Physics Help?!?! A very light object moving with a speed v collides?!!?
A very light object moving with speed v collides head-on with a very heavy object at rest, in a frictionless environment. The collision is almost perfectly elastic. The speed of the heavy object after the collision is A. slightly greater than v B. equal to v C. much less than v D. slightly less than v Thanks!
A 10.0 g object moving to the right at 23.0 cm/s makes an elastic head-on collision with a 15.0 g object.?
Let x and y be the velocities of the two objects afterward. Conserve momentum: (10)(23) + (15)(-28) = (10)x + (15)y Conserve energy: (10)(23^2) + (15)(28^2) = (10)x^2 + (15)y^2 Simplify to two equations: 2x + 3y = -38 2x^2 + 3y^2 = 3410 Substitute y = -(38 + 2x)/3 into second equation: 2x^2 + (1/3)(2x + 38)^2 = 3410 6x^2 + (4x^2 + 152x + 1444) = 10230 10x^2 + 152x - 8786 = 0 Then use quadratic formula: x = ( -152 +/- sqrt( 152^2 + 4(10)(8786) ) / 20 = ( -152 +- 612 ) / 20 = 23 or -38.2 whence y = -(38 + 2x)/3 = -28 or 12.8 Note that there are two solutions. One in which the same input numbers are returned corresponding to the "elastic collision" in which one ball passes seamlessly through the other, and the solution you really want: x = -38.2 and y = 12.8
Energy Loss of an Inelastic Collision?
Inelastic collisions do NOT preserve kinetic energy. In a perfectly inelastic collision, both objects stick and continue with the same final velocity, and the MOST KE is lost through heat. Call their final velocity "u". Momentum, of course, is conserved: pi = pf: mv = (m+M)u => u = mv / (m+M) Fraction of kinetic energy left: Kf/Ki = 1/2 (m+M) u^2 / (1/2 m v^2) = (m+M) (mv/(m+M))^2 / (m v^2) = m / (m+M) You wanted fractional loss: (Ki - Kf) / Ki = 1 - Kf/Ki = 1 - m / (m+M) = (m+M - m) / (m+M) = M / (m+M) As m gets small compared to M, this approaches 1. This makes sense, since if I ran into a brick wall, I would lose all my energy (and a lot of brain cells, too)
Perfectly elastic collision?
Cart A with a mass 9.9 kg is moving along with 30.5 m/s. It makes a head-on perfectly elastic collision with a stationary cart B with mass 4.6 kg. What is the velocity of cart A? I thought that since this is a perfectly elastic collision, all of the velocity would transfer from cart A to cart B, and cart A's velocity would then be 0. This is incorrect, according to my online quiz results. Why is that? Is there a formula I need to use to figure this out, instead?
Physics Questions Multiple Choice.?
1)F = m(v-u)/t = 0.25(29+29)/0.05 (since the two velocities are in different directions add the velocities) = 290N Answer - c 2)Here, Work done is the difference in mechanical energy.Since the mechanical energy does not change, Answer- c 3)By conservation of Mechanical energy, F = kx =(1/2)mv2 =50 * 0.01 Answer- a 4)Initial mechanical energy = mgh= 30*10*10= 3000J Final mechanical energy =(1/2)mv^2 = (30*12^2)/2 = 15 * 144 = 2160J Work Done = Final Energy- Initial Energy = 840J Answer is not included...Something wrong with the sum or my workings... 5)Answer - b (Change in energy is work done) 6)No idea 7)Answer = e(Since the masses are not given values cannot be added to the equation of conservation of linear momentum) 8)Answer - d
Is colliding with a brick wall travelling at 50mph equal to colliding head-on with another car also travelling at 50mph?
You have two basic answers so far, “the same, because in both cases your car goes from 50 mph to zero in a very short time,” and “hitting the other car is worse, because the combined speed is 100 mph, which is twice as bad as 50 mph”.Both answers are correct in different senses, although I think the former is better.How bad a collision is depends on the velocity of what you hit, its composition and its mass. Actually, “mass” is complicated here, because when your car hits either a car or a brick wall, lots of things are colliding, some of which absorb energy by deforming or breaking. But I take “brick wall” to mean “object of so much strength, rigidity and more mass than your car, that it does not move or deform significantly”. After all, a brick will is connected to the earth. If it’s a solid wall with a solid foundation, your car will not break or move it. If the brick wall is less solid or not well-anchored, then hitting it is not as bad for you (but it will be worse for the wall).So the brick wall brings your car from 50 mph to zero in the amount of time it takes for the metal of your car frame to deform enough to absorb all the energy or bounce.The natural assumption for your question is that the other car is identical to yours in both in both mass and composition. In that case it’s nearly identical to the solid brick wall. If the other car is less massive than yours, hitting the other car is better, because you might move from (say) 50 mph to 5 mph. If the other car is more massive than yours, hitting it is worse than the brick wall, because you’ll go from (say) 50 mph to 5 mph is the other direction, a total chance of velocity of 55 mph.Composition matters as well. If the other car is made of marshmallow, then you don’t mind hitting it. If it’s a sharpened vanadium steel spike, you’re in big trouble even if it’s not all that massive.
If 2 automobiles collide, they usually do not stick together. Does this mean the collision is elastic?
Sometimes they do. No. An elastic collision is an encounter between two bodies in which the total kinetic energy of the two bodies after the encounter is equal to their total kinetic energy before the encounter. Elastic collisions occur only if there is no net conversion of kinetic energy into other forms. Most of the energy of the auto collision results in heat as the various pieces of metal deforms under pressure. Very little of the energy appears as motion. Think about it, if a car rammed head on at 60 MPH into a brick wall and the collision were elastic, the car would wind up moving backwards at 60 MPH after the collision. You know that doesn't happen. And a collision with a brick wall at 60 is about the same as two autos colliding head on at 30 MPH each.