   Chapter 6, Problem 42P

Chapter
Section
Textbook Problem

An 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 (Fig. P6.42). 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. Figure P6.42

To determine

Initial speed of the bullet.

Explanation

Given Info: Mass of the bullet is m=8.00×103kg, Mass of the block is M=250×103 kg, initial velocity of the block is vB=0 ,height of the table is Δy=1.00 m, and distance moved by block is Δx=2.00 m.

The expression for projectile motion of the block-bullet combination is

Δy=v0yt+12ayt2                                                                          (1)

• Δy is the height of the table.
• v0y is the vertical speed of the block-bullet combination.
• t is the time taken by  block –bullet combination to reach the floor.
• ay is the acceleration due to gravity.

Substitute 1.00 m for Δy,0 for v0y,9.80 m/s2 for ay in equation (1)

1.00m=(0)t+12(9.80 m/s2)t21.00m=(4.9m/s2)t2t2=1.00m4.9m/s2t=0.452

Thus the time taken by block –bullet combination to reach the floor is 0.452s.

The expression for velocity of the bullet remains in the block is,

v0x=Δxt

• v0x is the horizontal speed of the block-bullet combination

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