Two particles having position vectors r→1=(3i^+5j^) meter and r→2=(−5i^−3j^) metres are moving with velocities υ→1=(4i^+3j^) and υ→2=(ai^+7j^) m/s. If they collide after 2 seconds, the value of a is:
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a
2
b
4
c
6
d
8
answer is D.
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Detailed Solution
Here, r→1=(3i^+5j^)m;r→2=(−5i^−3j^)m v→1=(4i^+3j^)ms−1;v→2=(ai^+7j^)ms−1 ∴ Δr→=r→1−r→2=(3i^+5j^)−(−5i^−3j^)=(8i^+8j^)m and Δv→=v→2−v→1=(a−4)i^+4j^ But Δv→=Δr→Δt ∴ (a−4)i^+4j^=(8i^+8j^)2 On solving, we get a=8