Question

0.05 kg bullet collides and sticks to a 2.5 kg stationary block suspended from a string. the bullet and block swings to a maximum height of 12 cm. what was the initial speed of the bullet?

Answer 1

Answer: The initial speed of the bullet will be 10.95 m/s.

Step-by-step explanation:

For the given system, the energy remains conserved. Firstly, the energy was in the form of kinetic energy of the bullet and after the collision, it got converted to the potential energy of block and bullet. Hence, the equation becomes:

`(1)/(2)m_1v_1^2+(1)/(2)m_1u_1^2=(m_1+m_2)gh`

where,

`m_1,u_1\text{ and }v_1` are the mass, initial velocity and final velocity of the bullet.

`m_2` = mass of the block

g = acceleration due to gravity

h = height reached by both ball and block

We are given:

`m_1=0.05kg\\m_2=2.5kg\\u_1=?m/s\\v_1=0m/s\\g=9.8m/s^2\\h=12cm=0.12m`

Putting values in above equation, we get:

`(1)/(2)(0.05)(0)+(1)/(2)(0.05)u_1^2=(0.05+2.5)* 9.8* 0.12\\\\u_1^2=119.952m^2/s^2\\\\u_1=10.95m/s`

Hence, the initial speed of the bullet will be 10.95 m/s.

Answer 2

first calculate the combine velocity of the bullet and stick, using the formula
kinetic energy = potential energy
0.5mv^2 = mgh
where m is the combined mass
v is the velocity
g is the acceleration due to gravity ( 9.81 m/s^2
h is the heigth

0.5 ( 0.05 + 2.5 kg) v^2 = ( 9.81 m/s^2 ) ( 0.05 + 2.5 kg) ( 0.12 m )
solve for v
v = 1.53 m/s
using momentum balance
M = mv
m1v1 + m2v2 = mv
where m1 is the mass of the bullet
v1 is the initial velocity of the bullet
m2 mass of the stick
v2 initial velocity of stick

(0.05 kg)(v1) + ( 2.5 kg)( 0 ) = ( 0.05 + 2.5 kg)( 1.53 m/s )
v1 = 78.25 m/s