Question

A bullet of mass 4.4 g strikes a ballistic pendulum of mass 2.9 kg. The center of mass of the pendulum rises a vertical distance of 8.3 cm. Assuming that the bullet remains embedded in the pendulum, calculate the bullet's initial speed.

Answer 1

844.9 m/s

Step-by-step explanation:

We are given that

Mass of bullet,m=4.4 g=
`4.4* 10^(-3) kg`

1kg=1000 g

Mass of pendulum=m'=2.9 kg

Vertical distance,y=8.3 cm=
`0.083 m`

1 m=100 cm

We have to find the initial speed of bullet.

According to law of conservation of energy

Change in kinetic energy=Change in potential energy

Initial kinetic energy=0,Initial potential energy=0

`K_f=P_f`

`(1)/(2)(m+m')v^2=(m+m')gy`

Where
`g=9.8 m/s^2`

`v^2=2yg`

`v=√(2gy)=√(2* 9.8* 0.083)=1.28 m/s`

It is final velocity after collision.

According to law of conservation of momentum

`mu+m'u'=(m+m')v`

Initial speed of pendulum,u'=0

`4.4* 10^(-3)u=(4.4* 10^(-3)+2.9)(1.28)`

`u=((4.4* 10^(-3)+2.9)(1.28))/(4.4* 10^(-3))`

u=844.9 m/s

Hence, the initial speed of bullet=844.9 m/s