A thin block of soft wood with a mass of 0.0720 kg rests on a horizontal frictionless surface. A bullet with a mass of 4.67 g is fired with a speed of 556 m/s at a block of wood and passes completely through it. The speed of the block is 18.0 m/s immediately after the bullet exits the block. (a) Determine the speed of the bullet as it exits the block.

Explanation:

Let the mass of the thin block be M = .072 kg .

mass of bullet m = 4.67 x 10⁻³ kg.

speed of bullet v = 556 m/s.

speed of block after billet exits the block = 18 m/s.

We shall apply law of conservation of momentum

momentum of bullet block system

.072 x 0 + 4.67 x 10⁻³  x 556 = .072 x 18 +  4.67 x 10⁻³  x V here V is velocity of bullet after exit

= 2.59652 = 1.296 + 4.67 x 10⁻³  x V

4.67 x 10⁻³  x V = 1.30052

V = .278 x 10³ m /s

= 278 m /s

force is mass*acceleration.

we have a very small mass: 11.4*0.001 kg

if we want 0.035n of force, we need an acceleration:

0.035 = 11.4*0.001*a

a = 0.035/(11.4*0.001) m/s^2