Question

( u_{1}=7.45 times 10^{5} mathrm{m} / mathrm{s}, u_{2}^{prime}=0 )
Now by taw of consewation of enelgy ( frac{1}{2} m u_{1}^{2}+frac{1}{2} m u_{2}^{2}=frac{k q^{2}}{gamma} )
Hese ( _{q}=e^{m}=1 cdot 67 times 10^{-27} mathrm{kg} )
( K=frac{1}{4 pi epsilon_{0}}=9 times 10^{9} mathrm{m} / mathrm{F} )
Puttug the values we get, ( frac{1}{2} times 1.67 times 10^{-27} timesleft(7.5 times 10^{5}right)^{2}+0 )
( =frac{9 times 10^{9} timesleft(1.6 times 10^{-19}right)^{2}}{gamma} )
on solumg this we get ( gamma=5 times 10^{-103} mathrm{m} )

# or -10 - 5x10-6 Or 5110 9x5x10-8 7-100+10=100m t oto i.e., r=4.7x10-2 m 1 2 - AN EN As here. F= so a=- mp i.e., acceleration is not constant during motion. ustration -12: A proton moves with a speed of 7.45 X 10% m/s directly towards a free proton originally at rest. Find the distance of closest approach for the two protons. Given (1/470 €)=9x10ºm/F; m, =1.67x10-21 kg and e=1.6x10-19 coulomb. lution: As here the particle at rest is free to move, wher due to

Solution