Skip to content
dronacharya.ai
← JEE Main 2017 · 9 Apr · Shift 1
PhysicsSingle correct+4 / -1Question 3 of 90

JEE Main 2017 · 9 Apr · Shift 1, Physics question 3

Two particles A and B of equal mass M are moving with the same speed vv as shown in the figure. They collide completely inelastically and move as a single particle C. The angle θ that the path of C makes with the X-axis is given by :

X-Y axes. A approaches the origin from below-left along a line at 30° to the negative Y-axis, B from below-right along a line at 45° to it; after the collision C moves off at angle θ to the X-axis.
Choose one option
Answer and worked solution

Official answer: A

From the figure, A moves at 30° and B at 45° to the Y-axis, both towards +Y:
v⃗A=v(12,32),v⃗B=v(−12,12).\vec v_A=v\left(\tfrac12,\tfrac{\sqrt3}2\right),\qquad\vec v_B=v\left(-\tfrac1{\sqrt2},\tfrac1{\sqrt2}\right).
Momentum is conserved, so C moves along v⃗A+v⃗B\vec v_A+\vec v_B:
tan⁡θ=32+1212−12=3+21−2.\tan\theta=\frac{\frac{\sqrt3}2+\frac1{\sqrt2}}{\frac12-\frac1{\sqrt2}}=\frac{\sqrt3+\sqrt2}{1-\sqrt2}.

Topics:Elastic and inelastic collisionsConservation of linear momentum
← Previous questionNext question →