JEE MAIN - Physics (2022 - 29th June Morning Shift - No. 2)
A body of mass M at rest explodes into three pieces, in the ratio of masses 1 : 1 : 2. Two smaller pieces fly off perpendicular to each other with velocities of 30 ms$$-$$1 and 40 ms$$-$$1 respectively. The velocity of the third piece will be :
15 ms$$-$$1
25 ms$$-$$1
35 ms$$-$$1
50 ms$$-$$1
Explanation
Given problem a body of mass $M$ explodes into three pieces of mass ratio $1: 1: 2$
$$ \therefore $$ Mass of fragments will be $x, x, 2 x$
Hence, $M=x+x+2 x=4 x \mathrm{~kg}$
As in the process of explosion no external forces are involved and explosion occurs due to internal forces. Thus, momentum of the system will be conserved.
$p_{\text {initial }}=p_{\text {final }}$
By law of conservation of momentum,
$$ M \times 0=\frac{M}{4} \times 30 \hat{i}+\frac{M}{4} \times 40 \hat{j}+\frac{2 M}{4} \vec{v} $$
Where $\vec{v}$ is the velocity of the third fragment.
$$ \frac{M \vec{v}}{2} =-\frac{M}{4}(30 \hat{i}+40 \hat{j})$$
$$ \Rightarrow $$ $$\vec{v} =-15 \hat{i}-20 \hat{j}$$
Thus, magnitude of $\vec{v}=|\vec{v}|=\sqrt{v_x^2+v_y^2}$ $=\sqrt{(-15)^2+(-20)^2}$
$$ |\vec{v}|=\sqrt{625}=25 \mathrm{~m} / \mathrm{s} $$
$$ \therefore $$ Mass of fragments will be $x, x, 2 x$
Hence, $M=x+x+2 x=4 x \mathrm{~kg}$
As in the process of explosion no external forces are involved and explosion occurs due to internal forces. Thus, momentum of the system will be conserved.
$p_{\text {initial }}=p_{\text {final }}$
By law of conservation of momentum,
$$ M \times 0=\frac{M}{4} \times 30 \hat{i}+\frac{M}{4} \times 40 \hat{j}+\frac{2 M}{4} \vec{v} $$
Where $\vec{v}$ is the velocity of the third fragment.
$$ \frac{M \vec{v}}{2} =-\frac{M}{4}(30 \hat{i}+40 \hat{j})$$
$$ \Rightarrow $$ $$\vec{v} =-15 \hat{i}-20 \hat{j}$$
Thus, magnitude of $\vec{v}=|\vec{v}|=\sqrt{v_x^2+v_y^2}$ $=\sqrt{(-15)^2+(-20)^2}$
$$ |\vec{v}|=\sqrt{625}=25 \mathrm{~m} / \mathrm{s} $$
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