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JEE Main 2014 · 12 Apr · Shift 1, Physics question 3

Three masses m, 2m and 3m are moving in xx-yy plane with speed 3u, 2u, and u respectively as shown in figure. The three masses collide at the same point at P and stick together. The velocity of resulting mass will be :

Axes x, y with P at the origin. Mass m moves along +x towards P with speed 3u; mass 2m comes in towards P from the upper right with speed 2u, along a line at 60° above the +x axis; mass 3m comes in towards P from the lower right with speed u, along a line at 60° below the +x axis.
Choose one option
Answer and worked solution

Official answer: D

Momenta of the three masses (all directed into P):

  • mm: 3mu i^3mu\,\hat i.
  • 2m2m: 4mu4mu along (−cos⁡60∘,−sin⁡60∘)(-\cos60^\circ,-\sin60^\circ), i.e. (−2mu, −23mu)(-2mu,\,-2\sqrt3mu).
  • 3m3m: 3mu3mu along (−cos⁡60∘,+sin⁡60∘)(-\cos60^\circ,+\sin60^\circ), i.e. (−32mu, 323mu)(-\tfrac32mu,\,\tfrac32\sqrt3mu).

Total: (−12mu, −32mu)\left(-\tfrac12mu,\,-\tfrac{\sqrt3}2mu\right). Dividing by 6m6m:
v⃗=u12(−i^−3j^).\vec v=\frac u{12}\left(-\hat i-\sqrt3\hat j\right).

Topics:Elastic and inelastic collisionsConservation of linear momentum
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