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Two particles having position vec(r_(1))=(3hat(i) + 5hat(j)) meter and vec(r_(2))=(-5hat(i)-3hat(j)) metre are moving with velocities vec(V_(1))=(4hat(i)+hat(j))m/s and vec(V_(2))=(ahat(i) + 7hat(j)) m/s. If they collide after 2 seconds, the value of a is : |
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Answer» Solution :Here `Deltavec(r)=vec(r_(2))-vec(r_(1))=-8hati-8hatj` As the particles are moving in the same DIRECTION, the relative velocity is GIVEN by the difference of two velocities. `vec(v_(R))=(a-4)hati + 4hatj` `|vec(v_(R))|=sqrt((a-4)^(2)+(4)^(2))` Also `|vec(v_(R))|=|Deltavec(r)|/t` `sqrt((a-4)^(2)+16)=(8sqrt(2))/2` `(a-4)^(2)=16` or a - 4 = 4 and a = 8 |
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