Question

# A particle of mass 2m is projected at an angle of 45$$^o$$ with the horizontal with a velocity of 20$$\sqrt{2}$$m/s. After 1 s of explosion, the particle breaks into two equal pieces.As a result of this one part comes to rest. The maximum height from the ground attained by the other part is $$(g = 10 m/s^2)$$

A
50 m
B
25 m
C
40 m
D
35 m

Solution

## The correct option is D 35 mGiven :  $$v =20\sqrt{2}$$  $$ms^{-1}$$ $$\therefore$$ Vertical component of velocity  $$u = vsin 45^o = 20$$ $$ms^{-1}$$Let the height attained by  $$2m$$ in 1 sec be  $$h_1$$$$\therefore$$    $$h_1 = u t - \dfrac{1}{2}gt^2$$$$h_1 =20 (1) - \dfrac{1}{2}(10)(1)^2$$            $$\implies h_1 = 15$$ mNow at point A, this breaks into two equal pieces such that one comes to rest and the other moves upward with velocity $$v'$$Applying conservation of momentum in y direction :$$(2m) u = 0 + m(v')$$             $$v ' = 2u = 2(20) =40$$   $$ms^{-1}$$Let the height above point A that the piece attains be  $$h'$$$$\therefore$$    $$0 - (v')^2 = 2 (-g) h'$$$$- 40^2 = 2 (-10) h'$$                $$\implies h' = 20$$  mThus total height attained by the piece       $$H = h_1+h' = 15+20 = 35$$  mPhysics

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