Posted by Megan on Tuesday, October 16, 2012 at 4:16pm.
Law of conservation of energy
KE =PE
(m1+m2) •u²/2 =kx²/2
u=sqrt{kx²/(m1+m2)}=.....
Law of conservation og linear momentum
m1v + 0= (m1+m2)u
v=(m1+m2)u/m1 =....
Work backwards, because you cannot solve it forwards since there are two unknowns.
(a) Start with the energy transferred to the compressed spring. The block transfers energy to the spring, that's why it compresses.
Ek = Ee
mv^2 = kx^2 (halves cancel)
v = sqrt[(kx^2)/m]
v = .310 m/s
(b) Momentum is conserved. The momentum before is the momentum of the bullet. After, it is the combined mass of bullet and block.
Let u be the initial velocity of bullet
mu = (m1+m2)v
u = 1.63 m/s
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