A projectile of mass m is fired with a velocity v0 at an angle - with the horizontal- At the highest position in its flight- it explodes into two fragments of masses m1 - m3 and m2 - 2m3- The fragment of mass m1 falls vertically with zero initial speed-a Find the initial velocity of the fragment m2-b Find the distance d at which the fragment m2 will land with respect to m1-

H=v_0^2sin^2θ2g Initial momentum at A =mv_0cosθ ∴ mv_0cosθ=2m3v ∴ v=32v_0cosθ Let m_2 takes t time to land after explosion H=12gt^2 ...

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Byju's Answer

Standard XII

Physics

Com in Perfectly Inelastic Collision

A projectile ...

Question

A projectile of mass $m$ is fired with a velocity $v_{0}$ at an angle $θ$ with the horizontal. At the highest position in its flight, it explodes into two fragments of masses $m_{1} = \frac{m}{3}$ and $m_{2} = \frac{2 m}{3}$ . The fragment of mass $m_{1}$ falls vertically with zero initial speed.
(a) Find the initial velocity of the fragment $m_{2}$ .
(b) Find the distance $d$ at which the fragment $m_{2}$ will land with respect to $m_{1}$ .

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Solution

$H = \frac{v_{0}^{2} {sin}^{2} θ}{2 g}$
Initial momentum at A
$= m v_{0} cos θ ∴ m v_{0} cos θ = \frac{2 m}{3} v ∴ v = \frac{3}{2} v_{0} cos θ$
Let $m_{2}$ takes t time to land after explosion
$H = \frac{1}{2} g t^{2} \Rightarrow \frac{v_{0}^{2} {sin}^{2} θ}{2 g} = \frac{g t^{2}}{2} \Rightarrow t = \frac{v_{0} sin θ}{g} ∴ d = \frac{3}{2} v_{0} cos θ t = \frac{3}{2} v_{0} cos θ \frac{v_{0} sin θ}{g} = \frac{3}{4} \frac{v_{0} sin 2 θ}{g}$

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H=v20sin2θ2gInitial momentum at A=mv0cosθ∴mv0cosθ=2m3v∴v=32v0cosθLet m2 takes t time to land after explosionH=12gt2⇒v20sin2θ2g=gt22⇒t=v0sinθg∴d=32v0cosθt=32v0cosθv0sinθg=34v0sin2θg