Question

a bomb of mass 20 kg initially at rest explodes into two pieces of mass of 10 kg m1 and m2 if the heavier one remain at rest and other move in opposite direction to each other with same speed of 10 m/s.then find the ratio of m1 to m2​

Answer 1

a bomb of mass 20 kg initially at rest explodes into three pieces of mass of 10 kg, $m_1$ and $m_2$. if the heavier one remains at rest and other move in opposite direction to each other with same speed of 10 m/s.then find the ratio of $m_1$ to $m_2$.

applying law of conservation of linear momentum,

$Mu=m_1v_1+m_2v_2+m_3v_3$

bomb of mass, M = 20kg initially at rest.

so, Mu = 0

a/c to question, heavier one remains ar rest. so, velocity of $m_3$ 10kg, $v_3$ = 0

and other two move in opposite directions with same speed 10m/s.

so, $v_1=-v_2=10m/s$

now, 0 = $m_1$(10)+$m_2$(-10)+10kg × 0

or, 0 = $(m_1-m_2)$ × 10

or, $\frac{m_1}{m_2}$ = 1

hence, ratio of $m_1$ to $m_2$ = 1 : 1

Answer 2

yes that is the correct answer