Calculate the velocity of the bullet before striking the block?
916585
Conservation of momentum so mv is constant.\n\nTotal final mass = 5333+14.6 = 5347.6g\n\nTotal final momentum = 5347 x 1.7 gm/s = 9090.92 gm/s\n\nBullet's initial momentum = 9090 gm/s\n\nMomentum/mass = velocity\n\n9090.92/14.6 = 622.666m/s
393875
The way I see it.... u calculate its impulse with the ground.... mass1*velocity1 (pre-shock)=m2*v2 (after shock) then caculate it's potential energy by newton's law of gravity using v2....(god I'm being a GEEK!)
157080
This is just the regular projectile motion problem (see your text) just with an added initial velocity in the x direction (the velocity of the ship).\n\nLet's take as our positive direction the direction the gun shoots. (That is, the ship is moving in the negative direction.) Thus, the initial x component of velocity of the projectile is the x-component of the guns velocity less the velocity of the ship. (I'm going to use Vo for the initial velocity, since using "not" as a subscript confuses me.)  And I'll use A for alpha to save space.\n\nVxo = Vo cos(A)- v\nVyo = Vo sin(A)\n\nOur equations of motion are\n\nX = Xo + Vox t + 1/2 ax t^2\nY = Yo + Voy t + 1/2 ay t^2\n\nTo find the range, we want when the projectile splashes back down again, i.e. Y = 0.  If we set Xo = 0, then X will be the range, i.e. the X position when Y = 0 again.  And there is no acceleration in the x direction, the y acceleration is -g, the acceleration of gravity.  Plugging all that in, we get\n\nR = Vxo t \nR = (Vo cos(A) - V) t                     (1)\n\n0 = Vyo t - g t^2/2\nVyo t = g t^2/2\nVyo = g t/2\n2 Vyo/g = t\n2 ( Vo sin(A)) / g = t                     (2)\n\nPlugging our expression for t (Eqn. 2) into Eqn. 1 yields\n\nR = (Vo cos(A) - V)2 ( Vo sin(A)) / g \n\nwhich is the formula you gave.\n\nTo find the max range, we can use calculus.  There may be a more straightforward way if you don't know calculus, but this works for me.\n\nThe range formula is a function R of variable A.  To find the max, take the derivative and set it equal to 0.  Thus\n\ndR/dA = 2 Vo/g [sin(A)(-Vo sin(A)) + (Vo cos(A) - V)cos(A)]\n       =2 Vo/g [-Vo sin^2(A) + Vo cos^2(A) - V cos(A)]\n\nusing a trig identity\n\n       =2 Vo/g[Vo cos^2(A) - Vo + Vo cos^2(A) - V cos(A)]\n       =2 Vo/g[2Vo cos^2(A) - V cos(A) - Vo]\n\nWhat we have here is a quadratic equation with the variable cos(A).  When we set it equal to 0, in order to find the minimum, we can use the quadratic formula.\n\ncos(A) = (- (-V) + SQRT[V^2 - 4(2Vo)(-Vo)])/2 (2Vo)\n        = (V + SQRT[V^2 + 8Vo^2])/4Vo
927051
The given final speed doesn't seem to compute. I get 14.8 m/s. \nMa*Va + Mb*Vb = (Ma+Mb)*Vf \nis the formula that applies to show conservation of momentum in a completely inelastic collision, which is what you have. The only explanation I could suspect is some cargo was spilled. \n\nTo continue the assignment - kinetic energy is given by\n(1/2)*m*v^2\nThere is no requirement that all the original kinetic energy be measurable as kinetic energy after a completely inelastic collision. But you should be able to point to where the missing energy went. Some of the energy probably went into cart damage. Perhaps the cargo slid around.
743381
Velocity(at time t) = Velocity (at time = 0) + Acceleration x Time\n\nV(t) = Vi + At\nV(t) = 32 ft/sec + (-32 ft/sec/sec)(t)\nV(1 sec) = 32 - 32 = 0 ft/sec\n\ny(t) = y(initial) + (V x t) + 1/2(A)(t^2)\ny(1 sec) = 0 + (32) +  (-16) = 16 feet\n\nTo hit ground, y = 0, so (V x t) = 1/2(A)(t^2)\n32t = 16t^2\n2 = t, t = 2 seconds\n\ny(max) occurs at zero velocity, which is at t = 1 sec, x = 16 feet
799487
Alright... physics is fun if you get the right problems :) .  \n\nThe collision is elastic... your book should have a conservation of momentum equation.  It states that p1+p2 = Ptotal\n\nYou have to use component vectors for the velocity after the collision.  A good thing to do would be to draw the problem.  It would make it a lot easier.  So use conservation of momentum.\n\nAnd there are two equations that would really help you:\n\nFinal velocity of block A= ((MB-MA)/(MA+MB))*VAinitial\n\nFinal velocity of block B= ((2MA)/(MA+MB))*VAinitial\n\nHelp with physics:\n\nALWAYS draw pictures epecially on your exams.  Get on the good side of the professor, go to a tutoring center if you have one, do many problems and memorize how to do the basic ones if you want an A on the test.
