Question

A body of mass m at rest explodes into three pieces, two of which of mass m/4 each are thrown off in perpendicular directions with velocity of 3 ms${}^{-1}$ and 4 ms${}^{-1}$ respectively. The third piece will be thrown off with a velocity of :

• A. 3 ms${}^{-1}$
• B. 2.5 ms${}^{-1}$
• C. 2 ms${}^{-1}$
• D. 1.5 ms${}^{-1}$

Answer 1

The correct option is B 2.5 ms${}^{-1}$

Since mass $m$ is divided in three pieces two of which has mass

$\frac{m}{4}$ each thus the ratio of masses is $1:1:2$ and the masses are

$m_1=m_2=\frac{1}{4}$ and $m_3=\frac{2}{4}=\frac{1}{2}$

Now, the two pieces of equal mass fly off perpendicular to other hence,

$(m_1{v}_1{)}^2+(m_2{v}_2{)}^2=(m_3{v}_3{)}^2$

$(\frac{1}{2}{v}_3{)}^2=(\frac{1}{4}3{)}^2+(\frac{1}{4}4{)}^2$

$(\frac{1}{2}{v}_3{)}^2=\frac{1}{16}((3{)}^2+(4{)}^2)$

$(\frac{1}{2}{v}_3{)}^2=\frac{25}{16}$

$\frac{1}{2}{v}_3=\sqrt{\frac{25}{16}}$

$v_3=\frac{5}{4}\times 2$

$v_3=2.5m{s}^{-1}$