Question

A 3 kg bomb explodes into 3 equal pieces A, B and C. A flies with a speed of 40 m/s and B with a speed off 30 m/s making an angle off ${90}^0$ with the direction of A. The angle made by the direction of C with that of A is

• A. $co{s}^{-1}\left(\frac{4}{5}\right)$
• B. $\pi -co{s}^{-1}\left(\frac{4}{5}\right)$
• C. $co{s}^{-1}\left(\frac{3}{5}\right)$
• D. $\pi -co{s}^{-1}\left(\frac{3}{5}\right)$

Answer 1

The correct option is B $\pi -co{s}^{-1}\left(\frac{4}{5}\right)$

Linear momentum is conserved before and after explosion.
The bomb is at rest before it explodes. So, the initial momentum is zero.
Let's say the piece A moves along positive x-axis, and the piece B along positive y-axis. And let's assume that the piece C is moving at an angle of $\theta$ with the negative x-axis at a speed of v.
Conserving along the x-axis.
$0=1\left(40\right)-1\left(vcos\theta \right)$
$\Rightarrow vcos\theta =40$
Conserving along the y-axis
$0=1\left(30\right)-1\left(vsin\theta \right)$
$\Rightarrow vsin\theta =30$
From these two equations,
$tan\theta =3/4$
$\Rightarrow cos\theta =4/5$
So,piece C makes an angle of $\pi -co{s}^{-1}\frac{4}{5}$ with piece A.
Option B.