# What is the speed of the block immediately after the bullet exits?

A 3.2kg block of wood sits on a table. A 3.0g bullet, fired horizontally at a speed of 400 m/s , goes completely through the block, emerging at a speed of 180 m/s .

By the law of momentum conservation:-
=>m1u1 + m2u2 = m1v1 + m2v2
=>0 + 3 x 10^-3 x 400 = 3.2 x v1 + 3 x 10^-3 x 180
=>v1 = 0.21 m/s [+ve indicate the direction is same as that of bullet]

Use Conservation of Momentum. The formula is m1v1 + m2v2 = m1vi + m2v2. m is the mass while v is the velocity. The left side are the initial values while the right side are the final values. Now just plug in the number, and before you do that just make sure all the masses have kilograms units and all the velocities have m/s units. This is what it will look like after you plug in the numbers:
(.003 kg)(400 m/s) + (3.2 kg)(0 m/s) = (.003 kg)(180 m/s) + (3.2 kg)(v)
1.2=.54 + 3.2v
v=1.2 – .54/3.2
v=.206 m/s

Net effective speed of bullet = (400 – 180), = 220m/sec.
(220 x .003kg) = .66.
(.66/3.2kg) = 0.206m/sec. velocity imparted to block.

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