A bullet of mass 6.00g is fired horizontally into a wooden block of mass 1.21kg
ID: 1288035 • Letter: A
Question
A bullet of mass 6.00g is fired horizontally into a wooden block of mass 1.21kg resting on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.190. The bullet remains embedded in the block, which is observed to slide a distance 0.300m along the surface before stopping.A bullet of mass 6.00g is fired horizontally into a wooden block of mass 1.21kg resting on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.190. The bullet remains embedded in the block, which is observed to slide a distance 0.300m along the surface before stopping.
What was the initial speed of the bullet?
Explanation / Answer
the work done by sliding the block+bullet along the table is equal to the KE at the beginning of the slide
W = F*d
F = mu m g = 0.190 (1.21 + 0.006) (9.81) = 2.266 N
so
W = 2.66N * 0.300m = 0.6799 J
then
KE = 1/2 m v^2
0.6799 = 1/2 (1.216) v^2
v^2 = 1.11834
v = 1.0575 m/s
(I am assuming that you want the speed of the bullet)
Now the momentum of the bullet just before collision is the same as the momentum of block+bullet just after collision (at the start of the slide)
m1v1 = m2v2
0.00600 v1 = 1.216 (1.0575)
v1 = 214.32 m/s or
speed of bullet is 214 m/s [nearly]
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