A bullet of mass 4.2 g strikes a ballistic pendulum of mass 1.0 kg. The center of mass of the pendulum rises a vertical distance of 18 cm. Assuming that the bullet remains embedded in the pendulum, calculate the bullet's initial speed.

449.020m/s

Explanation:

We have been provided with the following information:

Mass of bullet = Mb = 4.2g

Mass of pendulum = 1.0 = Md

Vertical distance = h = 18cm

Gravitational force = g = 9.8

We have kinetic energy converted to potential energy for the entire system

1/2(Mb+Md)V² = (Mb + Md)gh

We have V as the speed of system during collision

Mbv = (Mb+Md)V

We divide through by Mb

v = (Mb+Md/Mb)√2gh

4.2x10^-3 = 0.0042

18cm = 0.18m

(0.0042+1.0/0.0042)√2x9.8x0.18

= 1.0042/0.0042√3.528

= 239.095x1.878

= 449.020m/s

This is the answer to the question

Thank you!

the chart show the masses and velocities of two colliding objects that stick together after a collision.

according to the law of conservation of momentum, the momentum of the object after the collision is  

answer

1,500

kg • m/s.