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A particle A is projected with an initia...

A particle A is projected with an initial velocity of `60 m//s` at an angle `30^@` to the horizontal. At the same time a second particle B is projected in opposite direction with initial speed of `50 m//s` from a point at a distance of 100 m from A. If the particles collide in air, find (a)the angle of projection `alpha` of particle B, (b) time when the collision takes place and (c) the distance of P from A, where collision occurs. `(g= 10 m//s^2)`

Text Solution

Verified by Experts

(a) Taking x andy directins as shown in figure
Here , `vec(a)_(A) = - ghat(j)`
`vec(a)_(B) = ghat(j)`
`u_(Ax) = 60 cos 30^(@) = 30 sqrt(3)m/s: u_(Ay) = 60 sin 30^(@) = 30m//s`
`u_(Bx) = - 50 cos alpha and u_(By) = 50 sin alpha`

Relative acceleration between the two is zero as `vec(a_(A)) = vec(a_(B))`. Hence, the relative motion between the two is uniform. Condition of collision is that us should be along AB. this is possible only when `u_(Ay) = u_(By)` i.e., component of relative velocity along y-axis should be zero.
(or) ` 30 = 30 sin alpha " " therefore" alpha = sin^(-1) (3//5)`
(b) Now `|vec(u_(AB)) | =u_(ax) - u_(bx)`
` = (30 sqrt(3)+ 50 cos alpha)m//s = (30 sqrt(3) + 50 xx (4)/(5)) m//s = (30sqrt(3)+40)m//s`
Therefore , time of collision is `t = (AB)/(|vec(u_(AB))|)= (100)/(30sqrt(3)+40) = 1.09s`
(c) Distance of point P form A where collision taken place is .
`s= sqrt((u_(Ax)t)^(2) + (u_(Ay) t-(1)/(2)"gt"^(2))^(2)) = sqrt((30sqrt(3) xx 1.09)^(2) + (30 xx1.09 - (1)/(2)xx10xx1.09xx1.09)^(2)) therefore " " s = 62.64 m`
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