903139
A bullet of course.  A gun uses a small amount of kinetic energy for the trigger hammer to strike the expolsive cap in the shell but thereafter the explosive force is sent into the bullet, with and equal opposite reaction energy dissipated by providing recoil which the gun uses (if automatic) to load the next shell.\n\nI assume the trick in the question is to ignore the equal and opposite reaction rule but not all of the force thus generated is in the form of kinetic energy which relates to bodies in motion.
978760
Projectile Motion-  Examine the initial velocity and angle necessary to hit targets at different points.
757769
Here's how to solve this.\nThe velocity is a vector, and right-angle components of velocity operate independently.\nThe bullet is fired with 2000 ft/s horizontal velocity but zero vertical velocity. So you know the horizontal and vertical components of velocity at the instant the bullet leaves the barrel.\n\nHowever, once it leaves the gun's barrel, the bullet experiences no horizontal forces, but vertically it is acted upon by gravity.\n\nSo: 1) how long does it take for the bullet to fall 5 ft, starting with zero vertical velocity? (hint: s = 1/2 a t^2\n\n2) Once you figure out how long until the bullet hits the ground, then figure out how far the bullet has travelled horizontally during that time, given 2000 m/s initial horizontal velocity and no horizontal forces.\n\nDoes this help?
981397
This might be one of those questions that you have to answer with just using the formula, i.e. you will not get an actual answer, but an equation that if you knew the mass, length and force you could calculate the height dependant upon the force input, but thats pretty high level stuff, you would need so solve for angular acceleration and momentum, also in this sort of question you would be given either an impact force on a part of the rod (i.e. where the rod was hit) of you would be told the rod has some momentum. Then you could solve.\n\nI suggest you have another look at the question, I think you will find it says the rod is already moving.
709316
assuming second block is originally at rest\n\nconservation of momentum says\n\nv11m1=v2m2+v12m1  (v11 initial block1 velocity, v12 final block 1 velocity)\n\nnow 1/2 mv11^2 = 1/2m2v2^2+1/2m1v12^2\nor m1v11^2 = m2v2^2+m1v12^2\n\n2 equations and 2 unknowns
827544
1) use conservation of energy:\n\n1/2*m*vb^2=1/2*m*va^2+m*g*h\n\nm=2\nva=4\ng=9.8\nh=6\nv=sqrt(16+2*9.8*6)\n=11.56 m/s\n\n2.\nSame concept\nm*g*(bounce)=1/2*m*va^2+m*g*(drop)\nmass divides out\nbounce=va^2/(2*g)+drop\nva=6\ng=9.8\ndrop=4\nbounce=36/(2*9.8) + 4\n=5.84 m\n\n3)  since the mass isn't moving it's offset the 300N force\n\n4)\nAgain, use conservation of energy\nNote that the change in potential energy will subtract from the kinetic energy of the start \n\n1/2*m*vc^2=1/2*m*va^2-m*g*delta-h\nthe mass divides out\nva=8\ndelta-h=2\ng=9.8\nvc=sqrt(64-2*9.8*2)\n=4.98\n\nj
808209
Using one of the equations of motion\n\n(v*v) - (u*u) = 2*a*s\n\nwhere \nv is final velocity\nu is initial velocity\na is accleration\ns is displacement\n\n\nOR \nuse \n\nv * (dv / ds) = a
170531
1) V2^2=V1^2+2d*g ----> 0=47^2+2d*-9.8 ----> d=112.7 m\n2)remeber when mesuring the time we take the velocity on the Y axes and when mesuring range or distance we take velocity on the X axes got it?\nd=V1+0.5*g*t^2 and since the rock is thrawn horizontaly the velocity on the Y axes once it's thrawn is zero, by the way the decplacement would be minus cuz the rock is projecting downwards \n-3.5 m = 0 + 0.5 * -9.8 * t^2 -----> t=0.845 seconds\nd=V*t=47*0.845= 40 m\n3)this is a tough one you'd better draw it :\na.V2^2=V1^2+2d*g ----> 0= (47sin37)^2 + 2d*-9.8 ----> d=40.8m\nb.V2=V1+g*t ---> 0=47sin37+-9.8*t ---> t=2.886 seconds\nrange=V*cos37*t=47cos37=108 m\nc.V2=V1+g*t=47sin37+(-9.8*2)=8... m/sec (velocity on the Y axes)\nvelocity on the X axes is constant and equals=47cos37=37.5m/sec\nthe net velocity^2=37.5^2+8.685^2 ---->velocity of rock after 2 seconds of launch=38.5m/sec
