How Do You Solve a Frictionless Collision Where Two Masses Stick Together?

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In a frictionless collision involving Mass A (60 kg) moving west at 5.00 m/s and Mass B (20 kg) moving south at 20 m/s, the two masses stick together after the collision. The collision is inelastic, as kinetic energy is not conserved; some energy is lost during the process. The total kinetic energy before the collision is greater than after, confirming it is not elastic. The magnitude and direction of their combined velocity can be calculated using vector addition. This discussion highlights the distinction between elastic and inelastic collisions in physics.
justinbaker
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not even sure where to start on this one

Consider the following frictionless collision: Mass A (60 kg) moves west at 5.00 m/s and mass B (20 kg) moves south at 20 m/s. They collide and stick together.

1.) what are the magnitude and the direction of their velocity after they stick together? Give sketch and angle.

2.) Is any kinetic energy lost in the collision? If so how much?

3.) Is this an elastic collision? why or why not?




thanks for the help everyone
 
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well, according to my physics book:

Elastic Collision: One in which the total kinetic energy of the system after the collision is equal to the total kinetic energy before the collision.

Inelastic Collision: One in which the total kinetic energy of the system is not the same before and after the collision; if the objects stick together after colliding, the collision is said to be completely inelastic.


so...i think that should answer your 2nd and 3rd questions...i'd help with the first, but I'm pretty sure i'd screw that up, lol. I hope this helps. :o)
 
can anyone else help with the 1st one?
 
nvr mind i got it
 
Thread 'Variable mass system : water sprayed into a moving container'

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}...
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