Question

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 arm attached to a low friction pivot point) that is initially at rest. The bullet exits the block of wood with a speed of 180 m/s and the wooden block swings from its lowest point to a maximum height of 9.2 cm above that location. Determine the velocity of the bullet before it hit the ballistic pendulum. a. First, use conservation of energy to determine the velocity of the wooden block immediately after the collision, just as it began to swing upward. Vblock = m/s b. Determine the total momentum of the bullet-block system immediately after the collision. Pfinal = 0 . kg.m/s c. Now, determine the initial velocity of the bullet using conservation of momentum. Vbullet = m/s

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*$\triangle$h

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

V1 = V_block = 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 = V_bullet

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