordinate systems, rotational
Monday, October 6, 2008 at 12:24pm by bobpursley
From conservation of energy followuing impact, gH = V'^2/2 V' = 1.3795 m/s That is the velocity of the block when it starts swinging. From conservation of momentum prior to the impact, mV = (m+M)V' V = (3.3048/0.0048)*1.3795 = 949.8 m/s That should be rounded to 950 m/s
Monday, October 31, 2011 at 9:32am by drwls
can someone give me 5 examples for the conservation of the angular momentum , 5 for the conservation of the impulse , and 5 for the conservation of the mechanical energy , in sport , arts or in technique , or just give me a site to find these examples, but please xamples must ...
Sunday, April 10, 2011 at 5:05pm by auro
B: form conseravtion of momentum, which yields the angular velocities of 7 rad/sec that you used. D follows from conservation of energy. The work done equals the increase in kinetic energy.
Thursday, June 25, 2009 at 11:59pm by Count Iblis
Lacking "the figure below", I cannot visualize this problem. It looks like an exercise in using conservation of energy, with three types of energy: spring and gravitational potential, and kinetic.
Sunday, March 21, 2010 at 3:09pm by drwls
you have to use conservation of momentum, and conservation of energy on this. Remember the boy moves after he throws it also.
Sunday, September 11, 2011 at 6:39pm by bobpursley
Physics - Rube Goldberg Project
You might begin here: http://search.yahoo.com/search?fr=mcafee&p=Rube+Goldberg+Machine+to+demonstrate+projectile+motion%2C+conservation+of+energy+%26+conservation+momentum Sra
Tuesday, August 31, 2010 at 12:37pm by SraJMcGin
Conservation of energy
Try some of the following sites to see if any offer help: http://search.yahoo.com/search?fr=mcafee&p=slap+shot+in+hockey+according+to+conservation++of+energy Sra
Monday, June 13, 2011 at 8:11pm by SraJMcGin
some people refer to heat as "low grade energy. a)how is low grade energy different from high grade energy? b)give two example of high grade energy When heating a cup of tea, which would be preferable...a stove with a burner on top, or a heating pad? High grade energy is ...
Tuesday, May 8, 2007 at 4:41am by Emma
The compressed potential energy of the spring is (1/2) k X^2. This equals thesum of gravitational poteneital and kinetic energy at any altitude after it is launched. (1/2) k (3)^2 = (1/2)M V^2 + M g H V = 10 m/s when H = 5 m Solve for k For the additional height it travels, ...
Monday, November 3, 2008 at 12:08am by drwls
Henry's equation d = at^2 is not correct, but there may be a compensating error somewhere else, such as in assuming constant acceleration. I get the same answer, however. I prefer to use conservation of energy. Stored spring potential energy = kinetic energy when released (1/2...
Sunday, December 25, 2011 at 1:47pm by drwls
when you apply the brakes of a car its kinetic energy or energy of motion decreases. a)to what form of energy is the kinetic energy converted? b)where does the resulting energy end up? Do the brake drums on the wheels heat up? so does it convert into heat energy?and the ...
Tuesday, May 8, 2007 at 4:32am by Emma
Use conservation of energy, and make sure you include both the translational energy (1/2) M V^2 (due to linear motion of the center of mass), and the rotational kinetic energy (1/2) I w^2 = (1/2)I(V/R)^2. The sphere will go down the ramp faster because less of the energy gets ...
Saturday, November 21, 2009 at 9:12am by drwls
Calculate the vertical rise and apply conservation of energy. Work done = poetential energy increase. We will gladly critique your work
Friday, February 26, 2010 at 9:43am by drwls
Again,this cannot be solved without some assumption on energy, and in a head on collision, conservation of mechanical energy is out of the question. If you have other thoughts, I would like to hear them.
Sunday, November 14, 2010 at 7:01pm by bobpursley
(a) PE=mgh=mgL KE = mv²/2 Law of conservation of energy PE =KE mgL= mv²/2 v=sqrt(2gL)=... (b) Law of conservation of linear momentum mv= (m+m1)u u= m•v/(m+m1)=... (c) Law of conservation of energy KE1=PE1 (m+m1) •u²/2 =(m+m1) •g•h1 h1=(m+m1) •u²/2 • (m+m1...
