First slide

Mechanical energy conservation

Question

Two masses m and 2m are attached to two ends of an ideal spring and the spring is in the compressed state. The energy of spring is 60 joule. If the spring is released, then

Moderate

A

the energy of both bodies will be same
B

energy of smaller body will be 10J
C

energy of smaller body will be 20J
D

energy of smaller body will be 40 J

Solution

$F_{ext} = 0 From momentum conservation P_{1} + P_{2} = 0 m_{1} v_{1} + (- m_{2} v_{2}) = 0 \frac{m_{1}}{m_{2}} = \frac{v_{2}}{v_{1}} & \frac{v_{2}}{v_{1}} = \frac{1}{2} \frac{k_{1}}{k_{2}} = \frac{m_{1} v_{1}^{2}}{m_{2} {v_{2}}^{2}} = (\frac{1}{2}) {(\frac{2}{1})}^{2} = 2 N o w k_{1} + k_{2} = 60 \Rightarrow k_{1} + \frac{k_{1}}{2} = 60 = k_{1} = 40 J$

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