Question

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.)

Answer 1

2.58 mm

Step-by-step 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 momentum

m x u + 0 + ( m + M) x v

0.003 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 height

0.5 x ( M + m) x v² = ( M + m) x g x h

0.5 x 0.225 x 0.225 = 9.8 x h

h = 2.58 x 10^-3 m

h = 2.58 mm

Answer 2

The height raised by the blcok is 2.5 mm.

Step-by-step explanation:

Given that,

Mass of the bullet,
`m_1=3\ g=0.003\ kg`

Initial speed of the bullet,
`u_1=750\ m/s`

Mass of the block of wood,
`m_2=10\ kg`

It was at rest, initial speed of the blockof wood,
`u_2=0`

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,

`m_1u_1+m_2u_2=(m_1+m_2)V\\\\V=(m_1u_1+m_2u_2)/((m_1+m_2))\\\\V=(0.003* 750+0)/((0.003+10))\\\\V=0.224\ m/s`

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

`(1)/(2)mV^2=mgh\\\\h=(V^2)/(2g)\\\\h=((0.224)^2)/(2* 10)\\\\h=0.0025\ m\\\\h=2.5\ mm`

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