Qns On The Equations Of Momentum

  • Thread starter Thread starter Delzac
  • Start date Start date
  • Tags Tags
    Momentum
AI Thread Summary
The discussion centers on the equations of momentum, particularly the derivation of momentum conservation equations for elastic collisions. It explains that the assumptions V_2 ≠ U_2 and V_1 ≠ U_1 are necessary to avoid division by zero in the derivation, as setting these equal implies no collision occurs. The conservation of momentum for inelastic collisions is also mentioned, highlighting that when initial and final velocities are equal, no kinetic energy is lost. The relationship between relative velocities before and after a collision is emphasized, which is valid only for elastic collisions. Overall, the conversation clarifies the mathematical and physical implications of these equations in collision scenarios.
Delzac
Messages
389
Reaction score
0
hi, i read in a textbook that one equation for momentum is :

where U = intial velocity, V = Final velocity

U_1 + V_1 = V_2 + U_2 ----------- (1)

Equation (1) is derieve from Equation of " conservation of momentum " and " conservation of kinetic energy". and that they assume that

V_2 is not equal is U_2 and ,

V_1 is not equal to U_1.

why did they take this assumption? ( the book did not offer further explanation)

And what does it mean physically when u take

V_2 = U_2 and ,

V_1 = U_1. ??

+ Collision is Elastic +
 
Physics news on Phys.org
think about what u and v actually mean first, u is the initial velocity like you said and v is the final velcity.

i thought that the conservation of momentum is given by;

mu_{1}+mu_{2}=mv_{1}+mv_{2}

where u_{1} is the initial velocity of object 1 and u_{2} is the initial velocity of object 2, and likewise for the final velocity. this is conservation of momentum for inelastic collisons.

when v=u it means that no kinetic energy is lost during the collision.
 
That equation says that the relative velocity of approach before the collision equals the relative velocity of separation after the collision. It is derived by combining conservation of momentum and conservation of energy. (Thus it is only valid for elastic collisions.) In the derivation, one divides by terms equivalent to "V_2 - U_2" and "V_1 - U_1", so if V_2 = U_2 or V_1 = U_1, you'd be dividing by zero, which is a no-no.

But that's not very restrictive, since if V_2 = U_2 & V_1 = U_1 there would be no collision anyway. :wink:
 
Doc Al said:
That equation says that the relative velocity of approach before the collision equals the relative velocity of separation after the collision. It is derived by combining conservation of momentum and conservation of energy. (Thus it is only valid for elastic collisions.) In the derivation, one divides by terms equivalent to "V_2 - U_2" and "V_1 - U_1", so if V_2 = U_2 or V_1 = U_1, you'd be dividing by zero, which is a no-no.

But that's not very restrictive, since if V_2 = U_2 & V_1 = U_1 there would be no collision anyway. :wink:

lol I meant that I am a bit tired so not thinking straight
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top