A bullet of mass m is shot with a velocity u at an angle θ = 6 0 ∘ with the horizontal which strikes the box of mass 2m resting at the edge of a smooth surface 'OB' as shown in figure. Bulle

Question

A bullet of mass m is shot with a velocity u at an angle $θ = 6 0^{\circ}$ with the horizontal which strikes the box of mass 2m resting at the edge of a smooth surface 'OB' as shown in figure. Bullet strikes the box horizontally. After that bullet passes through the box and moves with velocity $u /4$ along horizontal direction. If bullet rebound elastically with wall AB. Then find the final position where bullet again strikes mass 2m.

TestHub · Accepted Answer

The bullet strikes horizontally (top of trajectory) with u cos 6 0 ∘ = 2 u ​ . Penetration momentum: m 2 u ​ = m 4 u ​ + 2 m v b ​ ⇒ v b ​ = 8 u ​ . Bullet reaches the wall at t 1 ​ = u /4 L ​ = u 4 L ​ ; the box is then at x = 8 u ​ t 1 ​ = 2 L ​ . After the elastic rebound the bullet moves left at 4 u ​ ; closing speed = 4 u ​ + 8 u ​ = 8 3 u ​ , gap = 2 L ​ , so t 2 ​ = 3 u 4 L ​ . Meeting point: x = 2 L ​ + 8 u ​ t 2 ​ = 2 L ​ + 6 L ​ = 3 2 L ​ . Answer: (C)

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