Question

A $6kg$ bomb at rest explodes into three equal pieces P, Q and R. If P lies with speed $30m/s$ and Q with speed $40m/s$ making an angle ${90}^o$ with the direction of P. The angle between the direction of motion of P and R is about

• A. ${143}^o$
• B. ${127}^o$
• C. ${120}^o$
• D. ${150}^o$

Answer 1

Answer: (B)

${127}^o$

As no external force is present in our system.

So Momentum is conserved:

Momentum of particle$P=30\times 2=60$

Momentum of particle$Q=40\times 2=80$

Resultant of Momentum of particle P and Q will be equal to Momentum of particle R

So,

$\left|R\right|$=$|P+Q|$=$\sqrt{(60{)}^2+(80{)}^2+2PQ\cos 90}$

$\left|R\right|=100$

$\left|mv\right|=100$

$\left|v\right|=50m/s$

To find angle between P and R

So, again the resultant of P and R will be equal to Q

Let the angle between P and R be $\theta$

Therfore

${\left|Q\right|}^2$=${\left|P\right|}^2$+${\left|R\right|}^2$ +$2PR\cos \theta$

$1600=900+2500+2\times 1500\times \cos \theta$

then

$\cos \theta$=$-\left(0.6\right)$

$\theta =127°$