Question

At given instant two particles are position vectors 4 icap - 4 j cap + 7 k cap metre and two icap + 2 j cap + 5 k cap metre respectively velocity of first particle v 0.4 i cap metre per second then velocity of second particle in metre per second if they collide after 10 seconds is

Answer 1

position of first particle , x1 = 4i 4j + 7k

position of 2nd particle, x2 = 2i + 2j + 5k

velocity of first particle, v1 = 0.4i

we have to find velocity of 2nd particle if they collides after 10sec.

let velocity of 2nd particle is v2.

so, relative velocity of particles × time taken = relative position of particles

or, (v1 - v2) × t = (x1 - x2)

or, {(0.4i) - v2 } × 10 = {(4i - 4j + 7k) - (2i + 2j + 5k)}

or, (0.4i - v2) × 10 = (2i -6j + 2k)

or, 0.4i - v2 = 0.2i - 0.6j + 0.2k

or, v2 = 0.4i - (0.2i - 0.6j + 0.2k)

or, v2 = 0.2i + 1j + 0.2k

hence, velocity of particle is (0.2i + 1j + 0.2k)