Question

A projectile of mass 4 kg is fired with a velocity 4 m/s at an angle ${45}^0$ with the horizontal. At the highest position in its flight, it explodes into two fragments of masses 1 kg and 3 kg. The heavier fragment falls vertically with zero initial speed. the horizontal distance between the fragments when they touch the ground is

Answer 1

intial velocity $u=20m/s$

angle of projection$={60}^0$

velocity at heighest point $=ucos\theta =\frac{u}{2}=10m/s$

since the system explodes due to internal forces momentum remains constant

$P_i=P_f$

$P_i=5\times 10=50\to 1$

$P_f=1\times V_1+4V_2\to 2$

$1=2⟹50=V_1+4V_2\to 3$

given kinetic energy doubled after explosion

${(K.E)}_f={2(K.E)}_i⟹{(K.E)}_i=\frac{m{v}^2}{2}=\frac{5\times {10}^2}{2}=\frac{500}{2}=250$

${(K.E)}_f=500⟹\frac{{V_1}^2+4{V_2}^2}{2}=500\to 4$

from solving $3$ and $4$ we get $V_1$ and $V_2$ as -10 & 15

time taken by the particle to reach the highest point $=\frac{T}{2}=\frac{usin\theta }{g}$

distance covered by both particles be ${}^{\prime }{d}^{\prime }$

let $g=10\frac{m}{s^2}$

$d_1=V_{1}t=\frac{10usin\theta }{g}=usin\theta$

$d_2=V_{2}t=\frac{15usin\theta }{g}=1.5usin\theta$

total distance between the particles when they reach the ground $d=d_1+d_2=2.5usin\theta$

$d=25\sqrt{3}=43.25m$