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A bullet of mass 8 g is fired into a 1.2 kg ballistic pendulum (a block of wood acting as the pendulum bob of a thin metal ar
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Answer #1

using energy conservation between initial and final position of pendulum after collision

KEi + PEi = KEf + PEf

PEi = 0

KEf = 0, at highest point velocity will be zero.

KEi = PEf

0.5*m1*V1^2 = m1*g*\triangleh

V1 = final speed of block just after collision = sqrt (2*g*\triangleh)

V1 = Vblock = sqrt (2*9.81*0.092) = 1.344 m/sec

Part B.

Momentum of bullet-block system after collision will be:

Pf = m1*V1 + m2*V2

m2 = mass of bullet = 8 gm = 8*10^-3 kg

V2 = speed of bullet just after it exits the block = 180 m/sec

Pf = 1.2*1.344 + 8*10^-3*180 = 3.0528 kg-m/sec

Pf = 3.05 kg-m/sec = momentum of bullet-block system after the collision

Part C.

Now using momentum conservation before and after collision

Since this completely inelastic collision, So

Pi = Pf

m1*u1 + m2*u2 = Pf

u1 = initial speed of block = 0 m/sec

u2 = Initial speed of bullet = ?

So,

u2 = (Pf - m1*u1)/m2

u2 = (3.0528 - 1.2*0)/(8*10^-3)

u2 = 381.6 m/sec = Vbullet

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