Question

A terriost throws a bomb of mass 6kg with 20m/s making an angle 60°with the horizontal if the bomb brust at the top of projectile motion into three pices 1kg 2kg and 3kg.The 3kg part falls freely under gravity.The 2kg part return back on same part and hit the terrorist then find the position where the 1kg part will land from the terrorist​

Answer 1

Answer:

As no external force is present in our system.

As no external force is present in our system.So Momentum is conserved:

As no external force is present in our system.So Momentum is conserved:Momentum of particle P=30×2=60

P=30×2=60Momentum of particleQ=40×2=80

P=30×2=60Momentum of particleQ=40×2=80Resultant of Momentum of particle P and Q will be equal to Momentum of particle R

P=30×2=60Momentum of particleQ=40×2=80Resultant of Momentum of particle P and Q will be equal to Momentum of particle RSo,

$|R| \: =∣P+Q∣= \: \sqrt{(60h) {}^{2 \: } + \: (80) {}^{2} 2pq \: \cos90 } \:$

$∣R∣=100</em></strong></p><p></p><p><strong><em>[tex]∣R∣=100$

$|mv| \: = \: 100$

$|v| \: = \: 50/s$

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

So, again the resultant of P and R will be equal to QLet the angle between P and R be θ

So, again the resultant of P and R will be equal to QLet the angle between P and R be θTherefore

$|Q| {}^{2} \: = \: |P| {}^{2} \: + \: |R| {}^{2} \: +2P \: Rcos \: θ$

1600=900+2500+2×1500×cosθ

1600=900+2500+2×1500×cosθthen

1600=900+2500+2×1500×cosθ

then

cosθ=−(0.6)

$θ=127°$