Answer:
Explanation:
Using the Condition That Initial Angular Velocity is equal to final angular velocity
total angular momentum is equal to angular momentum of the person + angular Momentum of the Platform
Note L= I ×ω
Final Angular momentum of the person is equal to the final angular momentum of the platform
Final Moment of Inertia of the person I_{per.f} = =70×(2.7)²=510.3Kgm2
Initial Angular Momentum L_{i} = Final Angular Momentum L_{f}
I_{plat.i} ×ω_{plat.i} + I_{per.i} ×ω_{per.i} = I_{plat.f} ×ω_{plat.f} + I_{per.f} ×ω_{per.f}
890(0.85) + 0 =890(ω_{plat.f}) +510.3(ω_{plat.f})
756.5 = 1400.3 (ω_{plat.f})
(ω_{plat.f}) =0.54rad/s
Answer:
(a)2.7 m/s
(b) 5.52 m/s
Explanation:
The total of the system would be conserved as no external force is acting on it.
Initial momentum = final momentum
⇒(4.30 g × 943 m/s) + (730 g × 0) = (4.30 g × 484 m/s) + (730 g × v)
⇒ 730 ×v = (4054.9 - 2081.2) =1973.7
⇒v=2.7 m/s
Thus, the resulting speed of the block is 2.7 m/s.
(b) since, the momentum is conserved, the speed of the bullet-block center of mass would be constant.
Thus, the speed of the bullet-block center of mass is 5.52 m/s.