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A shell projected from a level ground has a range R, it does not explode. At the highest point the shell explodes into two fragments having masses in the radio 1:3 with each fragment moving horizontally immediately after the explosion. If the lighter fragment falls at a distance R from the point of projection, the distance at which the other fragment falls from the point of projection is

\[\begin {array} {1 1} (a)\;2R & \quad (b)\;\frac{5R}{3} \\ (c)\;\frac{4R}{3} & \quad  (d)\;\frac{2R}{3} \end {array}\]

1 Answer

Using conservation of momentum $O = 2R \times M + 3 M (-x)$
$2 R \times M = 3 M \times x$
$x = \frac {2}{3} R$
$\begin{align*}\text{Distance from the point of projection is} & R+x \\ & = R + \frac{2R}{3} \\ & = \frac{5R}{3} \end{align*}$
answered Nov 21, 2013 by meena.p
edited Mar 23 by priyanka.c
 

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