Problem 7.43 Part A What is the initial velocity (assumed horizontal) of the bul
ID: 1773221 • Letter: P
Question
Problem 7.43 Part A What is the initial velocity (assumed horizontal) of the bullet? Express your answer using two significant figures. A bullet of mass m 0.0010 kg embeds itself in a wooden block with mass M 0.999 kg, which then compresses a spring (k = 140 N/m ) by a distance x = 0.050 m before coming to rest. The coefficient of kinetic friction between the block and table is -0.50. ValueUnits Submit My Answers Give Up Part B What fraction of the bullet's initial kinetic energy is dissipated (in damage to the wooden block, rising temperature, etc.) in the collision between the bullet and the block? Submit My Answers Give UpExplanation / Answer
a] Energy of spring compression
PS = 0.5kx^2
= 0.5*140*(0.05)^2
= 0.175 J
Energy lost to friction during spring compression
W2 = uMgx
= 0.5*(0.999+0.001)*9.8*0.05)
= 0.245
Kinetic energy of block plus bullet just after collison
KE2 = PS + W2 = 0.175 + 0.245 = 0.42
velocity of bullet plus block just after collision
KE2 = 0.5mv^2
v = sqrt(2*KE2/m)
= sqrt(2*0.42/(0.999+0.001)) = 0.92 m/s
use conservation of momentum to find initial bullet velocity
m(u) + M(0) = (M + m)v
u = (M + m)v / m
= (0.999 + 0.001)*0.92 / 0.001
= 920 m/s answer
b] kinetic energy of bullet before impact
KE1 = 0.5*0.001*920^2 = 423.2
fraction lost to impact
W1 / KE1 = (KE1 - KE2) / KE1
W1 / KE1 = (423.2 - 0.42)/423.2
= 0.999 = 99.9% answer