A 5 g bullet passes through a 1 kg wooden sign that was hanging at rest on a 0.2 m long rope. After passing through, the bullet loses half of its velocity, and the sign gains just enough velocity to swing around in a complete vertical circle. Assume that the rope is strong enough so that it won’t break, and assume that it doesn’t get any shorter as it swings around. What was the initial velocity of the bullet?

Question

A 5 g bullet passes through a 1 kg wooden sign that was hanging at rest on a 0.2 m long rope.  After passing through, the bullet loses half of its velocity, and the sign gains just enough velocity to swing around in a complete vertical circle.  Assume that the rope is strong enough so that it won’t break, and assume that it doesn’t get any shorter as it swings around.  What was the initial velocity of the bullet?

Elena · Answer

m1=0.005 kg, m2= 1 kg, R=0.2 m.
v=?

Law of conservation of momentum
m1•v = m1•(v/2) +m2•u
m1•v/2= m2•u,
u= m1•v/2•m2. (1)
Law of conservation of energy
KE =PE +KE1
m2•u²/2=m2•g•2R+m2•u1²/2
u²/2= g•2R+ u1²/2 . (2)

At the highest point
m2•a=m2•g
m2•u1²/R=m2•g.
u1=sqrt(gR)=
sqrt(9.8•0.2)=1.4 m/s. (3)

Substitute (1) in (2)
(m1•v) ² /(2•m2)² 2= g•2R+ u1²/2 .
v=(2•m2/m1)sqrt(4gR+u1²)=
=(2•1/0.005)sqrt(4•9.8•0.2+1.4²)=
=400•3.13=1252 m/s.
Check my calculations.

Jessica · Answer

it seems right, but the answer is 1120m/s?

Elena · Answer

Your answer (1120 m/s)is for the case when the block ( sign) is suspended by stiff rod of length R and of negligible mass. But your given data is "the rope"