Question

17.
Three interacting particles of masses 100.200 and
400gm have each a velocity 20m/sec and magnitude
along the positive direction of X-axis, Y-axis and
Z-axis. Due to the force of interaction the third
particle stops moving and the velocity of second
particle is (10j +5k) m/sec. Then the velocity of
the first particle is -
(A) 20 i + 20j + 70% (B) 101 + 101 +5k
(C) 20 i +70) +20k ()5i + 20 1 + 70​

Answer 1

Explanation:

kya patA question sahi se likho pehle

Answer 2

Answer: Options A is correct

Explanation: m1v1 + m2v2 +m3v3 = final momentum

100 * 20 = 100 * x

x =20m/s

200 * 20= 200 *10 + 100 * y

y =70m/s

400 *20 = 200 * 5 + 100 * z

z = 70m/s

20i +20j +70k

Option A