A bullet and a block (Momentum)

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SUMMARY

The discussion centers on calculating the initial speed of a bullet fired into a wooden block using principles of physics, specifically conservation of momentum and kinetic friction. A bullet with a mass of 7.00 g is embedded in a block of mass 1.29 kg, which slides 0.230 m before stopping due to a kinetic friction coefficient of 0.180. The internal energy loss due to friction is calculated to be 0.5262 J, leading to a block velocity of 0.9007 m/s. The initial speed of the bullet is determined to be 166.9 m/s, which, while mathematically correct, raises questions about its realism given the short sliding distance.

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A bullet of mass 7.00 g is fired horizontally into a wooden block of mass 1.29 kg resting on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.180. The bullet remains embedded in the block, which is observed to slide a distance 0.230 m along the surface before stopping.

What was the initial speed of the bullet?



Okay, first off, I know the block is sliding after the collision so that is internal energy.

\DeltaUinternal = Friction Force* Distance

=(12.7106N)(.180)(.230m) = 0.52621884J

From here, i do not know what to do.

Here was what I attempted:

KE=\DeltaUinternal

so

.5(1.297kg)v^2 = 0.52621884J

Velocity of the block was 0.9007m/s

Then I used Conservation of Momentum

Pi = Pf

.007kgV = (1.297kg)(.9007m/s)

V=166.9m/s


I don't know, seems slow for a bullet.
 
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Yes, it seems slow for a bullet, but then 23 cm doesn't seem very far for the block to slide. Your approach looks correct to me.
 

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