A bullet of mass 10g travelling horizontally at `200m//s` strikes and embeds in a pendulum bob of mass 2.0 kg.
(a) How much mechanical energy is dissipated in the collision?
(b) Assuming that `C_v` for the bob plus bullet is 3R, calculate the temperature increase of the system due to the collision. Take the molecular mass of the system to be `200g//mol.`

Answer

A, B, C

Explanation

(a) From conservation of linear momentum,
`0.01xx200=(2+0.01)v`
`:.` `v~~1m/s`
Mechanical energy dissipated in collision,
`=K_i-K_f`
`=1/2xx(0.01)(200)^2-1/2(2.01)(1)^2`
`~~199J`
(b) `n=m/M=(2000+10)/(200)=10.05`
Using,
`Q=nC_VDeltaT`
`:.` `DeltaT=(Q)/(nC_V)=(199)/(10.05xx3xx8.31)`
`=0.80^@C`