Question

A 0.0220 kg bullet moving horizontally at 400 m/s embeds itself into an initially stationary 0.500 kg block.

(a) What is their velocity just after the collision?

...............m/s

(b) The bullet-embedded block slides 8.0 m on a horizontal surface with a 0.30 kinetic coefficient of friction. Now what is its velocity?

.................m/s

(c) The bullet-embedded block now strikes and sticks to a stationary 2.00 kg block. How far does this combination travel before stopping?

................m

Step 1

Given mass of bullet m_{1} = 0.022 kg

Initial velocity of bullet u_{1} = 400 m/s

Mass of the block m_{2} = 0.5 kg

Initial velocity u_{2} = 0 m/s

Step 2

Part b:

We can find velocity of the block by writing the energy balancing equation.

(Initial kinetic energy of the block and bullet system ) - (work lost against friction ) = (Final kinetic energy of the block and bullet system)

Step 3

Part c:

Mass of block-bullet system m1 = 0.022+0.5 = 0.522 kg

Velocity of block-bullet system u1 = 15.39 m/s

Mass of stationa...

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