How Fast Was the Bullet Before Striking the Block?

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Homework Help Overview

The problem involves a bullet striking a block on a table, with the goal of determining the bullet's speed before impact. The scenario includes parameters such as the masses of the bullet and block, the coefficient of friction, and the distance the block moves after the bullet embeds itself.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss using conservation of momentum and energy to relate the bullet's speed to the motion of the block. There are questions about the correctness of applying conservation of momentum in this context and the accuracy of the mass conversion from grams to kilograms.

Discussion Status

Some participants have provided feedback on the method used, noting a potential error in the mass conversion. There is an ongoing exploration of the assumptions made in applying the conservation principles, but no consensus has been reached regarding the correctness of the approach.

Contextual Notes

Participants are working within the constraints of a homework problem, which may limit the information available for discussion. The specific values and relationships presented in the problem are being scrutinized for accuracy.

ritwik06
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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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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?
 
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.
 
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[/color]
 

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