- #1

- 162

- 2

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter whoareyou
- Start date

In summary, the conversation discusses the concepts of momentum and kinetic energy in collisions. It is explained that momentum depends on both speed and mass, while kinetic energy is a scalar quantity that can change depending on the type of collision. In an inelastic collision, momentum is conserved but kinetic energy is not, while in an elastic collision, both momentum and kinetic energy are conserved. The conversation also mentions that inelastic collisions can result in the generation of heat and sound, which can further decrease the total energy of the system.

- #1

- 162

- 2

Physics news on Phys.org

- #2

- 1,007

- 15

One can give better help if more details of the graph or problem are given.

- #3

- 162

- 2

But if no mass is lost, then it would stay constant right? Then if the speed decreases, so does momentum ... ? :?

- #4

- 133

- 2

The Kinetic energy is different. It is a scalar quantity and always positive. It can change depending on whether you have an elastic collision (energy is the same before and after) or if you have an inelastic collision (energy is different before and after).

- #5

- 162

- 2

- #6

- 1,007

- 15

- #7

- 162

- 2

^ But on the graph, the total kinetic energy of the system drops after the collision.

- #8

- 133

- 2

If the two things your are colliding stick together before and after the collision, then you would have [itex]KE_i=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2[/itex] while after [itex]KE_f=\frac{1}{2}(m_1+m_2)v_f^2[/itex] while the momentum would be [itex]p=p_i=m_1v_1+m_2v_2=p_f=(m_1+m_2)v_f[/itex] so you would get [itex]v_f=\frac{p}{m_1+m_2}=\frac{m_1v_1+m_2v_2}{m_1+m_2}[/itex] and inserting this into [itex]KE_f[/itex] we get [itex]KE_f=\frac{1}{2}(m_1+m_2)\frac{(m_1v_1+m_2v_2)^2}{(m_1+m_2)^2}=\frac{1}{2}\frac{(m_1v_1+m_2v_2)^2}{m_1+m_2}[/itex] which is clearly not equal to [itex]KE_i[/itex] in general.

If I take for example [itex]m_1=m_2=1 kg[/itex] and [itex]v_1=1 m/s, v_2=0 m/s[/itex] then we get [itex]p= 1 kg\cdot m/sec[/itex], [itex]KE_i=1/2 J[/itex] and [itex]v_f=1/2 m/sec[/itex] so that we finally get [itex]KE_f=\frac{1}{2}(2)\left(\frac{1}{2}\right)^2=1/4 J[/itex]. So [itex]KE_f<KE_i[/itex]

If I take for example [itex]m_1=m_2=1 kg[/itex] and [itex]v_1=1 m/s, v_2=0 m/s[/itex] then we get [itex]p= 1 kg\cdot m/sec[/itex], [itex]KE_i=1/2 J[/itex] and [itex]v_f=1/2 m/sec[/itex] so that we finally get [itex]KE_f=\frac{1}{2}(2)\left(\frac{1}{2}\right)^2=1/4 J[/itex]. So [itex]KE_f<KE_i[/itex]

Last edited:

- #9

- 162

- 2

- #10

- 1,007

- 15

A big mass M with initial speed u hits a stationary smaller mass m so that after the impact the two masses move together with speed v in the same direction as u.

Then by conservation of momentum: Mu = (M+m)v.

[itex]\frac{KE_{f}}{KE_{i}}[/itex] = [itex]\frac{\frac{1}{2}(M+m)v^{2}}{\frac{1}{2}Mu^{2}}[/itex]

But v = Mu/(M+m)

hence

[itex]\frac{KE_{f}}{KE_{i}}[/itex] = [itex]\frac{\frac{1}{2}(M+m)((Mu)/(M+m))^{2}}{\frac{1}{2}Mu^{2}}[/itex] = M/(M+m) which is less than 1.

That is althought the total KE decreased yet the momentum reamined constant.

- #11

- 162

- 2

- #12

- 162

- 2

I'm beginning to understand completely inelastic with the math ... thanks guys.

- #13

- 133

- 2

whoareyou said:

Then that would be an elastic collision, elastic collisions conserve both momentum and energy (they don't stick together), while inelastic collisions conserve momentum, but not energy. Doesn't this answer your problem? a.) the first Blue line is momentum, since its constant. b.) its an inelastic collision because the KE is different before than after.

- #14

- 162

- 2

is not equal to the total kinetic energy of the system before the collision. This is an

inelastic collision."

