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A 5.0g bullet moving at 100m//s strikes ...

A `5.0g` bullet moving at `100m//s` strikes a log. Assume that the bullet undergoes uniform deceleration and stops in `6.0cm`. Find (a) the time taken for the bullet to stop, (b) the impulse on the log and (c) the average force experienced by the log.

Text Solution

Verified by Experts

The correct Answer is:
A, B, C, D

(a) `v^2=u^2-2as`
`:. a=(u^2)/(2s)` `(v=0)`
`=((100)^2)/(2xx0.06)=8.3xx10^4m//s^2`
`v=u-at`
`0=u-at`
or `t=u/a=(100)/(8.3xx10^4)`
`=1.2xx10^-3s`
(b) `Impu lse=|Deltap|=mv_i`
`=(5xx10^-3)(100)`
`=0.5N-s`
(c) `Impu lse=(F_(av))t`
`:. F_(av)=(Impu lse)/(t)`
`=(0.5)/(1.2xx10^-3)`
`=417N`
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