CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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
loader
B
25 m
loader
C
40 m
loader
D
35 m
loader

Solution

The correct option is D 35 m
Given :  $$v =20\sqrt{2}$$  $$ms^{-1}$$
 $$\therefore$$ Vertical component of velocity  $$u = vsin 45^o  =  20$$ $$ms^{-1}$$
Let the height attained by  $$2m$$ in 1 sec be  $$h_1$$
$$\therefore$$    $$h_1 = u t - \dfrac{1}{2}gt^2$$
$$h_1 =20   (1) - \dfrac{1}{2}(10)(1)^2$$            
$$\implies  h_1 = 15$$ m
Now 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  $$h'$$
$$\therefore$$    $$0 - (v')^2 = 2 (-g) h'$$
$$- 40^2 = 2 (-10)  h'$$                $$\implies   h' = 20$$  m
Thus total height attained by the piece       $$H = h_1+h' = 15+20 = 35$$  m

483203_156684_ans.png

Physics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image