Calculating Acceleration and Velocity in Newton's Third Law Scenario

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SUMMARY

The discussion focuses on calculating the acceleration and final velocity of a paintball gun and paintball using Newton's Third Law. A paintball with a mass of 0.15 kg is fired at a velocity of 45 m/s after accelerating for 0.10 seconds. The acceleration of the paintball is calculated to be 450 m/s², resulting in a force of 67.5 N acting on it. Consequently, the paintball gun, with a mass of 5.5 kg, experiences a recoil acceleration of 12.27 m/s², leading to a final velocity of 1.2 m/s in the opposite direction.

PREREQUISITES
  • Understanding of Newton's Third Law of Motion
  • Basic knowledge of kinematic equations
  • Familiarity with force and mass calculations
  • Ability to manipulate equations for acceleration and velocity
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  • Explore practical applications of Newton's Third Law in various physical scenarios.
  • Learn about the concept of recoil in firearms and its implications in physics.
  • Investigate the effects of mass on acceleration in different systems.
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Homework Statement


A paintball with a mass of 0.15 kg is fired from a paintball gun that has a mass of 5.5kg. The paintball leaves the gun with a velocity of 45 m/s [N] having accelerated for 0.10s. Calculate the acceleration and the final velocity of the paintball gun.
 
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Here, you would use Newton's third law, and that the forces should equal. "For every action, there is an equal, but opposite reaction." The paintball gun forces the paintball in the +x direction, the ball pushes the gun in the -x direction. (this is why you feel recoil when shooting).

Find the force that's acting on the paintball.

Fball = mballaball

aball = (vf - vi)/t

This is all given to you in your question. simply solve for the Force. Since the total force must equal 0

Fball + Fgun = 0

Fball = - Fgun

Here, you can equate the force that's on the gun. Since you know the mass of the gun, you can find its acceleration. Then using the same equation as before,

agun = (vf - vi)/t

And just solve for the final velocity. (the initial should be zero as it starts from rest, before the gun is fired) (and the t value is the same as before. It's only accelerating as long as the ball enacts on the gun, so the time is 0.10s)
 
Rawrr! said:
Here, you would use Newton's third law, and that the forces should equal. "For every action, there is an equal, but opposite reaction." The paintball gun forces the paintball in the +x direction, the ball pushes the gun in the -x direction. (this is why you feel recoil when shooting).

Find the force that's acting on the paintball.

Fball = mballaball

aball = (vf - vi)/t

This is all given to you in your question. simply solve for the Force. Since the total force must equal 0

Fball + Fgun = 0

Fball = - Fgun

Here, you can equate the force that's on the gun. Since you know the mass of the gun, you can find its acceleration. Then using the same equation as before,

agun = (vf - vi)/t

And just solve for the final velocity. (the initial should be zero as it starts from rest, before the gun is fired) (and the t value is the same as before. It's only accelerating as long as the ball enacts on the gun, so the time is 0.10s)

When calculating the acceleration of the ball would I use 0 as the final velocity?
 
The final velocity would be the velocity it has when it leaves the barrel. So in your initial question, you say the ball leaves the gun with a velocity of 45 m/s . The initial velocity of the ball would be zero as both the gun and the ball are at rest, before the gun is shot.
 
Rawrr! said:
The final velocity would be the velocity it has when it leaves the barrel. So in your initial question, you say the ball leaves the gun with a velocity of 45 m/s . The initial velocity of the ball would be zero as both the gun and the ball are at rest, before the gun is shot.

aball= Vf-Vi/t
= 45-0/0.10
= 450m/s^2

Fball=ma
=(0.15)(450)
=67.5 N

Fgun= -67.5N

Fnet=ma
-67.5=(5.5)(-a)
agun=12.27 m/s^2

agun=Vf-Vi/t
12.27 =Vf-0 / 0.10
Vf = 1.2 m/s

Is this correct?
 
Looks correct! Just remember that the velocity of the gun is in the opposite direction of the ball. Just like when you shoot a bullet, the bullet goes one way, and the gun goes in the other (the recoil). (If it's going to be graded, it might be beneficial to write down the direction of the velocity, in case they look for that).
 

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