Question
A 6 kg bomb at rest explodes into three equal pieces P, Q and R. If P flies with speed 30 m/s and Q with speed 40 m/s making an angle 90° with the direction of P. The angle between the direction of motion of P and R is about
1. 143° 2. 127° 3. 120°
since initially bomb is in rest so initially momentum of the bomb = 0i+0j+0k
after explosion total momentum = 30mi+40mj+mv here v is a vector
now by momentum conservation 0 = 30mi+40mj+mv so v = -30i-40j
so vector along P is i and along R is -3i-4j
so by A.B = |A||B|cosθ
we get i.[-3i-4j] = 1*5*cosθ
so -3/5 = cosθ so θ = 127
Read 1 Solution.
As no external force is present in our system.
So Momentum is conserved:
Momentum of particleP=30×2=60
Momentum of particleQ=40×2=80
Resultant of Momentum of particle P and Q will be equal to Momentum of particle R
So,
∣R∣=∣P+Q∣=(60)2+(80)2+2PQcos90
∣R∣=100
∣mv∣=100
∣v∣=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 θ
Therfore
∣Q∣2=∣P∣2+∣R∣2 +2PRcosθ
1600=900+2500+2×1500×cosθ
then
cosθ=−(0.6)
θ=127°