Question

A bullet with a mass mb=11.5 g is fired into a block of wood at velocity vb=265 m/s. The block is attached to a spring that has a spring constant k of 205 N/m. The block and bullet continue to move, compressing the spring by 35.0 cm before the whole system momentarily comes to a stop. Assuming that the surface on which the block is resting is frictionless, determine the mass of the wooden block.

Answer 1

Step-by-step explanation:

mass of bullet, mb = 11.5 g

velocity of bullet, vb = 265 m/s

Spring constant, K = 205 N/m

compression in spring, Δd = 35 cm = 0.35 m

Let m be the mass of wooden block.

Let v is the velocity of wooden block.

Use conservation of momentum

mb x vb = ( m + mb) x v

0.0115 x 265 = ( m + 0.0115) x v

`v = (3.05)/(m + 0.0115)` ... (1)

Use conservation of energy

`(1)/(2)K * \Delta d^(2)=(1)/(2)\left ( m + m_(b) \right )v^(2)`

`205* 0.35* 0.35=\left ( m + m_(b) \right )* \left ( (3.05)/(m + m_(b)) \right )^(2)`

`25.1125* (m + 0.0115)=9.3025`

m = 0.36 kg

Answer 2

0.358326032852 kg

Step-by-step explanation:

`m_1=m_b` = Mass of bullet = 11.5 g

`v_1=v_b` = Velocity of bullet = 265 m/s

x = Displacement of spring = 35 cm

v = Velocity of the system

`m_2` = Mass of the block

As the momentum of the system is conserved we have

`m_1u_1=(m_1+m_2)v\\\Rightarrow v=(m_1u_1)/(m_1+m_2)`

As the energy in the system is conserved we have

`(1)/(2)(m_1+m_2)v^2=(1)/(2)kx^2\\\Rightarrow (m_1+m_2)\left((m_1u_1)/(m_1+m_2)\right)^2=kx^2\\\Rightarrow m_2=(m_1^2u_1^2)/(kx^2)-m_1\\\Rightarrow m_2=(0.0115^2* 265^2)/(205* 0.35^2)-0.0115\\\Rightarrow m_2=0.358326032852\ kg`

The mass of the block is 0.358326032852 kg