Conservation of Momentum in 2D Collision

Steps:

• Resolve initial momentum: only mass M has velocity $\vec{u}$ (assume along x-axis), so total $\vec{p_i} = M \vec{u}$.
• From diagram (view from above), post-collision: mass M moves at 3 m/s, 37° to original direction; mass 4M moves at 4 m/s, 53° opposite.
• Resolve post-collision momenta: x-component: $M(3\cos37^\circ) + 4M(-4\cos53^\circ) = M u$; y-component: .
• Solve x-equation: $3\cos37^\circ - 16\cos53^\circ = u$, yielding $u = 5$ m/s (y balances automatically for elastic or given paths).

Why D is correct:

• Momentum conservation in x-direction directly gives $u = 5$ m/s, matching the vector sum of final velocities.

Why the others are wrong:

• A: Underestimates by ignoring y-components' x-projections.
• B: Matches one final speed but neglects angle contributions.
• C: Equals sum of speeds but violates vector momentum balance.

Final answer: D