A bullet of mass m= 2g is fired at a block of wood that is mounted as a ballistic pendulum and has a mass of?

Question

M= 1 kg. After the bullet hits the stationary block and is embedded in it, the maximum rise of the pendulum is observed to be h =0.1 m.

a. Find the initial velocity, V, of block and bullet after the collision.

b. Find the initial velocity, v, of the bullet.

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Answer 1

m*v=(m+M)V (conservation of momentum)
1/2mv^2=(m+M)g*h (accepting that none of the bullet´s energy is transformed by friction into heat and not potential energy)
v=sqrt(( 2* (m+M)/m *g *h))=31.336m/s
.002*31.34 =1.002 *V so V= 0.061 m/s

Answer 2

1) m*v = (m+M)*V→ v = (m+M)*V/m
2) ½(m+M)V² = (m+M)gh → V = √(2gh) = √(2*9.8*.1) =

a. V = 1.4 m/s

From eq 1), v = (.002+1)*1.4/.002 =

b. v = 701.4 m/s