A 12.0-g bullet fired horizontally into a 100-g wooden block that is initially at rest on frictionless horizontal surface and connected to a spring having spring constant 150 N/m. The bullet becomes embedded in the block. If the bullet-block system compresses the spring by a maximum of 80.0 cm, what was the speed of the bullet at impact with the block?(273 m/s)

Explanation:

As the given collision is inelastic in nature. Therefore, momentum will be conserved. Also, both the block and bullet have the same final velocity.

Therefore, formula will be as follows.

Also,

where,  k = spring constant

            x = compression of spring

          = mass of bullet

          = mass of wooden block

Putting the given values into the above formula as follows.

                = 273.2 m/s

                            = 273 m/s (approx)

Therefore, we can conclude that speed of the bullet is coming with 273 m/s at impact with the block.

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