Answer:
A) 21.2 kg.m/s at 39.5 degrees from the x-axis
Explanation:
Mass of the smaller piece = 200g = 200/1000 = 0.2 kg
Mass of the bigger piece = 300g = 300/1000 = 0.3 kg
Velocity of the small piece = 82 m/s
Velocity of the bigger piece = 45 m/s
Final momentum of smaller piece = 0.2 × 82 = 16.4 kg.m/s
Final momentum of bigger piece = 0.3 × 45 = 13.5 kg.m/s
since they acted at 90oc to each other (x and y axis) and also momentum is vector quantity; then we can use Pythagoras theorems
Resultant momentum² = 16.4² + 13.5² = 451.21
Resultant momentum = √451.21 = 21.2 kg.m/s at angle 39.5 degrees to the x-axis ( tan^-1 (13.5 / 16.4)
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Answer:
D. Friction
Explanation:
Friction is a force that opposes motion. So a perpetual motion machine can never be built because it is impossible to eliminate frictional force. It can only be reduced
Answer:
<h2>The angular velocity just after collision is given as</h2><h2>
![\omega = 0.23 rad/s](https://tex.z-dn.net/?f=%5Comega%20%3D%200.23%20rad%2Fs)
</h2><h2>At the time of collision the hinge point will exert net external force on it so linear momentum is not conserved</h2>
Explanation:
As per given figure we know that there is no external torque about hinge point on the system of given mass
So here we will have
![L_i = L_f](https://tex.z-dn.net/?f=L_i%20%3D%20L_f)
now we can say
![m_1v_1\frac{L}{2} = (m_2L^2 + m_1(\frac{L}{2})^2)\omega](https://tex.z-dn.net/?f=m_1v_1%5Cfrac%7BL%7D%7B2%7D%20%3D%20%28m_2L%5E2%20%2B%20m_1%28%5Cfrac%7BL%7D%7B2%7D%29%5E2%29%5Comega)
so we will have
![0.49(1.89)(0.45) = (2.13(0.90)^2 + 0.49(0.45)^2)\omega](https://tex.z-dn.net/?f=0.49%281.89%29%280.45%29%20%3D%20%282.13%280.90%29%5E2%20%2B%200.49%280.45%29%5E2%29%5Comega)
![\omega = 0.23 rad/s](https://tex.z-dn.net/?f=%5Comega%20%3D%200.23%20rad%2Fs)
Linear momentum of the system is not conserved because at the time of collision the hinge point will exert net external force on the system of mass
So we can use angular momentum conservation about the hinge point