Question

A 10-g bullet moving horizontally with a speed of 1.8 km/s strikes and passes through a 5.0-kg block initially at rest on a horizontal frictionless surface. The bullet emerges from the block with a speed of 1.0 km/s. What is the kinetic energy of the block immediately after the bullet emerges

Answer 1

Answer:

$K_2=6.4J$

Explanation:

According to the principle of conservation of momentum, we have:

$\Delta p=0\\p_i=p_f\\m_1v_1_i +m_2v_2_i=m_1v_1_f+m_2v_2_f$

Here 1 is for the bullet and 2 is for the block. Since the block is initially at rest $v_i_2=0$. Solving for $v_2_f$ and replacing the given values:

$v_2_f=\frac{m_1(v_1_i-v_1_f)}{m_2}\\v_2_f=\frac{10*10^{-3}kg(1.8\frac{km}{s}-1\frac{km}{s})}{5kg}\\v_2_f=0.0016\frac{km}{s}*\frac{1000m}{1km}=1.6\frac{m}{s}$

The kinetic energy of the block is given by:

$K_2=\frac{m_2(v_f_2)^2}{2}\\K_2=\frac{5kg(1.6\frac{m}{s})^2}{2}\\K_2=6.4J$