Question

A bullet of mass 2.9 g strikes a ballistic pendulum of mass 4.7 kg. The center of mass of the pendulum rises a vertical distance of 15 cm. Assuming that the bullet remains embedded in the pendulum, calculate the bullet's initial speed.

Answer 1

Given mass of bullet (m) = 2.9×10^-3 Kg = 0.0029 Kg

mass of pendulum (M) = 4.7 kg.

Let speed of combined mass bullet(m) and pendulum (M) is = V

And given hight the center of mass rise(H) = 0.15 m

Using conserved of energy,

Kinetic energy = potential energy

½(m+M)V² = (m+M)*g*H

V = √{2g*H}

Putting values :

V = √{2×9.8×0.15} = 1.71 m/s

Now using conservation of momentum for bullet and pendulum system :

if v be the initial speed of bullet then :

initial momentum = final momentum

mv + M*0 = (m+M)V

v = (m+M)V/m = (0.0029 +4.7)*1.71 / 0.0029 = 2780.62 m/s

Bullet's initial speed is = 2780.62 m/s = 2.78 ×10³ m/s

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