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

- #3

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

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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).

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

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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]

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- #10

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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.

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

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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.

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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?

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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.

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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

vela

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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.

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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?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.

- #22

vela

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V and v aren't the same variable.

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Oh, I am completely sorry. I did not notice you capitalized v in the first part. I thought you just rewrote the same thingV and v aren't the same variable.

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