Change in Kinetic Energy in two different reference frames (one of them non-inertial) [duplicate]

Question

Let's say an object of mass 10 kg is fired from a cart of mass 90 kg. The object and the cart, of total mass 100 kg were initially moving together with a speed of 10 m/s. Then, the object is fired by some means from the cart such that it's velocity increases to 15 m/s (w.r.t ground), in the same direction. Finding the change in kinetic energy from ground and cart frame yields different values.

Calculating the kinetic energy difference in ground and cart frame of reference,

From Ground frame: $$m_1 = 10 kg, \ m_2 = 90 kg,\ u_1 = 10 m/s , \ u_2 = 10 m/s, v_1 = 15 m/s, v_2 = ?$$ Applying momentum conservation, $$ v_2 = 85/9 \ \ m/s$$ $$ \bigtriangleup KE = 1/2 ( \ 10 \times 15^2 + 90 \times (85/9)^2 - 100 \times 10^2 )$$ $$ = 1250/9 J \ = 138.888 J $$

From Cart frame (non-inertial): $$m_1 = 10 kg, \ m_2 = 90 kg,\ u_1 = 10 m/s , \ u_2 = 0 m/s, v_1 = (15 - 85/9) m/s = 50/9 m/s, v_2 = 0 m/s$$ $$ \bigtriangleup KE = 1/2 \times 10 \times (50/9)^2 $$ $$ = 154.321 J $$

There is a discrepancy of change in kinetic energy calculated from inertial and non-inertial frame. If one takes an inertial frame going with 10 m/s in the direction of the cart, the calculation from the frame will same change in kinetic energy as that given from the ground frame. What correction does the calculation from non-inertial frame need?

(Note: This problem deals with actual calculation of kinetic energy changes in a non-inertial frame. Here the pseudo-force is not constant/tractable, only impulse is.)

The question in K.E. with different frames is about calculating KE in different inertial frames. That question was resolved by saying that kinetic energy may be different for different inertial frames, but work energy theorem (related to change in kinetic energy) still holds. In this question,

1. We have gone a step ahead and actually applied the work-energy theorem, rather than just calculate kinetic energy from different frames.
2. The query here is about how to apply the work energy theorem for a non-inertial frame, to account for the change in kinetic energy.

Answer 1

In noninertial case you computed kinetic energy at two different frames. In 1D case, when you boost the frame by $\Delta V$ in x direction the kinetic energy changes:

$$E_k=\sum_i \frac{1}{2}m_iv_i^2 \rightarrow E'_k=\sum_i \frac{1}{2}m_i(v_i-\Delta V)^2=E_k-\frac{1}{2}\Delta V\sum_i m_i v_i+ \frac{1}{2}M\Delta V^2,$$ where $M=\sum_i m_i.$ In your case, the initial velocities were 0 so the middle term is zero and the last term gives you 15.432J. To get energy difference, you thus need to subtract this energy, which is basicaly initial energy from frame of the cart after firing the object: $154.321-15.432=138.89J$

Answer 2

What correction does the calculation from non-inertial frame need?

The calculation from the non-inertial frame does not need any correction. It is an accurate calculation. Note that the mechanism that launched the object still used 139 J and the KE changed by 154 J, so in this non-inertial frame energy is not conserved.

Energy is not necessarily conserved in non-inertial frames.

To understand why, we can appeal to Noether’s theorem, which basically says that every conserved quantity is related to a continuous symmetry in the laws of physics. In particular, the conservation of energy is related to the time-translation symmetry of the laws of physics.

In this particular non-inertial frame the laws of physics include an inertial force (a.k.a. a fictitious force). This inertial force turns on briefly during the firing of the object and counteracts the force between the cart and the object in order to keep the cart stationary. This inertial force is not time translation symmetric so according to Noether’s theorem energy is not conserved in this frame. Indeed, the action of this fictitious force suddenly changes the KE of the earth and the rest of the universe without any energy input.

Note, that energy is conserved in some non-inertial frames. For example, a rotating frame or a constantly accelerating frame. Both of these have inertial force laws which are constant in time, so although they are non inertial they still have the required symmetry per Noether’s theorem. In these frames there is a potential energy, and any work done by the inertial force is accompanied by a corresponding decrease in potential energy.