91Ó°ÊÓ

(In a totally inelastic collision, the two objects stick together after the collision. We have two objects with the same mass, m, and the same initial velocity, v. Let's denote the angle between their initial velocities as θ. After the collision, they move together at half their initial velocity, which is (v/2).)

(We can use the conservation of momentum to solve the problem, which states that the total momentum before the collision is equal to the total momentum after the collision. Let \(P_x\) and \(P_y\) be the components of the total momentum in the horizontal and vertical directions before the collision.)

(We can compute the component of the total momentum in the horizontal direction, before the collision, as \(P_x = mv\cos\theta + mv\cos{(180^\circ-\theta)}\). Similarly, the component of total momentum in the vertical direction, before the collision, is \(P_y = mv\sin\theta - mv\sin{(180^\circ-\theta)}\). After the collision, the total momentum in both horizontal and vertical directions is \(P_x' = 2m(v/2)\cos\theta'\) and \(P_y' = 2m(v/2)\sin\theta'\). Here, θ' is the angle between the final velocity and the horizontal axis.)

(Using the conservation of momentum, set the total momentum before the collision equal to the total momentum after the collision in both horizontal and vertical directions: \[P_x = P_x'\] \[mv\cos\theta + mv\cos{(180^\circ-\theta)} = 2m(v/2)\cos\theta'\] \[P_y = P_y'\] \[mv\sin\theta - mv\sin{(180^\circ-\theta)} = 2m(v/2)\sin\theta'\])

(We are looking for the angle θ between the initial velocities. Divide both equations by m and multiply them: \[(v\cos\theta+v\cos{(180^\circ-\theta)})(v\sin\theta-v\sin(180^\circ-\theta))=(v\cos\theta')^2+(v\sin\theta')^2\] Now, use the trigonometric identities: \[(2v^2\cos\theta\cos(180^\circ-\theta))(2v^2\sin\theta\sin(180^\circ-\theta))=(v^2\cos^2\theta'+v^2\sin^2\theta')\] \[4v^4(\sin\theta\cos\theta\cos(180^\circ-\theta))^2=(v^2)^2\] Simplify and find the relation between θ and θ': \[(2\sin\theta\cos\theta\cos(180^\circ-\theta))^2=1\] Since \(\sin(180^\circ-\theta)=-\sin\theta\) and \(\cos(180^\circ-\theta)=-\cos\theta\): \[(2\sin\theta\cos\theta\cos^2\theta-2\sin^2\theta\cos\theta)^2=1\] Now, use the identity \(\sin^2\theta+\cos^2\theta=1\) and solve for θ: \[\sin^2\theta\cos^2\theta=\frac{1}{4}\] \[\sin2\theta=\pm\frac{1}{2}\] Hence, the only possible angle between the initial velocities is: \[\theta = 120^\circ\] Therefore, the correct answer is (A).)