"When two objects stick together during a collision, as is the case with the putty

balls, we have a completely inelastic collision. The decrease in total kinetic energy

in a completely inelastic collision is the maximum possible."

So those mathematical processes you implemented above work for both kinds of inelastic collisions?

- #15

- 133

- 2

- #16

- 15,462

- 689

- #17

- 162

- 2

- #18

- 15,462

- 689

Anticipating your question, so why isn't kinetic energy conserved? The answer is that the law of conservation of energy says that energy is conserved. There is no law of conservation of kinetic energy. There are lots of other forms of energy besides the kinetic energy of the colliding objects. The crack of the bat when a baseball player hits the ball: That sound is energy. The collision between ball and bat is not an elastic collision. The crunching and bending that occurs during an automobile accident indicate that that collision also is not elastic. The sounds, the bending and breaking, and the heat generated by the collision are all examples of kinetic energy being converted to some other form of energy.

Momentum, on the other hand, is momentum.

- #19

- 1,007

- 15

Let mass M with velocity 2u hit mass m which was moving with velocity u. Now let us choose the ratio of the masses M and m so that after impact mass M moves with velocity 1.5u and the other mass m moves with velocity v.

By conservation of momentum one can find v interms of M, m and u.

Then I suggest that the OP will try to find the ratio final KE/initial KE.

- #20

Staff Emeritus

Science Advisor

Homework Helper

Education Advisor

- 15,988

- 2,629

Before the collision, the total momentum was 0 = mV + m(-V). After the collision, the momentum is also 0 = mv + m(-v). Momentum is conserved, but kinetic energy isn't.

- #21

- 133

- 2

vela said:Before the collision, the total momentum was 0 = mV + m(-V). After the collision, the momentum is also 0 = mv + m(-v). Momentum is conserved, but kinetic energy isn't.

Maybe better to use that after the collision the momentum is 0=mv'+m(-v'), since if the velocity v was the same before and after, how could KE not be conserved?

- #22

Staff Emeritus

Science Advisor

Homework Helper

Education Advisor

- 15,988

- 2,629

V and v aren't the same variable.

- #23

- 133

- 2

vela said:V and v aren't the same variable.

Oh, I am completely sorry. I did not notice you capitalized v in the first part. I thought you just rewrote the same thing

- #24

Staff Emeritus

Science Advisor

Homework Helper

Education Advisor

- 15,988

- 2,629

- #25

- 162

- 2

- #26

- 162

- 2

vela said:

Before the collision, the total momentum was 0 = mV + m(-V). After the collision, the momentum is also 0 = mv + m(-v). Momentum is conserved, but kinetic energy isn't.

Is this relationship valid even if the speed of m1 is v1 and the speed of m2 is v2 (ie. the speeds are different)?

- #27

Staff Emeritus

Science Advisor

Homework Helper

Education Advisor

- 15,988

- 2,629

You can always switch to the center-of-mass frame where the total momentum is 0. In this frame, if the two masses are equal, the initial speeds will end up being equal. For example, suppose you have two objects with the same mass. Mass 1 is moving to the right at 2 m/s toward mass 2 which is at rest. The center of mass is therefore moving to the right at 1 m/s. In the center-of-mass frame, mass 1 is moving to the right at 1 m/s and mass 2 is moving to the left at 1 m/s. This is the same situation I described in my earlier post.

If the masses are different, the initial speeds will be different but the momenta will be equal and opposite. Say object 1 has mass m, and object 2 has twice the mass of object 1. Assume object 1 is moving to the right at 3 m/s while object 2 is at rest in the lab frame. Again, the center of mass will be moving at 1 m/s to the right. In the center of mass frame, object 1 is moving at 2 m/s to the right, and object 2 is moving at 1 m/s to the left. The momentum of object 1 is p

If you go back to the lab frame, you'd find that object 1 is still moving to the right but at a speed of 0.5 m/s. Object 2 will also be moving to the right at a speed of 1.25 m/s. If you calculate the momenta and energies in this frame, you'll again find momentum is conserved and kinetic energy has decreased.

Share:

- Replies
- 9

- Views
- 120

- Replies
- 6

- Views
- 570

- Replies
- 7

- Views
- 737

- Replies
- 12

- Views
- 522

- Replies
- 1

- Views
- 277

- Replies
- 4

- Views
- 478

- Replies
- 19

- Views
- 973

- Replies
- 2

- Views
- 329

- Replies
- 5

- Views
- 773

- Replies
- 2

- Views
- 562