Newton's Second Law for a System of Particles

• peaceandlove
In summary, the initial velocity of a shell shot at an angle of 56° with the horizontal is 29 m/s. At the top of its trajectory, the shell explodes into two fragments of equal mass. One fragment falls vertically while the other continues traveling through the air along the y-axis. By using the conservation of momentum, the horizontal distance traveled by the other fragment can be calculated to be 119.35124 m.
peaceandlove

Homework Statement

A shell is shot with an initial velocity 0 of 29 m/s, at an angle of θ0 = 56° with the horizontal. At the top of the trajectory, the shell explodes into two fragments of equal mass. One fragment, whose speed immediately after the explosion is zero, falls vertically. How far from the gun does the other fragment land, assuming that the terrain is level and that air drag is negligible?

Homework Equations

Equation 1: R=((v0^2)/g)*sin(2θ0)

The Attempt at a Solution

R gives the horizontal range of the projectile, however, due to the change in mass and velocity of the shell, I divided the R by 1/2 to get the distance traveled by the shell before its explosion. After that, I'm not sure what to do.

peaceandlove said:
A shell is shot with an initial velocity 0 of 29 m/s, at an angle of θ0 = 56° with the horizontal. At the top of the trajectory, the shell explodes into two fragments of equal mass. One fragment, whose speed immediately after the explosion is zero, falls vertically. How far from the gun does the other fragment land, assuming that the terrain is level and that air drag is negligible?

Hi peaceandlove!

What is the speed of the shell just before it explodes?

What equation can you use to find the speed of the other fragment just after the explosion?

Why hello tiny-tim!

I have no clue what the speed of the shell is right before it explodes, so I'm just going to say 29 m/3.

I don't know if this would be the right equation to use, but according to (m1+m2)(v0)(cos (θ0))=(m1v1)+(m2v2) where v1=0 and (m1+m2)=1, the velocity of the fragment that continues to travel through the air along the y-axis is 32.4332 m/s.

peaceandlove said:
I have no clue what the speed of the shell is right before it explodes

ok, let's do that first …

what is its vertical component of speed right before it explodes?

and what is its horizontal component of speed right before it explodes?

I figured it out! I found the coordinates of the point where the shell explodes and the velocity of the fragment that did not fall straight down. Since no horizontal forces act, the horizontal component of the momentum is conserved. Since one fragment has a velocity of zero after the explosion, the momentum of the other equals the momentum of the shell before the explosion. I then used this information to determine where the fragment lands by analyzing a projectile launched horizontally at time t=0 and got 119.35124 m.

Excellent!

Woohoo!

1. What is Newton's Second Law for a System of Particles?

Newton's Second Law for a System of Particles states that the acceleration of a system of particles is directly proportional to the net force acting on the system and inversely proportional to the total mass of the system. In other words, the greater the force applied to a system of particles, the greater the acceleration will be, and the greater the mass of the system, the smaller the acceleration will be.

2. How is Newton's Second Law for a System of Particles different from Newton's Second Law for a Single Particle?

The main difference between the two laws is that Newton's Second Law for a System of Particles takes into account the combined effect of all the forces acting on a system, rather than just one single particle. This allows for a more accurate prediction of the motion of a system of particles.

3. How is the net force calculated in Newton's Second Law for a System of Particles?

The net force on a system of particles is calculated by adding up all the individual forces acting on each particle in the system. This includes both external forces, such as applied forces, and internal forces, such as forces between the particles in the system.

4. Can Newton's Second Law for a System of Particles be applied to non-rigid systems?

Yes, Newton's Second Law for a System of Particles can be applied to both rigid and non-rigid systems. However, for non-rigid systems, the mass of the system must be calculated as the sum of the individual masses of each particle, rather than assuming a constant mass for the entire system.

5. How is Newton's Second Law for a System of Particles related to the concept of inertia?

Newton's Second Law for a System of Particles is closely related to the concept of inertia, which is the tendency of an object to resist changes in its state of motion. In this law, the mass of the system plays a key role in determining the amount of inertia, as a larger mass will result in a greater resistance to changes in motion.

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