Question

Assume: The bullet penetrates into the block

and stops due to its friction with the block.

The compound system of the block plus the

bullet rises to a height of 0.13 m along a

circular arc with a 0.23 m radius.

Assume: The entire track is frictionless.

A bullet with a m1 = 30 g mass is fired

horizontally into a block of wood with m2 =

4.2 kg mass.

The acceleration of gravity is 9.8 m/s2 .

Calculate the total energy of the composite

system at any time after the collision.

Answer in units of J.

Taking the same parameter values as those in

Part 1, determine the initial velocity of the

bullet.

Answer in units of m/s.

Answer 1

To solve this problem we will start considering the total energy of the system, which is given by gravitational potential energy of the total of the masses. So after the collision the system will have an energy equivalent to,

`E_T = (m_1+m_2)gh`

Here,

`m_1`= mass of bullet

`m_2`= Mass of Block of wood

The ascended height is 0.13m, so then we will have to

PART A)

`E_T = (m_1+m_2)gh`

`E_T = (0.03+4.2)(9.8)(0.13)`

`E_T = 5.389J`

PART B) At the same time the speed can be calculated through the concept provided by the conservation of momentum.

`m_1v_1+m_2v_2 = (m_1+m_2)v_f`

Since the mass at the end of the impact becomes only one in the system, and the mass of the block has no initial velocity, the equation can be written as

`m_1v_1 =(m_1+m_2)v_f`

The final velocity can be calculated through the expression of kinetic energy, so

`E_T = KE = (1)/(2) (m_1+m_2)v_f^2`

`v_f = \sqrt{(2E_T)/(m_1+m_2)}`

`v_f = \sqrt{(2*5.389J)/(0.03+4.2)}`

`v_f = 1.5962m/s`

Using this value at the first equation we have that,

`m_1v_1 =  (m_1+m_2)v_f`

`v_1 =((m_1+m_2)v_f)/(m_1)`

`v_1 = ((0.03+4.2)(1.5962))/(0.03)`

`v_1 = 225.06m/s`