Transforming momentum between inertial reference frames

AI Thread Summary
The discussion focuses on transforming momentum between two inertial reference frames: one for a bug and one for a Mack truck. In the truck's frame, the system momentum is expressed as m_Bv_BT, while in the bug's frame, it is -m_Tv_BT. Participants note that removing m_Bv_BT from the expression seems unnecessary, as the key point is the difference in momentum between the two frames. The conversation highlights the importance of understanding the transformation process and the consistency of results derived from both frames. Ultimately, the discussion emphasizes that while the absolute values of momentum differ, the differences between them should remain consistent across frames.
bkraabel
Messages
26
Reaction score
0

Homework Statement


A bug of inertia m_B collides with the windshield of a Mack truck of inertia m_T \gg m_B at an instant when the relative velocity of the two is \boldsymbol v_{BT}.
(a) Express the system momentum in the truck’s reference frame, then transform that expression
to the bug’s reference frame, and in so doing remove m_B\boldsymbol v_{BT} from the expression. (Remember, in the bug’s reference frame, the bug is initially at rest and the truck is moving.)
(b) Now express the system momentum in the bug’s reference frame, then transform that expression to the truck’s reference frame, and in so doing remove m_T\boldsymbol v_{BT} from the expression.
(c) Is there something wrong here? How can we change the momentum by a small amount m_Bv_{BT} doing the transformation one way and by a large amount m_Tv_{BT} doing the transformation the other way?

Homework Equations


Take the bug's direction as the positive direction. System momentum in bug frame is
\boldsymbol p_{sys,B}=-m_T\boldsymbol v_{BT}
System momentum in truck frame is
\boldsymbol p_{sys,T}=m_B\boldsymbol v_{BT}

The Attempt at a Solution


I can see that the magnitude of the momentum is much larger in the bug frame, but I don't get the part about removing m_B\boldsymbol v_{BT}. It doesn't seem necessary or even possible. I understand that the absolute magnitude of the momentum in different inertial reference frames is not important. What is important is the difference between momenta in two inertial frames. This difference should be the same in the two frames.
 
Physics news on Phys.org
No special rules given for momentum transformation, just the regular Galilean transformation rules for transforming velocity between different frames. But applying a Galilean transformation just gives you the equations we've already written above.
 
Well yes - but you asked about the process.
You can just write down the equation by inspection because, basically, you know the outcome in advance... or you can start with one and formally apply the transformation step-by-step and demonstrate you get the same thing.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top