Tuesday, October 16, 2012 at 4:29pm by Elena
Are you supposed to assume the process is frictionless? I can't imagine a frictionless box on a ramp. Conservation of energy would be a good way to attack the problem. Subtract the potential energy increase from the work done by the person pushing. If there is no friction, the...
Thursday, December 9, 2010 at 1:53am by drwls
Why would anyone attach a rocket to a spring? I suggest using conservation of energy for this problem. Power delivered to Rocket = (Thrust)*(distance travelled, X) After you subtract the spring potential energy (1/2)kX^2 , the remainder is the kinetic energy of the rocket.
Thursday, November 25, 2010 at 12:36am by drwls
help with physics please!
I don't know what you mean by Us = 2.0 J This looks like a mass on a spring problem for which conservation of energy should be used. The sum of potential and kinetic energy equals a constant. At the equilibrium position, the kinetic energy is (1/2)M V^2 = (1/2)(2.5)(0.55)^ = 2...
Thursday, June 18, 2009 at 8:53am by drwls
Use momentum conservation to get the final velocity. Then calculate the new mechanical energy. See for yourself if mechanical energy is lost or gained.
Monday, November 23, 2009 at 7:36pm by drwls
Try applying conservation of energy. The potential energy of the compressed spring is (1/2) k X^2 where k is the spring constant and X = 0.05 m Show work for futher assistance.
Thursday, March 4, 2010 at 12:45am by drwls
Potential energy at the topmost point = mg(2r) Kinetic energy at dip bottommost point= 1/2 m (v^2) so conservation of energy means mg 2r = 1/2 m(v^2) or 4rg= v^2 or v = 2sqrt(rg)
Wednesday, October 21, 2009 at 8:53pm by Vipster
Physics - Rube Goldberg Project
I need to make a Rube Goldberg Machine and it must demonstrate projectile motion, conservation of energy and conservation momentum, any idea on what to make and how to make it? Help please
Tuesday, August 31, 2010 at 12:37pm by Shaila
The distance than the block rises with the ball inside will tell you the velocity right after impact (V2), using conservation of energy. V2 = sqrt (2gH); H = 0.145 m Once you have determined V2, get the original speed of the ball (V1) by using conservation of momentum. m V1...
Sunday, March 28, 2010 at 9:03pm by drwls
Conservation of momentum leads to .2kg*8.2m/s=.2kg*(-3.9)m/s + Mass*V Solve for V in therms of all the other. Then put that into this for V conservation of energy. 1/2 .2 *8.2^2=1/2 .2*(3.9)^2 + 1/2 MV^2 then solve for mass, after that, go back and solve for V
Thursday, November 12, 2009 at 3:37pm by bobpursley
Use conservation of energy. The loss of gravitational potential energy will equal the potential energy stored in the spring. 0.05 kg * 9.8 m/s^2 * 2 meters = (1/2) k X^2 X^2 = 0.014m^2 X = 0.12 m I have neglected the gravitational potential energy change during the small ...
Monday, March 16, 2009 at 11:48pm by drwls
Use a conservation of energy method. Initial KE + gravitational PE loss = spring potential energy. (1/2) (m + M) Vo^2 + (m + M) g H = (1/2) k X^2 Vo is the initial velocity after getting onto the bobsled. It is less than 14.7 m/s, because there will be some inelastic energy ...
Saturday, October 31, 2009 at 9:26pm by drwls
conservation of kinetic energy
Saturday, October 16, 2010 at 5:10pm by Anonymous
Potential energy at the top of the hill = mgh = 10*9.81*5.8 J =569 J Heat energy produced = 414 J If we assume conservation of energies, energy at the bottom of the hill in the form of kinetic energy =(1/2)mv² = 569+414 =983 Solve for v: v=sqrt(2*983/10) =14.0 m/s
Thursday, December 9, 2010 at 1:57pm by MathMate
Use conservation of linear momentum to compute the speed V of the bob of the pendulu immediatly after impact. Neglect mass loss. 0.01*370 = 0.01*100 + 2.0*V Solve for V. V = 0.005*(270) = 1.35 m/s Then, use conservation of mechanical (kinetic + potential) energy to compute how...
Friday, November 18, 2011 at 8:37am by drwls