Question

In the figure here, a 11.5 g bullet moving directly upward at 1210 m/s strikes and passes through the center of mass of a 7.5 kg block initially at rest. The bullet emerges from the block moving directly upward at 720 m/s. To what maximum height does the block then rise above its initial position?

Answer #1

Initial kinetic energy of the bullet = 1/2mu^2

Or, Ki = [1/2*11.5*10^-3*(1210)^2]

Or, Ki = 8418.575 J

Final kinetic energy of the bullet = 1/2mv^2

Or, Kf = [1/2*11.5*10^-3*(720)^2] J

Or, Kf = 2980.8 J

Loss in kinetic energy = Ki - Kf

= (8418.575 - 2980.8) J

= 5437.78 J

Thus,

Gain in potential energy of the block = 5437.78 J

Let h be the height reched by the block above its initial position due to this gain in potential energy, then, Mgh = 5437.78, M is the mass of the block

Or, h = 5437.78/Mg = 5437.78/7.5*9.8

Or, h = 73.98

Therefore, required height = 73.98 m

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