Two particles A and B are moving with constant velocities v1 and v2- At t - 0- v1 makes an angle -1 with the line joining A and B and v2 makes an angle -2 with the line joining A and B- Find their velocity of approach-

Velocity of particle A along line AB is #160; #160; V_A = v_1cosθ_1 #160; #160; #160; (towards right)Velocity of particle B along line AB is ...

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Standard XII

Physics

Relative Position

Two particles...

Question

Two particles A and B are moving with constant velocities $v_{1}$ and $v_{2}$ . At t = 0, $v_{1}$ makes an angle $θ_{1}$ with the line joining A and B and $v_{2}$ makes an angle $θ_{2}$ with the line joining A and B. Find their velocity of approach.

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Solution

Velocity of particle A along line AB is $V_{A} = v_{1} cos θ_{1}$ (towards right)
Velocity of particle B along line AB is $V_{B} = v_{2} cos θ_{2}$ (towards left)
Velocity of approach $V_{a p p r o a c h} = V_{A} + V_{B} = v_{1} cos θ_{1} + v_{2} cos θ_{2}$

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Two particles A and B are moving with constant velocities v1 and v2. At t = 0, v1 makes an angle θ1 with the line joining A and B and v2 makes an angle θ2 with the line joining A and B. Find their velocity of approach.

Velocity of particle A along line AB is VA=v1cosθ1 (towards right)Velocity of particle B along line AB is VB=v2cosθ2 (towards left)Velocity of approach Vapproach=VA+VB=v1cosθ1+v2cosθ2

Two particles A and B are moving with constant velocities $v_{1}$ and $v_{2}$ . At t = 0, $v_{1}$ makes an angle $θ_{1}$ with the line joining A and B and $v_{2}$ makes an angle $θ_{2}$ with the line joining A and B. Find their velocity of approach.

Velocity of particle A along line AB is $V_{A} = v_{1} cos θ_{1}$ (towards right)
Velocity of particle B along line AB is $V_{B} = v_{2} cos θ_{2}$ (towards left)
Velocity of approach $V_{a p p r o a c h} = V_{A} + V_{B} = v_{1} cos θ_{1} + v_{2} cos θ_{2}$