A particle of mass 0.01 kg moving horizontally with velocity 20 m/s strikes a stationary wedge of mass 0.05 kg near the apex of the wedge and comes to rest immediately after the collision. The wedge i

Question

A particle of mass 0.01 kg moving horizontally with velocity 20 m/s strikes a stationary wedge of mass 0.05 kg near the apex of the wedge and comes to rest immediately after the collision. The wedge is free to move on the smooth floor. Second collision of particle with wedge occurs at B. If the length AB of wedge is $\frac{10 K}{5}$ meter, then find K . Take $θ = cot^{- 1} (2)$ . Neglect the toppling of wedge.

TestHub · Accepted Answer

First collision (momentum): 0.01 × 20 = 0.05 V ⇒ V = 4 m/s (wedge). The particle rests at the apex and falls vertically while the face recedes at 4 m/s. The face reaches its vertical line when 4 t = d cot θ = 2 d , with d = 2 1 ​ ( 10 ) t 2 = 5 t 2 . 4 t = 10 t 2 ⇒ t = 0.4 s , d = 5 ( 0.4 ) 2 = 0.8 m = h (hits at B). A B = s i n θ h ​ = 0.8 5 ​ = 5 ​ 10 K ​ ⇒ K = 0.4 . Answer: 0.40.

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