Question

A particle of mass $2m$ is projected at an angle of $45^\circ $ with horizontal with a velocity of $20\sqrt{2} \, m \, s^{- 1}$ . After $1 \, s$ explosion takes place and the particle is broken into two equal pieces. As a result of the explosion, one part comes to rest. The maximum height attained by the other part is (Take $g=10 \, m \, s^{- 2}$ )

• A. $35m$
• B. $40 \, m$
• C. $15 \, m$
• D. $20 \, m$

Answer 1

Answer: (A)

Applying conservation of linear momentum at the time of the collision, or
at $t=1 s$
$m \overrightarrow{ v }+ m (0)=2 m (20 \hat{ i }+10 \hat{ j })$
$\therefore \quad \overrightarrow{ v }=40 \hat{ i }+20 \hat{ j }$
At 1 sec, masses will be at height:
$h _{1}= u _{ y } t +\frac{1}{2} v _{ y } t ^{2}=(20)(1)+\frac{1}{2}(-10)(1)^{2}=15 m$
After explosion other mass will further rise to a height:
$h _{2}=\frac{ u _{ y }^{2}}{2 g }=\frac{(20)^{2}}{2 \times 10}=20 m : u _{ y }=20 m / s$ just after the collision.
$\therefore $ Total height $h=h_{1}+h_{2}=35 m$