What are the differing consequences of two different objects having the same momentum but differing kinetic energies if they were to collide into a wall, all else held equal?

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If they both have the same momentum, would they move the wall ( and the planet earth) by the same amount, but the one with higher kinetic energy has the higher potential to do more deformation to the wall, depending on comparative material properties?

In: Physics

3 Answers

Anonymous 0 Comments

In a collision, the energy has to go somewhere and the momentum has to go somewhere. The momentum has to end up as the momentum of something, and the energy has to end up as some form of energy.

In an elastic collision, kinetic energy is also conserved. This would be like if two impossibly perfect billiard balls or two elementary particles collided and bounced off each other.

In an inelastic collision, some or all of the kinetic energy is converted into other forms of energy, which typically involves breaking things. Cars are designed to be as inelastic as possible in a collision, channeling a lot of the kinetic energy into sound and heat and deformation of the crumple zones in the car body. If it’s done well, the cars will stick together after the collision, which is the most kinetic energy you can lose while conserving momentum.

A wall, in physics problems, is an object of infinite mass that doesn’t really move. So the object hitting it will deform, and possibly damage the wall, to the extent that all the kinetic energy is used up. More energy means more damage.

Anonymous 0 Comments

That’s the general gist.

One example is shooting a rifle.

The gun and bullet have equal but opposite momentum, but the bullet has a few orders of magnitude more energy. The rifle gives you a bit of a kick, the bullet can easily cause catastrophically lethal damage to someone at the other end.

Anonymous 0 Comments

Momentum (p) is equal to mass times velocity (m×v).

Or p=m×v

Kinetic Energy (KE) is equal to one half the mass times the square of the velocity (1/2×m×v^2)

Or KE=1/2×m×v^2

If two objects have the same momentum but differnet kinetic energies, then we can write it as m1×v1=m2×v2 and m1×v1^2 < m2×v2^2 (since both sides are multiplied by 1/2 we can ignore that.) By solving that m1=(m2×v2)/v1 and that v1=(m2×v2)/m1 and then substituting those respectively then we can see that v1<v2 and m2<m1.

So, the object with greater kinetic energy will have less mass and more speed. Now, since both are hitting identical walls, their mass will increase by the same amount, and the speed will decrease by the same amount. But, the object with greater kinetic energy will still have greater kinetic energy. And therefore the wall will be deformed more.