if the momentum of a body increses by 20%,then by how much percent does its Kinetic energy increses?
477496
For nonrelativistic motion,\nE=p^2/2m        (check it by expansion)\nso ∆E=2p∆p/2m  =p∆p/m\n ∆E/E = (p∆p/m)(2m/p^2)\n      =2(∆p/p)\nBut ∆p/p = 0.2      (given)\ntherefore,  ∆E/E =0.4  or 40%
771568
Basically you have the right idea. Use momentum (mass x velocity) for the motion and the sudden change in speed and/or direction (acceleration) and you get force x time = kinetic energy trying to rip you apart or squash you like a bug.
14955
(a) it's an inelastic collision.  An inelastic collision is when they stick together, and an elastic collision is when they bounce off of each other\n(b) p=mv  momentum= mass x velocity.  p= 30 m/s x 500kg = 15,000 kg*m/s\n(c) well, momentum is conserved, so the we would use the formula from before and the momentum(p) from before, except that now the mass is different because it is the combined masses of the two vehicles (300+500=800) , so you can solve for the new velocity.\np=mv  15000 = 800 x v\n\ndivide both sides by 800 to get  v=15000/800 = 18.57 m/s
985678
Incoming momentum = p1 = mv1\nOutgoing momentum = p2  = mv2\nChange in momentum = p2 - p1 = m(v2-v1)\nThis change occurs in time, t.\nRate of change of momentum = m(v2-v1)/t = impressed force, f.\nm=0.149 kg, v1 = 43.4 m/s, v2 = -48.7 m/s, t = 1.1x10^-3 sec\n\nf = m(v2-v1)/t\nYou do the arithmetic.
714752
This problem is a conservation of momentum problem:\n\n(a) In an elastic collision, the formulas for calculating the final velocities are derived from both the conservation of momentum formula and the conservation of kinetic energy formula. Check out my source for more information, but for now just trust me that they are valid:\n\nv1 = (v1(m1 - m2) + 2m2v2)/(m1 + m2)\nv1 = (1*(4 - 7.2) + 0)/(4 + 7.2) = -0.3 m/s\n\nv2 = (v2(m2 - m1) + 2m1v1)/(m1 + m2)\nv2 = (0 + 2*4*1)/(4 + 7.2) = 0.7 m/s\n\n(b) In an inelastic collision, just use the conservation of momentum formula:\n\nm1v1' + m2v2' = (m1+ m2)v\n\n4*1 = (4+7.2)*v\nv = 4/(4+7.2) = 0.36 m/s\n\nI hope this helps!
925506
You need to consider conservation of momentum and conservation of energy.\n\nConservation of Momentum says that the momentum before tha collision is equal to the momentum after the collision.\n\nm*v1 = (m + M)*v2\n\nwhere m is the mass of the bullet, M is the mass of the block, v1 is the speed of the bullet, and v2 is the speed of the bullet/block system. Therefore,\n\nv2 = m*v1/(m + M)\n\nTo find out how high the bullet/block system rises, consider conservation of energy. The kinetic energy of the bullet/block is converted to potential energy as the system rises so that\n\n0.5*(m + M)*v2^2 = (m + M)*g*h\n\nwhere g is the gravitation acceleration and h is the height to which the bullet/block rises. Solving for h\n\nh = v2^2/(2*g).
256338
a) when it it closest to the gravitational source\nb) when the potential energy is minimum (same as a)\nc) speed = |velocity|\nd) momentum(p)= mv\ne) KE = (mv^2)/2\nf) usualy the potential energy is defined as 0 at infitity\ng) conservation of energy\nh) when it it closest to the gravitational source\ni) conservation of angular momentum\n\nThat should be enouph to figure out the answers... but think about it and draw some pictures so that you know it is right.
860846
Force = 60 N.\nMass = 3 kg\nTime = 10 s\n\nAn object's change in momentum, Δp equals the Impulse, I on the object. I = ∫F dt = Ft. \n\nΔp = I = Ft = (60 N) * 10 s = 600 Ns\n\nAssuming the object started at rest,\np (initial) = mv = (3 kg) (0 m/s) = 0 Ns\nΔp = 600 Ns\np (final) = 600 Ns\n\np (final) = mv = (3 kg) * v(final)\nv(final) = (600 Ns) / (3 kg) = 200 m/s\nΔv = v(final) - v (initial) = 200 m/s - 0 m/s = 200 m/s\n\nThe object's change in momentum was 600 Ns (kg * m/s).\nThe object's change in velocity, assuming it started from rest, was 200 m/s.
