Concervation of Momentum: Bullet and Block of Mass

In summary, using conservation of momentum, the speed of the block after the bullet exits can be found by solving the equation (.0197kg)(445m/s) + (.854kg)(0) = (.0197kg)v'1 + (.854kg)v'2, where v'1 is the velocity of the bullet and v'2 is the velocity of the block. The values for v'1 and v'2 can be solved for using the known values of the mass and velocity of the bullet and the final velocity of the block.
  • #1
sweetpete28
80
0
A bullet of mass m = 19.7 g traveling horizontally at speed vo = 445 m/s strikes a block of mass M = 854 g sitting on a frictionless, horizontal table. The bullet comes out the other side of the block at speed v = 415 m/s. What is the speed of the block after the bullet exits the block?

Using conservation of momentum, I have the following equation but there are two unknowns (v'1 and v'2) so I need 2 equations but am kind of lost as to what should be my second equation? Can someone please help?

m1v1 + m2v2 = m1v'1 + m2v'2

(.0197kg)(445m/s) + (.854kg)(0) = (.0197kg)v'1 + (.854kg)v'2

8.7665 = .0197v'1 + .854v'2
 
Physics news on Phys.org
  • #2
sweetpete28 said:
A bullet of mass m = 19.7 g traveling horizontally at speed vo = 445 m/s strikes a block of mass M = 854 g sitting on a frictionless, horizontal table. The bullet comes out the other side of the block at speed v = 415 m/s. What is the speed of the block after the bullet exits the block?

Using conservation of momentum, I have the following equation but there are two unknowns (v'1 and v'2) so I need 2 equations but am kind of lost as to what should be my second equation? Can someone please help?

m1v1 + m2v2 = m1v'1 + m2v'2

(.0197kg)(445m/s) + (.854kg)(0) = (.0197kg)v'1 + (.854kg)v'2

8.7665 = .0197v'1 + .854v'2

You know v1', it is given in the problem.
 
  • #3
the velocity on the right side of the equation is 415, it is given in the problem. you can solve for V2 using the same equation
 

FAQ: Concervation of Momentum: Bullet and Block of Mass

What is the conservation of momentum?

The conservation of momentum is a fundamental principle in physics that states that the total momentum of a closed system remains constant, regardless of any internal changes or external forces acting on the system.

How does the conservation of momentum apply to a bullet and block of mass?

In the case of a bullet and block of mass, the conservation of momentum means that the total momentum of the system before and after the collision remains the same. This means that the momentum of the bullet before it hits the block is equal to the combined momentum of the bullet and block after the collision.

What factors affect the conservation of momentum in this scenario?

The conservation of momentum in a bullet and block of mass scenario is affected by the masses of the objects involved, as well as their velocities before and after the collision. The angle of the collision and the elasticity of the objects can also play a role in the conservation of momentum.

What happens if the bullet and block have different masses?

If the bullet and block have different masses, the conservation of momentum still applies. The heavier object will experience a smaller change in velocity compared to the lighter object, but the total momentum of the system will remain constant.

Is the conservation of momentum always applicable in real-life scenarios?

In ideal conditions, the conservation of momentum is always applicable. However, in real-life scenarios, there may be external forces such as friction and air resistance that can affect the conservation of momentum. These factors must be taken into account when analyzing the conservation of momentum in real-world situations.

Back
Top