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 $(g=10m/s^2)$

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

Answer 1

The correct option is D 35 m

Given : $v=20\sqrt{2}$ $m{s}^{-1}$

$\therefore$ Vertical component of velocity $u=vsin{45}^o=20$ $m{s}^{-1}$

Let the height attained by $2m$ in 1 sec be $h_1$

$\therefore$ $h_1=ut-\frac{1}{2}g{t}^2$

$h_1=20\left(1\right)-\frac{1}{2}\left(10\right)(1{)}^2$

$⟹h_1=15$ m

Now at point A, this breaks into two equal pieces such that one comes to rest and the other moves upward with velocity $v^{\prime }$

Applying conservation of momentum in y direction :

$\left(2m\right)u=0+m\left(v^{\prime }\right)$

$v^{\prime }=2u=2\left(20\right)=40$ $m{s}^{-1}$

Let the height above point A that the piece attains be $h^{\prime }$

$\therefore$ $0-(v^{\prime }{)}^2=2(-g)h^{\prime }$

$-{40}^2=2(-10)h^{\prime }$ $⟹h^{\prime }=20$ m

Thus total height attained by the piece $H=h_1+h^{\prime }=15+20=35$ m