Can someone help me with these 2 physics questions?

i dont know about #1 but i think number 2 should be a correct explanation

 

#1. Momentum (if I remember correctly) is simply mv. Collision results in transfer of that momentum between objects. So they exchange energy, however the momentum exchanged is not equal, nor is the resulting momentum after collision.

#2. Sounds right. The bullet drives the gun with an equal and opposite force. Determining the velocity of the gun would only require setting both sides of the equation (gun and bullet) equal to each other in terms of mass and velocity.

 

1. m1v1 + m2v2 = m1v2 + m2v1...I think.. Conservation of momentum anyhow.

2. Same thing...momentum is always conserved.

 

Sarte is right for #1. For an isolated system, the total momentum of the system before and after the collision is conserved. Whether it is elastic, inelastic, or totally inelastic just helps you to solve by eliminating some of the variables (v2 would be the same for inelastic), or using Kinetic energy conservation (for elastic).

 

1.) in an inelastic collision (both cars stick together) the equation is m1v1 + m2v2 = (m1 + m2)vfinal. if it's an elastic collision then m1v1 + m2v2 = m1v1final + m2v2final and kinetic energy is also conserved.

2.) momentum is always conserved, so firing a bullet is basically the same as an elastic collision.

 

do your own homework, man!

 

[1] Yes

(a) Let's say there are NO momentum changes. That is, the momentum change for car 1 is zero, that is dM1=0. Similarly, the momentum change for car 2, dM2 is also zero. { dM1 simply stands for deltaM1 or change of momentum for car 1 }

(b) Now, let's say the momentum change of car 1 is not zero; let's say it's some value G, so dM1 = G
Since total momentum has to be conserved, the sum of the change in momentum has to be zero, that is dM1+ dM2 = 0. From this, it's easy to see that dM2 = -G, that is "equal but opposite." Even for case (a) the momentum change, albeit zero, is still equal and opposite.

[2] Yes, there is a problem with this explanation. The linear momentum of the bullet is NOT conserved. Consider a bullet, sitting in the chamber of a gun, happy, "at rest." Then some jerk pulls on the trigger, and accelerates it to over 200 meters/sec. Assume that the bullet's mass remains very much the same. Obviously, there's a huge change in momentum for the bullet.

So, what's the dilly yo? Oh, wait, you're from Hong Kong... nevermind...

The "linear momentum of the bullet DOES NOT 'have' (sic.) to be conserved." The total linear momentum of "THE WHOLE SYSTEM" has to be conserved. For the system you may want to consider the gun and the bullet, and maybe even the person holding the gun. That's reason the gun "kicks" when fired: it's the momentum "balancing itself." If you remember from problem [1], dM1 = -dM2
So, the change in linear momentum for the bullet is equal and opposite for the change in momentum for everything that stays behind.

Try to imagine what happens when the bullet finally lodges into someone's gluteus maximus. Is the momentum conserved?

P.S. Archery is cool. You should do archery. Or hockey.