Question

A bullet of mass 4.3 g strikes a ballistic pendulum of mass 1.9 kg. The center of mass of the pendulum rises a vertical distance of 12 cm. Assuming that the bullet remains embedded in the pendulum, calculate the bullet's initial speed.

Answer 1

since the bullet remains embedded in the pendulum, the collision is perfectly inelastic.

Before collision
Initial speed of bullet = u m/s;

mass of bullet = 4.3*10^-3 kg
Initial speed of pendulum = 0;

mass of pendulum = 1.9 kg
Hence,
initial momentum of bullet-pendulum system p(i)
= 4.3*10^-3 kg*u + 1.9 kg*0
= (4.3*10^-3u) kgm/s

Just after collision

Final speed of bullet = final speed of pendulum = v m/s
Total mass of bullet and pendulum = 1.9 kg + 4.3*10^-3kg = 1.9043 kg
Hence,
final momentum of bullet-pendulum system p(f)
= 1.9043*v
= (1.9043v) kgm/s

The string attached to the pendulum will not have the time to react and hence there is no external force on the bullet-pendulum system.
Thus, apply conservation of linear momentum.
p(f) = p(i)
=> 1.9043v = 4.3*10^-3u
=> u = 442.82 v ------------- (1)

Now apply principle of conservation of mechanical energy.
KEi + PEi = KEf + PEf
=> (0.5)(1.9043)(v^2) + 0 = 0 + (1.9043)(9.81)(0.12)
=> v = sqrt(2.35) = 1.53 m/s

Substitute this value into equation (1),
u = 442.82*1.53 = 677.51 m/s