A bullet and a block (Momentum)

[Chandasouk]

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.

[tex]\Delta[/tex]U_internal = Friction Force* Distance

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

From here, i do not know what to do.

Here was what I attempted:

KE=[tex]\Delta[/tex]U_internal

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.

 

[AEM]

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.

 

[kerimek]

Your calculation is correct. The initial speed of the bullet was 166.9 m/s. This may seem slow for a bullet, but it is important to remember that the bullet was embedded in the block and therefore, the mass of the system increased significantly. This decrease in velocity is also due to the friction force acting on the block, which decreases the overall momentum of the system. It is also possible that the bullet was not fired at its maximum speed, or that there were other factors at play that affected its velocity. Nonetheless, your calculations are accurate and provide a reasonable answer.