## A 3-gram bullet is fired horizontally into a 10-kilogram block of wood suspended by a rope from the ceiling. The velocity of the bullet just

A 3-gram bullet is fired horizontally into a 10-kilogram block of wood suspended by a rope from the ceiling. The velocity of the bullet just before the collision is 750 m/s. The block swings in an arc, rising a height h from its lowest position. What is h in millimeters? (Hint: You must apply conservation of momentum to the collision to find out how fast the block+bullet move together before you can use conservation of Energy to find h.)

## Answers ( )

Answer:2.58 mm

Explanation:mass of bullet, m = 3 g = 0.003 kg

mass of block M = 10 kg

initial velocity of bullet, u = 750 m/s

Let they move together after collision with velocity v.

Use conservation of momentumm x u + 0 + ( m + M) x v0.003 x 750 = ( 10 + 0.003) x v

v = 0.225 m/s

Let the both reach at height h.

use conservation of energy

Kinetic energy at the bottom = Potential energy at height0.5 x ( M + m) x v² = ( M + m) x g x h0.5 x 0.225 x 0.225 = 9.8 x h

h = 2.58 x 10^-3 m

h = 2.58 mmAnswer:The height raised by the blcok is 2.5 mm.

Explanation:Given that,

Mass of the bullet,

Initial speed of the bullet,

Mass of the block of wood,

It was at rest, initial speed of the blockof wood,

The block swings in an arc, rising a height h from its lowest position. The momentum remains consered. Let V is the velocity of the block +bullet system together. So,

Now using the conservation of energy to find the height h raised by the blck. So,

So, the height raised by the blcok is 2.5 mm.