Why Does Kinetic Energy Increase More with Higher Initial Velocities?

[jbriggs444]

Or it could just be that I have not reached the place in my studies where I am ABLE to calculate what you are asking. If I could then I would. Everyone has to start somewhere.

This is a basic homework thread, not a masterclass in differential calculus.

No calculus is involved. Just addition, subtraction, multiplication and division. No fancy formulas that you do not already know. The one useful insight is the one that @TSny gave in post #8.

Rather than considering a continuous exhaust stream produced from the starting speed to the ending speed (which does require calculus and leads to the Tsiolkovsky rocket equation) go back to post #8 and review.

Let me reproduce that idea here.

We will work with symbols. One could, instead, substitute in fixed values, e.g. "A 10 kg rifle shoots a 10 g bullet rearward at 500 meters per second relative to the recoiling rifle . The rifle begins moving at either 1 meter per second or 10 meters per second. Perform an energy analysis". In my opinion, the algebra is more clear with symbols rather than numbers. But some people feel more comfortable with numbers.

You have a large mass M, a small mass m and a massless spring between them. They are lined up on a horizontal axis. You are free to imagine carts with frictionless wheels, pucks on ice, boats on a lake, ships in space or a rifle and a bullet.

You have a massless cord tied around all three holding them together. The spring is compressed and contains a certain amount of potential energy. That energy is sufficient to result in a change in velocity $\Delta V$ in the large mass M. Of course, this will result in a change of $\Delta v$ in the velocity of the small mass m.

The entire assembly is initially moving together in a straight line with the M in front the the m behind. The initial velocity is $V_i = v_i$. You cut the cord and examine the results.

Determine the final velocities $V_f$ of $M$ and $v_f$ of $m$.

Determine the initial and final kinetic energy of just $M$. What is the difference between the two? Does it depend on $v_i$? [This is the part that you've done previously].

Determine the initial and final kinetic energy of just $m$. What is the difference between the two? Does it depend on $v_i$?

Determine the initial total kinetic energy. Determine the final total kinetic energy. Compare the two. What is the difference between the two? Does that difference depend on $v_i$?

Can you perform the calculations to determine these results? How far can you get? If you show your work, we can help when you get stuck.

Edit: I've edited the above a few times to make the algebra simpler. I hope you did not begin work on one of the tougher-to-get-started-on versions.

 

[digalumps]

Thank you very much I will work on this now.

 

[digalumps]

Ok my first working out. An infinitely light person is sitting on top of a 'stationary' 10kg rocket in space

KE = 0 (v=0)

Then person throws a 1kg bowling ball of the back at 1 m/sec

Conservation of momentum. m1v1=m2v2

Thus the rocket moves off in opposite direction at 0.1 m/sec

KE=1/2 mv squared.

Bowling ball: 0.5 x 1 x 1 x 1 = 0.5 Joules

Rocket: 0.5 x 10 x 0.1 x0.1 = 0.05 joules

Why are the two KEs not the same (but opposite directions) given the conservation of energy?

 

[PeroK]

Why are the two KEs not the same (but opposite directions) given the conservation of energy?

If there were conservation of energy then nothing could change. You have zero KE before and you would need zero KE after. The energy comes from the propulsion mechanism: in this case energy stored in the muscles.

There is equal and opposite momentum, but KE is not a vector, so there is no concept of equal and opposite KE. The KE's are not of equal magnitude unless the masses are the same.

 

[digalumps]

Thanks that makes sense - basically I have added 0.5 + 0.05 =0.55 joules from my muscles. Thank you.

 

[gleem]

The rocket scenario is complex so why not look at another similar one Consider a body with a loaded spring mechanism that ejects a small mass in the opposite direction of motion imparting an impulse. You can configure the mechanism so as to produce the same impulse for different velocities.

 

[PeroK]

The rocket scenario is complex so why not look at another similar one Consider a body with a loaded spring mechanism that ejects a small mass in the opposite direction of motion imparting an impulse. You can configure the mechanism so as to produce the same impulse for different velocities.

There's nothing complex about the rocket scenario if you consider a sequence of finite expulsions. You only need to study one example and compare the change in KE in the rest frame of the rocket and one other frame.

 

[gleem]

The OP created an impulse out of nowhere despite the reference to a rocket. No account of the ejected mass is addressed. The Op assumed the energy needed to produce the impulses were the same.

Consider this simpler scenario. The impulses are created by striking the moving masses M with another mass m so as to produce an impulse of 1 kgm/sec. Clearly m must be moving faster to impart a given impulse to the faster M than the slower M thus requiring more energy on the part of m.

 

[etotheipi]

Maybe some of this mess below will be helpful... tranform into the CoM frame of the rocket, and consider a finite expulsion of a mass $\delta m$ from the rocket whose final mass is then $M - \delta m$. The internal energy converted to kinetic is frame invartiant and I'll just call it $\delta E$ because why not$$0 = v_1\delta m + v_2(M-\delta m) \implies v_1 = -\frac{v_2(M-\delta m)}{\delta m}$$ $$\delta E = \frac{1}{2}v_1^2\delta m + \frac{1}{2}v_2^2 (M - \delta m) = \frac{1}{2}\frac{v_2^2 (M-\delta m)^2}{\delta m^2} \delta m + \frac{1}{2}v_2^2 (M-\delta m) = \frac{1}{2}v_2^2 (M-\delta m) \left( \frac{M}{\delta m}\right)$$i.e. the rocket gains $\frac{\delta m}{M}$ of the available energy in the CoM frame. $v_2$ is just the increment in the rocket velocity $\delta v$$$\delta v = \sqrt{\frac{2\delta E}{M - \delta m} \frac{\delta m}{M}}$$N.B. To convert into the continuous case, divide through by $\delta t$ and let $\delta m \rightarrow 0$ and $\delta t \rightarrow 0$,$$a = \sqrt{\frac{-2\dot{E}\dot{M}}{M^2}}$$Return to the discrete case and Galileailean boost all velocities by $U$, you can see that the (unsimplified... because it's too hot today for that) change in kinetic energy of the rocket will be $$
\begin{align*}
\Delta T_{r} = \frac{1}{2} (M - \delta m)\left(U + \sqrt{\frac{2\delta E}{M - \delta m} \frac{\delta m}{M}} \right)^2 - \frac{1}{2}U^2(M - \delta m) \\ \\=\frac{1}{2}(M - \delta m) \times 2U \sqrt{\frac{2\delta E}{M - \delta m} \frac{\delta m}{M}} + \frac{1}{2}(M - \delta m) \frac{2\delta E}{M - \delta m} \frac{\delta m}{M}
\end{align*}
$$And for the fuel,$$
\begin{align*}
\Delta T_{f} = \frac{1}{2} \delta m\left(U - \sqrt{\frac{2\delta E}{\delta m} \frac{M - \delta m}{M}} \right)^2 - \frac{1}{2}U^2 \delta m \\ \\ = - \frac{1}{2} \delta m \times 2U \sqrt{\frac{2\delta E}{\delta m} \frac{M - \delta m}{M}} + \frac{1}{2} \delta m \frac{2\delta E}{\delta m} \frac{M - \delta m}{M}
\end{align*}
$$Now you can check that when you add these, the terms on the left cancel and the total change in kinetic energy $\Delta T_f + \Delta T_r$ is still indeed equal to $\delta E$, even though now the rocket gains kinetic energy and the fuel loses kinetic energy. Maybe you can also play around with these expressions in the limit $\delta m \ll M$