Momentum conservation

In summary: using conservation of momentum and energy equations, with a slight correction for the bullet's mass.
  • #1
ritwik06
580
0

Homework Statement



A bullet(mass 100 g) strikes a block of 4 kg kept on a table. The friction coefficient of table-block is 0.25. The bullet gets embedded and moves a distance of 20 m. Find speed of bullet.

Homework Equations


The Attempt at a Solution


Using Conservation of momentum,
speed of block + bullet= v/401 (v is the speed of the bullet)
now equating Energy;
Work one by friction = kinetic energy of system
0.25*4.01*10*20=0.5*4.01*(v/401)^2
v=4010
am i right?
 
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  • #2
ritwik06 said:

Homework Statement



A bullet(mass 100 g) strikes a block of 4 kg kept on a table. The friction coefficient of table-block is 0.25. The bullet gets embedded and moves a distance of 20 m. Find speed of bullet.

Homework Equations





The Attempt at a Solution


Using Conservation of momentum,
speed of block + bullet= v/401 (v is the speed of the bullet)
now equating Energy;
Work one by friction = kinetic energy of system
0.25*4.01*10*20=0.5*4.01*(v/401)^2
v=4010
am i right?

Is applying conservation of momentum correct here?
 
  • #3
ritwik06 said:

Homework Statement



A bullet(mass 100 g) strikes a block of 4 kg kept on a table. The friction coefficient of table-block is 0.25. The bullet gets embedded and moves a distance of 20 m. Find speed of bullet.

Homework Equations





The Attempt at a Solution


Using Conservation of momentum,
speed of block + bullet= v/401 (v is the speed of the bullet)
now equating Energy;
Work one by friction = kinetic energy of system
0.25*4.01*10*20=0.5*4.01*(v/401)^2
v=4010
am i right?
Your method is fine, but you missed a decimal point.
100 grams is 0.1kg, not .01kg.
 
  • #4
ritwik06 said:

Homework Statement



a bullet(mass 100 g) strikes a block of 4 kg kept on a table. The friction coefficient of table-block is 0.25. The bullet gets embedded and moves a distance of 20 m. Find speed of bullet.

Homework Equations





The Attempt at a Solution


using conservation of momentum,
speed of block + bullet= v/401 (v is the speed of the bullet)
now equating energy;
work one by friction = kinetic energy of system
0.25*4.01*10*20=0.5*4.01*(v/401)^2
v=4010
am i right?

solved
 

What is momentum conservation?

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

Why is momentum conservation important?

Momentum conservation is important because it allows us to predict the motion and interactions of objects in a system, and to understand how forces affect the overall motion of a system. It also helps us to understand and analyze collisions, explosions, and other dynamic events.

What factors affect momentum conservation?

Momentum conservation is affected by the mass and velocity of objects in a system, as well as any external forces acting on the system. In a closed system, the total momentum will remain constant as long as there are no external forces present.

Does momentum conservation apply to all types of motion?

Momentum conservation applies to all types of motion, including linear, rotational, and angular motion. It is a universal principle that applies to all physical systems.

How is momentum conserved in a collision?

In a collision between two objects, momentum is conserved as long as there are no external forces acting on the system. This means that the total momentum of the objects before the collision will be equal to the total momentum after the collision, even if the individual momentums of the objects change.

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