The equation (1/2)(mv0)^2 = 1/2(M+m0)gh is invalid for conservation of energy due to the dimensional inconsistency of the terms involved; specifically, (1/2)mv0 represents momentum, not energy. The discussion clarifies that while momentum is conserved in the collision between the dart and the block, kinetic energy is not conserved, indicating that the collision is inelastic. The total momentum remains zero before and after the collision, but the kinetic energy drops to zero post-collision, confirming the inelastic nature of the interaction.
PREREQUISITESStudents of physics, educators teaching mechanics, and anyone interested in understanding the principles of momentum and energy conservation in collision scenarios.
Your question is whether the collision between the dart and block is elastic or not?j04015 said:
Because ##\frac{1}{2}mv_0## has dimensions of momentum and not energy.j04015 said:
Whoops, typo. I meant (1/2)(mv0)^2kuruman said:
PeroK said:
If the collision wasn't elastic the entire problem doesn't make sense.PeroK said:
That statement is false!j04015 said:
I see the issue now. Momentum is conserved but not energy. Thanks!PeroK said:
You can see that if you consider what's going on in the center of mass frame. Before the collision, both dart and block move with opposite momenta. Total momentum is zero and the kinetic energy is non-zero. After the collision, the dart and the block are at rest. Total momentum is zero (conserved) and total kinetic energy is also zero (not conserved).j04015 said:
The conservation of energy equations will not be compatible with conservation of energy in any frame if you assume that the dart sticks. However, I agree that considering the com frame makes it very explicit.kuruman said: