Two particles having position vectors r ₁=(3 i +5 j ) metres and r ₂=(-5 i -3 j ) metres are moving with velocities v ₁=(4 i +3 j ) m / s and v ₂=(a i +7 j ) m / s . If they collide after 2 ~s , then the value of a is :

Step-by-step solution

r ₁=(3 i +5 j ) m r ₂=(-5 i -3 j ) m v ₁=(4 i +3 j ) m / s and v ₂=(a i +7 j ) m / s Time taken in collision, t= r v ...(i) where, r= resultant position vector = r ₁- r ₂=8 i +8 j v= resultant velocity v ₂- v ₁=(a-4) i +4 j From Eqs. (i), we get 2= 8 i +8 j (a-4) j +3 j [ t=2 ~s ](a-4) i +4 j =4 i +4 j a-4=4a=8