Question

a 3.6 kg block of wood sits on a frictionless table. a 3.0 g bullet, fired horizontally at a speed of 540 m/s , goes completely through the block, emerging at a speed of 230 m/s . part a what is the speed of the block immediately after the bullet exits? express your answer with the appropriate units.

Answer 1

The speed of the block immediately after the bullet exits is 150 m/s.

Step-by-step explanation:

To determine the speed of the block immediately after the bullet exits, we can use the principle of conservation of momentum. Since there is no external force acting on the system, the total momentum before the bullet exits is equal to the total momentum after the bullet exits.

The total initial momentum is given by the product of the mass of the bullet and its initial speed, while the total final momentum is the product of the mass of the block and its final speed.

mathematically, we have:

(mass of bullet * initial speed of bullet) = (mass of block * speed of block after bullet exits)

Plugging in the given values, we can solve for the speed of the block:

(3.0 g * 540 m/s) = (3.6 kg * speed of block after bullet exits)

540 g*m/s = 3.6 * speed of block after bullet exits

speed of block after bullet exits = (540 g*m/s) / (3.6 kg)

By simplifying the units, we get:

speed of block after bullet exits = 150 m/s