Momentum = (mass) times (velocity) = 0.25 x 40 = 10 kg-m/sec
The idea is to use conservation of momentum. Calculate the total momentum before the collission, add it up, then calculate the combined velocity after the collision, based on the momentum.
Answer:1. ball falls from a height of 1.41m, so what is its momentum when it hits the floor? Well, assuming the ball falls on earth:mom = m.vacceleration = 9.8m/s/swhere m=mass of ball; V=velocity of ball when it hits the groundHow do we know what velocity it travels at when it hits the ground? We use uniform acceleration formulas (because the force due to grav is constant):v=u+a.t; S=u.t+(1/2).a.t^2; V^2=u^2 + 2.a.swhere s=displacment; u=inital velocity; v=final velocty; a=acceleration; t=timeSo which one applies? what do we know about the ball, and what do we want to find out?we know balls mass, inital velocity, which we'll say is 0. We know the acceleration is 9.8m/s/s.We want to know it final velocity, so we'll use V^2 = u^2 + 2.a.su=0soV^2 = 2.a.sthereforeV^2 = 2.(9.8m/s/s).(1.41m) = 27.636soV= root(27.636) = 5.26m/s (3.s.f)we know v, now we can work out momentum:v.m = (5.26m/s).(41.8/1000)kg (notice I have to put it in kg, not grams)0.2197kgm/sNow, this is the momentum the ball has at the moment it touches the ground, and because we know that the force exerted is equal to the rate of change of momentum and we know this change takes 0.012seconds to occur, and that the ball moves with the same velocty in the opposite direction:total change in mom = 2.(0.2197kgm/s) = 0.4394kgm/s, which takes 0.012sectherefore force = (0.4394kgm/s) / (0.012sec)36.62 newtons!Hopfully someone will jump in if I made some error somewhere.
cgs means centimeter,gram,second. cgs unit of velocity is cm/s.
A 60 gram bullet fired from a gun with 3150 joules of kinetic energy has a velocity of 324.04 meters per second or 1,063.12 feet per second. (This is about 725mph).
Momentum is mass times velocity. A bullet could theoretically have the same momentum as a moving truck if the bullet's speed is great enough. But practically, no--a bullet going that fast in the atmosphere would break up or burn up instantly. In outer space, it would be possible, but it would be hard to get the bullet up to that speed. Bullets already travel very fast (a fast bullet can go 4,000 feet per second, which is 2,700 miles per hour), but they are very light (a 250 grain bullet = 0.036 pounds). If a truck weighs 10 tons and is going 55 miles per hour, for instance, that 250 grain bullet would have to travel 30 million miles per hour to have the same momentum. Of course, the trivial answer is yes--both can have zero momentum if neither is moving!
The idea is to use conservation of momentum. Calculate the total momentum before the collission, add it up, then calculate the combined velocity after the collision, based on the momentum.
use the Impulse-Momentum theorem where (Fnewtons)(tseconds)=change in momentum change in momentum=Pf-Pi; P=mv
Their difference is that a 15 gram ball is lighter than a 60 gram ball.
You because the product of your mass and velocity will be higher than that of a bullet
Answer:1. ball falls from a height of 1.41m, so what is its momentum when it hits the floor? Well, assuming the ball falls on earth:mom = m.vacceleration = 9.8m/s/swhere m=mass of ball; V=velocity of ball when it hits the groundHow do we know what velocity it travels at when it hits the ground? We use uniform acceleration formulas (because the force due to grav is constant):v=u+a.t; S=u.t+(1/2).a.t^2; V^2=u^2 + 2.a.swhere s=displacment; u=inital velocity; v=final velocty; a=acceleration; t=timeSo which one applies? what do we know about the ball, and what do we want to find out?we know balls mass, inital velocity, which we'll say is 0. We know the acceleration is 9.8m/s/s.We want to know it final velocity, so we'll use V^2 = u^2 + 2.a.su=0soV^2 = 2.a.sthereforeV^2 = 2.(9.8m/s/s).(1.41m) = 27.636soV= root(27.636) = 5.26m/s (3.s.f)we know v, now we can work out momentum:v.m = (5.26m/s).(41.8/1000)kg (notice I have to put it in kg, not grams)0.2197kgm/sNow, this is the momentum the ball has at the moment it touches the ground, and because we know that the force exerted is equal to the rate of change of momentum and we know this change takes 0.012seconds to occur, and that the ball moves with the same velocty in the opposite direction:total change in mom = 2.(0.2197kgm/s) = 0.4394kgm/s, which takes 0.012sectherefore force = (0.4394kgm/s) / (0.012sec)36.62 newtons!Hopfully someone will jump in if I made some error somewhere.
None. Assuming they are falling with the same conditions, they accelerate equally. But the 200 gram object has the greatest terminal velocity therefore reaching a higher velocity before resting at a constant speed.
Before the shot, total momentum of the rifle/bullet system is zero. Momentum is conserved, so must total zero after the shot. Magnitude of momentum = m V (mass, speed); we'll take care of direction independently. Momentum of the rifle: m V = (3.8) x (2.4) = 9.12 kg-m/sec backwards. We need momentum of the bullet = 9.12 kg-m/sec forward m V = 9.12 ===> V = ( 9.12 / m ) = ( 9.12 / 0.013 ) = 701.54 m/s forward
cgs means centimeter,gram,second. cgs unit of velocity is cm/s.
Sure. That's a perfectly good unit of momentum. So is (any unit of mass) divided by (any unit of speed).
The mass and velocity of an object do not determine its wavelength: it could be travelling in a straight line!
A 60 gram bullet fired from a gun with 3150 joules of kinetic energy has a velocity of 324.04 meters per second or 1,063.12 feet per second. (This is about 725mph).
450-480 fps but I mould use at least .25 or more.