Conservation Of Momentum: Paintball Firing

[DeathCrawls]

Homework Statement

A student hold a 2kg air rifle loosely and fires a paint ball that has a mass of .001 kg. The paintball is fired at 150 m/s. What is the recoil velocity of the rifle? If the 48kg student holds the rifle more firmly, so that it rests against there body, what is the new recoil velocity?

Homework Equations

P=mv

The Attempt at a Solution

I have no idea, I know your supposed to make initial and final but I don't know how to work out the problem.



Can you please tell me how you did this

 

[rock.freak667]

Conservation of momentum: In a given direction, momentum before = momentum after.

Momentum = mass*velocity

So before, what is the total momentum?

After what is the momentum of the bullet when fired and if the recoil velocity is 'v', what the momentum of the person when the bullet is fired??

 

[DeathCrawls]

I got the total momentum as being .15 from 150x.001
Then i got the other side to be 2v+.15
But wen I try to solve it comes out as 0?

the equation i got was .15=2v+.15

 

[rock.freak667]

I got the total momentum as being .15 from 150x.001
Then i got the other side to be 2v+.15
But wen I try to solve it comes out as 0?

the equation i got was .15=2v+.15

The other side (2v+0.15) kgm/s is correct. For the left side, before the gun is fired, it isn't moving! So it has no velocity and hence the momentum is?

 

[DeathCrawls]

0! wow, thank you high school -.-

 

[rock.freak667]

0! wow, thank you high school -.-

Just remember, your velocity should be negative since the recoil velocity's direction is opposite to direction of the bullet.

 

[DeathCrawls]

Can i ask you some other question that I need help on?

 

[rock.freak667]

Can i ask you some other question that I need help on?

Sure, assuming it is momentum related, otherwise, you might ask something I don't know much about.

 

[DeathCrawls]

1.A 20 kg child running at 4 m/s jumps into the lap of an adult sitting in a office chair with wheels. The adult plus the chair have mass of 60 kg. The collision of course is inelastic. Determine the velocity after the collision and determine how much energy is lost.

I F
20 60 20 60
v=4 v=0 v=0 v=?
80 0 0 60v
80=60v?
v=1.33333333...
energy lost was 2.7777777...
Is that right?

2. Two clowns on skateboards are standing motionless on a smooth surface two meters apart. Clown B 950 kg) tosses a 5 kg medicine ball to clown A (70kg) at 10 m/s. find a) velocity of clown B after throwing the ball, and b) the velocity of the clown A after catching the ball.

No idea how to do this one :(

 

[DeathCrawls]

Last one (FYI) this is for a review that will not be graded but I just need help for a final that I will be having tmrw, trying to get all the practice in I can.

A .5 kg super ball is dropped onto the floor. It is going 20 m/s when it hits and is going 18 m/s as it comes up off the floor. a)what is the impulse of the ball? b) if the super ball bounce is in the contact with the ground for .0025 seconds, determine the amount of force acting on the ball.

Like the previous one I tired and epicly failed

 

[wbandersonjr]

I F
20 60 20 60
v=4 v=0 v=0 v=?
80 0 0 60v
80=60v?
v=1.33333333...
QUOTE]

The right side is incorrect. The momentum is simply the momentum of the combined child-adult-chair, so it should be (m-adult + m-child)*v-final


Also, the energy that is being gained or lost is kinetic energy. You need to determine the initial and final kinetic energies of the child.

 

[DeathCrawls]

The right side is incorrect. The momentum is simply the momentum of the combined child-adult-chair, so it should be (m-adult + m-child)*v-final


Also, the energy that is being gained or lost is kinetic energy. You need to determine the initial and final kinetic energies of the child.[/QUOTE]

so the ending velocity should be 1?
making the lost 3?

 

[wbandersonjr]

2. Two clowns on skateboards are standing motionless on a smooth surface two meters apart. Clown B 950 kg) tosses a 5 kg medicine ball to clown A (70kg) at 10 m/s. find a) velocity of clown B after throwing the ball, and b) the velocity of the clown A after catching the ball.

In a) it is simply an inelastic collision being run in reverse. Instead of the ball and clown colliding, they are separating. So instead of the final side of the conservation of momentum equation having the combined mass*v, now you will have the initial side having the combined mass and the final side having the individual momentums added together.

In b) it is simply an inelastic collision between the ball and clown A

 

[wbandersonjr]

so the ending velocity should be 1?
making the lost 3?

Yes final velocity is 1m/s

The energy lost is not 3 J though. Remember kinetic energy is 1/2 m*v^2. You need to figure out how much kinetic energy the child had initially and then the child's final kinetic energy.

 

[DeathCrawls]

Yes final velocity is 1m/s

The energy lost is not 3 J though. Remember kinetic energy is 1/2 m*v^2. You need to figure out how much kinetic energy the child had initially and then the child's final kinetic energy.

woooooo what does that mean, never even heard of that equation

 

[DeathCrawls]

In a) it is simply an inelastic collision being run in reverse. Instead of the ball and clown colliding, they are separating. So instead of the final side of the conservation of momentum equation having the combined mass*v, now you will have the initial side having the combined mass and the final side having the individual momentums added together.

In b) it is simply an inelastic collision between the ball and clown A

sooo can you put that in like noob terms or so?

 

[wbandersonjr]

Kinetic energy is the energy associated with movement. it is calculated using mass and velocity. So the child's initial kinetic energy is .5*m-child*[v-initial(squared)]. The final kinetic energy is .5*m-child*[v-final(squared)]. The final will be less than the initial. The reason is because initially the adult had zero kinetic energy (was not moving) and the child had kinetic energy. After they collided, the child essentially gave some amount of kinetic energy to the adult (and chair), so in the end the child lost kinetic energy and the adult gain kinetic energy (both are moving, the child slower than initially). The cool thing is that the total kinetic energy before and after is equal (energy is conserved!).

 

[wbandersonjr]

sooo can you put that in like noob terms or so?

Take your conservation of momentum equation for an inelastic collision and flip it around so that the combined mass*velocity is on the initial side because clown B and the ball are initially together. Bascially it is the same as the paintball question, with a clown and ball instead of a gun and paintball.

The second part is simply an inelastic collision in which the ball collides with Clown A

 

[gneill]

Kinetic energy is the energy associated with movement. it is calculated using mass and velocity. So the child's initial kinetic energy is .5*m-child*[v-initial(squared)]. The final kinetic energy is .5*m-child*[v-final(squared)]. The final will be less than the initial. The reason is because initially the adult had zero kinetic energy (was not moving) and the child had kinetic energy. After they collided, the child essentially gave some amount of kinetic energy to the adult (and chair), so in the end the child lost kinetic energy and the adult gain kinetic energy (both are moving, the child slower than initially). The cool thing is that the total kinetic energy before and after is equal (energy is conserved!).

Careful, this is an inelastic collision. Kinetic energy is not conserved in an inelastic collision.

Also, the problem did not specify that it was the difference in kinetic energy of the child in particular that was of interest, rather it just said "Determine the velocity after the collision and determine how much energy is lost." So the kinetic energy after collision should include the child and the adult+chair.

 

[wbandersonjr]

Thanks for catching my error.