Question 108: Centre of Mass and Rotational Motion

Question

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 :

Accepted Answer

Given problem a body of mass $M$ explodes into three pieces of mass ratio $1: 1: 2$
$ 	herefore $ 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.
$p_{	ext {initial }}=p_{	ext {final }}$
By law of conservation of momentum,
$
M 	imes 0=\frac{M}{4} 	imes 30 \hat{i}+\frac{M}{4} 	imes 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}
$

Answer

15 ms
$-$1

Answer

25 ms
$-$1

Answer

35 ms
$-$1

Answer

50 ms
$-$1

Centre of Mass and Rotational Motion

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 :

Choose the correct answer: