An 8.66g bullet is fired into a 260g block that is initially at rest at the edge

Question

An 8.66g bullet is fired into a 260g block that is initially at rest at the edge of a table of h = 1.07 m height (see the figure below).

The bullet remains in the block, and after the impact the block lands d = 1.81 m from the bottom of the table. How much time does it take the block to reach the ground once it flies off the edge of the table?

What is the initial horizontal velocity of the block as it flies off the table? (assume this to be in the positive direction)

Determine the initial speed of the bullet.

Explanation / Answer

Let initial velocity be v of the bullet and final velocity of bullet + block system be v' so .00866v = .26866v' Given it falls down a height of 1.07 m so 1.07 = .5*9.8*t^2 t = .467 sec So it takes .467 sec to reach the ground Horizontal distance traveled by the system of block + bullet = 1.81 m time take for this movement = .467 sec so speed imparted in horizontal direction = 3.875 m/s so v' = 3.875 m/s Initial speed of the bullet hence will be v = .26866v'/.00866 = 120.24 m/s

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