Grade 11Mechanics

Two particles are projected horizontally from the same elevated point in opposite directions with velocities 4m/s–¹ and 9m/s–¹ respectively. At the moment when their velocity vectors are mutually perpendicular, the separation between them is (g=10m/s–²)

Khyati

8 Years agoGrade 11

1 Answer

HIMANSHU SHEKHAR DAS

8 Years ago

let us solve this question using vectors.

let the first particle be thrown horizontally in +x direction, so u₁=4 i m/s

and for the other particle so velocity is u₂= -9 i m/s

let us find their velocity vectors after t seconds, by using the formula v=u+a*t.

as we know that the body’s horizontal velocity will not be changed because no force acts on horizontal only vertical velocity will be changed.

so vertical velocity gained by first body u=0, a=-10m/s², v₁= -10t m/s

and v₂ also equal to -10t m/s

so the velocity vector of v₁ and v₂ after t seconds is:

V₁= 4i – 10t j, V₂= -9i -10t j;

we know that the dot product of 2 vectors is 0 when they are perpendicular to each other.

so here the velocity vectors are perpendicular, so their dot product should also be 0

by doing dot product and equating it with 0 we get, t=0.6 sec,

so the displacement vectors at time 0.6 sec will be,

for the first particle:

S_x=4*0.6=2.4

S_y=-0.5*a*t²=-1.8 m

so S₁=2.4i-1.8j

similarly we can find S₂and it comes out to be:

S₂= -5.4i-1.8j

so the separation between them at the time is 2.4+5.4=6.8 m

(taking only x direction displacement as the y direction displacement is same.

so the answer is 6.8m

thanks

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Two particles are projected horizontally from the same elevated point in opposite directions with velocities 4m/s–¹ and 9m/s–¹ respectively. At the moment when their velocity vectors are mutually perpendicular, the separation between them is (g=10m/s–²)

Khyati

1 Answer

HIMANSHU SHEKHAR DAS

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