Question

A bullet of mass m = 8.00 g is fired into a block of mass M = 250 g that is initially at rest at the edge of a table of height h = 1.00 m. The bullet remains in the block, and after the impact the block lands d = 2.00 m from the bottom of the table. Determine the initial speed of the bullet.

Answer #1

**after collsion**

**for the bullet + block system**

**along vertical**

**initial velocity voy = 0**

**acceleration ay = -9.8 m/s^2**

**displacement y = -1 m**

**y = voy*t + (1/2)*ay*t^2**

**-1 = - (1/2)*9.8*t^2**

**t = 0.452 s**

**along horizontal**

**x = v*t**

**v = x/t = 2/0.452 = 4.42 m/s**

**speed of bullet + block , v = 4.42 m/s**

**from momentum conservation**

**momentum before collsion = momentum after
collision**

**m*vo = (M + m)*v**

**vo = (M + m)*v/m**

**vo = (0.25 + 0.008)*4.42/0.008**

**vo = 142.545 m/s <<<<------answer**

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