Question

A .050 kg bullet with a velocity of 150m/sec is shot into a 3.0 kg ballistic pendulum. How high will the pendulum rise after the bullet is imbedded?

Answer 1

Given
`m_p->" Mass of the ballistic pendulum"=3.0kg`

`m_b->" Mass of the bullet "=0.05kg`

`v_p->" Initial velocity of the ballistic pendulum"=0m/s`

`v_b->" Initial velocity of the bullet before it is embedded"=150m/s`

`"Let "v_s->" velocity of the bullet pendulum system after"`
`" the bullet is embedded"`

By law of conservation of linear momentum we can write

`(m_p+m_b)xxv_s=m_pxxv_p+m_bxxv_b`

`=>v_s=(m_pxxv_p+m_bxxv_b)/(m_p+m_b)=(3xx0+0.05xx150)/3.05=2.459m/s`

Let the initial KE of the system lifts the system at maximum height h, where its KE becomes zero and it gains only PE.
By the law of conservation of mechanical energy this PE will be equal to the initial KE of the system.

So

`(m_p+m_b)xxgxxh=1/2(m_p+m_b)xxv_s^2`
`"where "g->"acceleration due to gravity"`

`h=v_s^2/(2xxg)=2.459^2/(2xx9.8)~~0.31m`