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Question: Velocity of a bullet changes from u to v after passing through a board of thickness d. Force of resi...

Velocity of a bullet changes from u to v after passing through a board of thickness d. Force of resistance is directly proportional to the velocity. Time of motion of bullet in the board is given by

A

d(u−v)uvlog⁡e(u/v)\frac { d ( u - v ) } { u v \log _ { e } ( u / v ) }

B

duvlog⁡e(u/v)\frac { d u } { v \log _ { e } ( u / v ) }

C

dvulog⁡e(u/v)\frac { d v } { u \log _ { e } ( u / v ) }

D

d(u−v)uvlog⁡e(v/u)\frac { d ( u - v ) } { u v \log _ { e } ( v / u ) }

Answer

d(u−v)uvlog⁡e(u/v)\frac { d ( u - v ) } { u v \log _ { e } ( u / v ) }

Explanation

Solution

force of resistance, f = - cv2

i.e., m dvdt=−cv2\frac { \mathrm { dv } } { \mathrm { dt } } = - \mathrm { cv } ^ { 2 } i.e., mdvv2\frac { \mathrm { mdv } } { \mathrm { v } ^ { 2 } }

i.e., ∫uvdvv2=∫0t−cmdt\int _ { u } ^ { v } \frac { d v } { v ^ { 2 } } = \int _ { 0 } ^ { t } \frac { - c } { m } d t

i.e., t = (u−vuv)mc\left( \frac { u - v } { u v } \right) \frac { m } { c } …………… (i)

Also m dvdt\frac { \mathrm { dv } } { \mathrm { dt } } ds = - kv2 ds

i.e., mvdv = - kv2 ds

i.e., mdv = - kvds

i.e. ∫uvmdvv=−c∫0dds\int _ { \mathrm { u } } ^ { \mathrm { v } } \mathrm { m } \frac { \mathrm { dv } } { \mathrm { v } } = - \mathrm { c } \int _ { 0 } ^ { \mathrm { d } } \mathrm { ds }

i.e. loge vu=−cdm\frac { \mathrm { v } } { \mathrm { u } } = - \frac { \mathrm { cd } } { \mathrm { m } } i.e. loge uv=cdm\frac { \mathrm { u } } { \mathrm { v } } = \frac { \mathrm { cd } } { \mathrm { m } }

i.e. d = mclog⁡euv\frac { \mathrm { m } } { \mathrm { c } } \log _ { \mathrm { e } } \frac { \mathrm { u } } { \mathrm { v } } ……… (ii)

From (i) and (ii) t = d(u−v)uv(log⁡euv)\frac { d ( u - v ) } { u v \left( \log _ { e } \frac { u } { v } \right) }