Question

A 0.0125 kg bullet strikes a 0.240 kg block attached to a fixed horizontal spring whose spring constant is 2.25*10^3N/m and sets it into oscillation with amplitude of 12.4 cm. What was the initial speed of the bullet if the two objects move together after impact?

Answer 1

The value is
`u_1 =  236 \ m/s`

Step-by-step explanation:

From the question we are told that

The mass of bullet is
`m_b  =  0.0125 \  kg`

The mass of the block is
`M_B  =  0.240 \  kg`

The spring constant is
`k  =  2.25*10^(3) \  N/m`

The amplitude is
`A= 12.4 \ cm  =  0.124 \ m`

Generally according to the conservation of momentum is

`m_b u_1 + M_B  u_2 =  (m_b + M_B) v`

given that the block was at rest we have that

`m_b u_1  =  (m_b + M_B) v`

Now the angular velocity of the both bodies is mathematically represented as

`w =  \sqrt{(k)/(M_B  + m_b) }`

`w = \sqrt{( 2.25*10^(3))/( 0.0125 + 0.240 ) }`

`w =  94.4 \  rad/s`

Given that the system after collision set into oscillation

The maximum linear velocity of the system after impact is mathematically represented as

`v  = A * w`

`v  =  94.4 *0.124`

`v  =  11.7 \  m/s`

From above equation

`u_1  =  ((m_b +  M_B ) * v)/(m_b)`

=>
`u_1 =  ((0.240 + 0.0125) *  11.7)/( 0.0125)`

=>
`u_1 =  236 \ m/s`