906577
Hi!\n(A) \nBy definition di=dm * X^2 , where di is the amount added to I when the mass dm is added to the system. I~Inertia . In this case for the particle X=d.\n\nEach mass m=0.85 kg add to the system (delta I)1 = m * d^2.\n\nEach rod has an Inertia of (delta I)2 = (1/3) M * d^2 . This is because for a rod you have to calculate:\n             S { dm * x^2 = (M/d)*dx * x^2 } ==> (M/d) * (1/3)(X^3) with X=d. => (1/3)M * d^2  \n\nso the total inertia of the system is I= 2*[ m*d^2+ (1/3) M*d^2] because we have 2 rods and tho particles.:\n    I= 7.84 * 10^(-3) kg / m^2.    Where ^ is used for power!\n\n(B) kinetic energy= (1/2) Inertia  * (angularspeed)^2 =3.528 /10^4\n\nIf you want to know more try the link below!
1010100
ok, so if we're assuming that it was a head on elastic collision, and all the energy was transferred then there are two possible situations: if the 25lb ball was sliding then the total energy transferred will be equal to the amount it had originally (since we assumed elastic collision) that is equal to 1/2mv^2...where mass would usually be measured using the metric system. \nif the ball was rolling then it has not only translational energy (1/2mv^2) but also rotational energy which is equal to 1/2Iw^2 where I = inertia of a ball...if it is a solid sphere the moment of inertia is equal to 2/5mr^2 (m = mass, r = radius) and w (omega) = rotational speed in radians per second...which is equal to velocity/radius. so the amount of energy transferred would be equal to 1/2Iw^2 + 1/2mv^2.
1110309
The Rate of change of momentum of a body is equal to the resultant force acting on the body, and takes place in the direction of the force.\n\nMomentum = mass * velocity\n\nThe masses are the same on Earth and the moon.\n\nMomentum would be equal!
666681
1/2 * mass * velocity^2.
1099052
21 %\n\nbcoz range is proportional to the square of velocity \n\n*********************************************************************\n\nR prpotional to square (u)\n\nlet u0 initial velocity ,r0 inital range\n\nu1 velocity with 10 % increase ,r1 new range\n\nlet u0=10  >>> then r0=100 (proportional to 100)\n\nwith 10% increase in velocity\n\nu1=11 >>>> r1 =121\n\nthus %age increase =21 %\n\n\n*********************************************************************
708999
Find the inertia.  Find the change in inertia.  Use conservation of momentum to  find the new speed. etc,.
762251
If 2 billiard balls of mass m (ma and mb) are perfectly elastic and ball A approaches from the left with velocity 2 m/s and hits ball B which has velocity 1 m/s, and is also going to the right, what's the status after the collision. \n\nA is going to the right with velocity 1 m/s and B is going to the right with velocity 2 m/s.\n\nma*2 m/s + mb*1 m/s = ma*1m/s + mb*2m/s\nmomentum is conserved
792153
‘impulse’ or momentum\n=m*(v2-v1)=0.42*(-17-4.1)=-8.862 kg*m/s
136454
I will give you a direction to solve the problem...\n\nFirst I will do the problem from a rectilinear formulation(using force, acceleration, velocity) then transfer it to a curvilinear formulation(using Torque, angular accelaration, angular velocity).\n\ngiven F=ma (or F=m(v/t) or more specifically F=m(dv/dt) or even more specifically F=d(mv)/dt).\nWhere F=force, m=mass, v=velocity, and t=time.\n\ngives you Ft=mv (conservation of energy=>conservation of momentum or the impulse-momentum equation).\n\nThis states that it will take a certain amount of force for a certian period of time to produce a certain change in velocity of an object assuming that the mass remains constant.\n\nTransmitting to curvilinear motion force becomes torque, v becomes rv(radius time velocity or w), mass becomes moment of inertia (or I) and t stays the same.