Question

A 0.100kg arrow is shot into a 0.400kg wood at rest. After the impact, the block and arrow moves with a common velocity of 3.0m/s. Calculate the velocity of the arrow just before entering the wood.

Answer 1

Let us consider the conservation of momentum law. Before the impact the total momentum was

$p_1 = m_av_a + m_bv_b = 0.100\,\mathrm{kg}\cdot v_a + 0.400\,\mathrm{kg}\cdot 0, \\ p_1 = 0.100\,\mathrm{kg}\cdot v_a.$

After the impact the total momentum is

$p_2 = (m_a+m_b)\cdot v_{\text{tot}} = 0.500\,\mathrm{kg}\cdot3.0\,\mathrm{m/s} = 1.50\,\mathrm{kg\cdot m/s}.$

Since $p_1=p_2,$ $\;\; v_a = \dfrac{p_2}{0.100\,\mathrm{kg}} = \dfrac{ 1.50\,\mathrm{kg\cdot m/s}}{0.100\,\mathrm{kg}} = 15\,\mathrm{m/s}.$