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Question

(A) Let the mirror be in the $yz$-plane. The particle is moving in the $xy$-plane. The horizontal component of the velocity is $v_x = v \cos(60^\circ)

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(A) Let the mirror be in the $yz$-plane. The particle is moving in the $xy$-plane. The horizontal component of the velocity is $v_x = v \cos(60^\circ) = 10 	imes 0.5 = 5 \, m/s$ towards the mirror.For a plane mirror,the velocity of the image $v_i$ is related to the velocity of the object $v_o$ by the relation $v_{ix} = -v_{ox}$ (assuming the mirror is stationary).Here,the object velocity component towards the mirror is $v_{ox} = 5 \, m/s$.Therefore,the velocity of the image component towards the mirror is $v_{ix} = -5 \, m/s$ (the negative sign indicates it is moving away from the mirror in the coordinate system,but relative to the mirror,the image moves towards it).The velocity of approach between the object and its image is given by $v_{app} = v_{ox} - v_{ix} = 5 - (-5) = 10 \, m/s$.

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