Question

A plastic block of mass 0.20.2kg is initially at rest on a horizontal frictionless surface. A bullet of mass 1212 g is fired with an initial speed of 550550m/s towards the block. The bullet passes completely through the block and exits the other side with a speed of 20.020.0m/s. What percent of the initial kinetic energy of the bullet block system is lost during the collision?

Answer 1

Answer:

94.29619%

Explanation:

$m_1$ = Mass of bullet= 0.012 kg

$m_2$ = Mass of block = 0.2 kg

$u_1$ = Initial Velocity of bullet= 550 m/s

$u_2$ = Initial Velocity of block = 0 m/s

$v_1$ = Final Velocity of bullet = 20 m/s

$v_2$ = Final Velocity of block

As the linear momentum of the system is conserved

$m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\Rightarrow v_2=\dfrac{m_1u_1+m_2u_2-m_1v_1}{m_2}\\\Rightarrow v_2=\dfrac{0.012\times 550+0.2\times 0-0.012\times 20}{0.2}\\\Rightarrow v_2=31.8\ m/s$

Change in kinetic energy is given by

$\Delta K=\dfrac{1}{2}(m_1u_1^2-m_1v_1^2-m_2v_2^2)\\\Rightarrow \Delta K=\dfrac{1}{2}(0.012\times 550^2-0.012\times 20^2-0.2\times 31.8^2)\\\Rightarrow \Delta K=1711.476\ J$

Percentage change is given by

$\dfrac{\Delta K}{\dfrac{1}{2}m_1u_1^2}\times 100=\dfrac{1711.476}{\dfrac{1}{2}0.012\times 550^2}\times 100=94.29619\ \%$

The percent of the initial kinetic energy of the bullet block system is lost during the collision is 94.29619%