Bullet embedded in block: finding distance, time, and energy loss

[nathan8]

Homework Statement

40-g Bullet is fired with a velocity of magnitude v0=700 m/s into a 5kg block of wood. Coefficient of kinetic friction = 0.6.
a) how far will the block move

b) how long after impact does it come to rest

c) percentage of energy lost on impact

(block has no velocity until bullet is completely embedded in block)

Homework Equations

The Attempt at a Solution

mass bullet: m = 40g = .04kg
mass block: M = 5kg
velocity bullet = 700 m/s
uk = 0.6


KE of bullet
1/2*0.04*700^2 = 9800J

Conservation of Momentum Work Energy

mv+0 = (M+m)v -uk(m+M)gd=-1/2(m+M)v^2
0.04(700) = 5.04v ukgd=1/2v^2

v = 5.56m/s 0.6(9.81)d=0.5(5.56)^2
d = 2.63

a) 2.63m

b) D=v*t t=d/v
t = 2.63/5.56
t= 0.473s

c) kinetic energy bullet = 9800J
Kinetic energy of block and bullet = 77.9J

%= 77.9/9800 == 0.07%

I am most confused on the percentage of energy lost?

 

[ehild]

Take care. The momentum is conserved during the impact, as we can assume it so fast that the force of friction does practically no work during that time. After the impact, there is force on the block, so its momentum will change.

Momentum is not equal to energy.

Energy is not conserved. ehild

 

[zgozvrm]

Also, take care with the units. Bullet are generally measured in grains (abbreviated "gr"), not grams (abbreviated "g").

40 g (grams) would be a rather large bullet, but not out of the realm of possibility. A .50 BMG shell will have a bullet weighing anywhere from about 42 g to 52 g which equates to around 647 gr to 800 gr.

By contrast, a .44 magnum bullet can weigh around 340 gr (22 g)

 

[nathan8]

The weight of the bullet is 40 grams, sounds large to me as well.

Will work energy equations work better here? Or am I on the right track?

 

[gneill]

If you determine the acceleration of the block+bullet (due to friction), then for part b you have initial velocity, final velocity (zero) and acceleration. A simple kinematic equation relates them with time.