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Two particles having position verctors v...

Two particles having position verctors `vecr_(1)=(3hati+5hatj)` metres and `vecr_(2)=(-5hati-3hatj)` metres are moving with velocities `vecv_(1)=(4hati+3hatj)m//s and vecv_(2)=(alphahati+7hatj)m//s`. If they collide after 2 seconds, the value of `alpha` is

A

2

B

4

C

6

D

8

Text Solution

Verified by Experts

The correct Answer is:
D
Relative position `(Deltavecr) = vecr_(2) - vecr_(1) =-8hati-8hatj`
`|Deltavecr| =8sqrt2`
Relative velocity `vecv_("rel") = vecv_(2) -vecv_(1) = (alpha -4) hati + 4hatj`
`|vecv_("rel")|= sqrt((alpha-4)^2+16)" "....(i)`
Now, `|vecv_("rel") |=(|Deltavecr|)/(t)=(8sqrt2)/2= 4sqrt2 " "...(ii)`
By (i) and (ii) –
`(alpha -4)^2 + 16 = 16 xx 2 implies alpha - 4=4`
`implies alpha = 8`
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