955270
I won't do it for you, but here is what you need to do:\n\nFind the height at which the explosion occurs = h.\n\nFind the horizontal component of momentum at the instant before the explosion (which is actually the entire momentum, since there is no velocity in the vertical direction at that instant!) = Mtot x Vhoriz =m1 (Mtot is the entire shell).......(1)\n\nNow, the total momentum the instant after the explosion is:\n\n(M1 x 0) + (M2 x V2) .......(2)(since the large part of the shell stops travelling horizontally, and is already not travelling vertically) Where V2 is the horizontal velocity of the small part of the shell.\n\nSince momentum is conserved, these two quantities are equal, so:\n\nMtot x Vhoriz = M2 x V2, so you can calculate V2, the new horizontal velocity of the small part of the shell.\n\nNow, simply calculate the time taken for the shell to travel vertically to the ground using s = 1/2at^2, and from V2, you can calculate the horizontal distance travelleed by the small part of the shell...presto!
703031
Well the first thing to do in a problem like this is extract information from it.\n1.) Weight of Block = 10kg\n2.) Velocity of block(Vi) = 6m/s\n3.) Final velocity of block(Vf) = 0m/s\n4.) Friction acting on block = 20N The friction is always opposite of the direction of the block.\n\nSecond:\nWhat are we calculating? Distance traveled by block! DISTANCE.  We need a formula that will give us distance as an answer! So...we have final velocity, initial velocity, and distance....lets see...how about this one...\n vf^(2) = vi^(2) + 2*a*d \n\nThis equation has Vf, Vi, and D.  \nVf = 0 m/s\nVi = 6 m/s\nD = What we are trying to find.\nThe only thing missing in this equation that we have to find is (a) or acceleration. Hmm.\n\nThere is actually acceleration on  the block.  Friction is acting on the block in the opposite direction; therefore the block is decelerating.  We can use F=ma to find the force of deceleration.\nFriction Force = 20N\nMass = 10kg\nAcceleration = ?\n\nUse the following equation:\nF=ma\n1.) -20N = 10kg * a\n2.) a = -2 m/s (Yes it's negative b/c we are decelerating!)\n\nNow we go back to the original problem!\n vf^(2) = vi^(2) + 2*a*d \n\nThus\n(0^2) = (6^2) + (2)*(-2)*(d)\n0 = 36 - 4d\n-36 = -4d.\nDistance = 9meters! (Make sure to get the units correct!)\nI think I got this right...if not tell me what I did wrong pls\n\n-----------------------------------------------------------------------------------\nThe guy who did it using Energy equations forgot to add one thing....\nmass x (velocity)^2 = Force x distance\nso 10kg * (6m/s)^2 = 20N * x\nso (10 * 36)/20 = 18 meters\n\nThis is another way to do it.  He is using the energy equations to calculate energy in the block. When the block is in motion there is Kinetic Energy in it.  When the block has no motion, there is zero Kinetic Energy in it.  However, the individual that did this messed up the equation as follows:\nKe = (1/2) * Mass * Velocity^2\n\nThus the answer is not 18 but 9 meters!!!!!\nHere is the correction:\n1.) .5*mass x (velocity)^2 = Force x distance\n2.) .5*10kg*(6m/s)^2 = 20N * x distance\n3.) (5 * 36)/20 = distance&#92;\n4.) Distance =9 meters!
735311
the potential energy of the ball:\nPE = mgh = (.6 kg)(g)(.7m)\nwhen the ball collides with the block all the Potential Energy has been converted to Kinetic Energy so,\nPE = KE = (.6 kg)(g)(.7m) = (1/2)(m)(v^2)        (v^2 is v squared)\nall that kinetic energy is transfered to the block so\nKE(block) = 4.12 Newtons\n\nin one dimension the solution is that all momentum and kinetic energy is transfered to the block such that,\n(a) ball speed after collision = 0 m/s\nand \n(b) block speed after collision\n(1/2)(m)(v^2) = 4.12 Newtons\n(1/2)(2.5 kg)(v^2) = 4.12 Newtons\nv = 1.8 m/s