How Fast Was the Bullet Before Striking the Block?

[ritwik06]

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?

 

[ritwik06]

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?

 

[PhanthomJay]

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]

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]