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Question : 32 of 160
Marks:
+1,
-0
Solution:
r1‌=(3+5)mr2‌=(−5−3)mv1‌=(4+3)m∕ s‌ and ‌v2‌=(a+7)m∕ sTime taken in collision,
t=‌ . . . (i)
where,
r= resultant position vector
‌=r1−r2‌=8+8v= resultant velocity
v2−v1=(a−4)+4From Eqs. (i), we get
2‌=‌‌‌‌‌∵t=2s](a−4)+4‌=4+4a−4‌=4a‌=8
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