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 tyre direction of motion of P and R is about
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As no external force is present in our system.
So Momentum is conserved:
Momentum of particle P = 30 x 2 = 60 (i)
Momentum of particle Q = 40 x 2 = 80 (j)
Resultant of Momentum of particle P and Q will be equal to Momentum of particle R
So |R| = | P + Q | = Root (60)² + (80)² + 2 PQ Cos 90
| R | = 100
| m v| = 100
|v| = 50 m/s
To Find angle b/w P & R
So, Again the resultant of P and R will be equal to Q:
Let the angle b/w P and R be Theta
Therefore
|Q|² = |P|²+|R|² + 2 PR Cos Theta
1600 = 900 + 2500 + 2 x 1500 x Cos Theta
then
Cos theta = -(.6)
Theta = 127 Degree
Thus, the answer is 127 degrees.
So Momentum is conserved:
Momentum of particle P = 30 x 2 = 60 (i)
Momentum of particle Q = 40 x 2 = 80 (j)
Resultant of Momentum of particle P and Q will be equal to Momentum of particle R
So |R| = | P + Q | = Root (60)² + (80)² + 2 PQ Cos 90
| R | = 100
| m v| = 100
|v| = 50 m/s
To Find angle b/w P & R
So, Again the resultant of P and R will be equal to Q:
Let the angle b/w P and R be Theta
Therefore
|Q|² = |P|²+|R|² + 2 PR Cos Theta
1600 = 900 + 2500 + 2 x 1500 x Cos Theta
then
Cos theta = -(.6)
Theta = 127 Degree
Thus, the answer is 127 degrees.
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