AS S\"Own in tre figure below, a bullet is fired Bt ans passes through a ece af
ID: 1778889 • Letter: A
Question
AS S"Own in tre figure below, a bullet is fired Bt ans passes through a ece af target psper suspended by a massless strng The bullet has a mass m, a speea v befor-the collson wth the tarqet, and a speed (0.526)v after passing through the rget. -he C:llision s ineastic and d ng the colision, the-mount of retenergy lost CV the bullet and paper is equal to (0.333 MO-pe that is, 0.333 o tra kinetic erarqV of the nullet before the colls on Determine the mass of the target and the speed v oft-e target the i-stant after the :ollisidn in terms of the mass m o-- re bulet end speec v of the bulet aefore the C:llision, Expr-S5 your answers to at least 3 decimas.) b) Alber colisionExplanation / Answer
for bullet
initial velocity vi = v
final velocity vf = 0.526 v
kinetic before collision Ki = (1/2)*m*vi^2 = (1/2)*m*v^2
kinetic after collision Kf = (1/2)*m*vf^2 + (1/2)*M*V^2 = (1/2)*m*(0.56*v)^2 + (1/2)*M*V^2
loss of kinetic energy dK = Ki - Kf
dK = (1/2)*m*v^2 - (1/2)*m*(0.56*v)^2 - (1/2)*M*V^2
dK = (1/2)*v^2*(0.6864m - M)
dK/Ki = (0.6864m - M)/m
0.333 = (0.6864m - M)/m
M = 0.353 m <<<---ANSWER
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from momentum conservation
m*v = M*V + m*0.56v
V = (m/M)*0.44 v
V = (m + 0.353m*0.44 v
V = 0.559 v