A bullet of mass 5 g moving at a speed of 200 m/s strikes a rigidly fixed wooden plank of thickness 0.2 m normally and passes through it losing half of its kinetic energy. If it again strikes an identical rigidly fixed wooden plank and passes through it, assuming the same resistance in the two planks, the ration of the thermal energies produced in the two planks is
A bullet of mass 5 g moving at a speed of 200 m/s strikes a rigidly fixed wooden plank of thickness 0.2 m normally and passes through it losing half of its kinetic energy. If it again strikes an identical rigidly fixed wooden plank and passes through it, assuming the same resistance in the two planks, the ration of the thermal energies produced in the two planks is
The correct option is A 1:1
Here we assume that all the work done on the bullet inside the wood is converted into thermal energy. Thermal energy produced in the first case is
H1 = 12(5 × 10−3)(200)2 J
Since the resistance in the second plank is the same as the first and also the thickness is equal, the same amount of work will be done on the bullet as in the first case and hence the thermal energy H2 = H1. The ratio is 1:1.
1:1
Here we assume that all the work done on the bullet inside the wood is converted into thermal energy. Thermal energy produced in the first case is
H1 = 12(5 × 10−3)(200)2 J
Since the resistance in the second plank is the same as the first and also the thickness is equal, the same amount of work will be done on the bullet as in the first case and hence the thermal energy H2 = H1. The ratio is 1:1.
