A) mAV/mB

B) mBV/mA

C) mAV/(mA+mB)

D) (mA+mB)V/mA

Makenna Booker
2022-07-18
Answered

A car (mass mA) is driving at velocity V, when it smashed into an unmoving car (mass mB), looking bumpers. Both cars move together at the same velocity. The common velocity will be by

A) mAV/mB

B) mBV/mA

C) mAV/(mA+mB)

D) (mA+mB)V/mA

A) mAV/mB

B) mBV/mA

C) mAV/(mA+mB)

D) (mA+mB)V/mA

You can still ask an expert for help

yatangije62

Answered 2022-07-19
Author has **16** answers

From linear momentum conservation-

${P}_{1}={P}_{f}\phantom{\rule{0ex}{0ex}}P=mV\phantom{\rule{0ex}{0ex}}{m}_{A}V=({m}_{A}+{m}_{B}){V}^{\prime}\phantom{\rule{0ex}{0ex}}{V}^{\prime}=\frac{{m}_{A}V}{({m}_{A}+{m}_{B})}\phantom{\rule{0ex}{0ex}}V=\text{initial velocity}\phantom{\rule{0ex}{0ex}}{V}^{\prime}=\text{final velocity}$

So, common velocity will be mAV/(mA+mB) option (C)

${P}_{1}={P}_{f}\phantom{\rule{0ex}{0ex}}P=mV\phantom{\rule{0ex}{0ex}}{m}_{A}V=({m}_{A}+{m}_{B}){V}^{\prime}\phantom{\rule{0ex}{0ex}}{V}^{\prime}=\frac{{m}_{A}V}{({m}_{A}+{m}_{B})}\phantom{\rule{0ex}{0ex}}V=\text{initial velocity}\phantom{\rule{0ex}{0ex}}{V}^{\prime}=\text{final velocity}$

So, common velocity will be mAV/(mA+mB) option (C)

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