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A 5.00-g bullet is shot through a 1.00-kg wood block suspended on a string 2.00 m long. The center of mass of the block rises a distance of 0.38 cm. Find the speed of the bullet as it emerges from the block if its initial speed is 450 m/s.

Short Answer

Expert verified

The emerging speed of bullets after the collision is 435m/s .

Step by step solution

01

Determination of speed of the wooden block after the collisionGiven Data:

The initial speed of the bullet is" width="9" height="19" role="math">u=450m/s

The height raised by the block is h=0.38cm=0.0038m

The length of the string is l=2m

The mass of the wooden block is: M=1kg

The mass of the bullet is: M=5g=0.005kg

02

concept

The velocity of the wooden block is calculated by using raised height and then applying momentum conservation for the emerging speed of a bullet.

The speed of the wooden block after the collision is given as:

vw=2gh

Here, g is the gravitational acceleration.

Substitute all the values in the above equation.

vw=29.8m/s0.0038mvw=0.0745m/s

03

Determination of emerging speed of bullet after collision

The emerging speed of bullet after the collision is given as:

mu=Mvw+mv

Substitute all the values in the above equation.

0.005kg450m/s=1kg0.0745m/s+0.005kgvv=435m/s

Therefore, the emerging speed of bullet after collision is .

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