Question

A2.0kg arrow is fired at 10m/s into a stationary 10 kg block of ice as shown below: the arrow sticks into the block of ice and they start moving . What is the velocity of the arrow/ice combination after the collison

Answer 1

the velocity is 1.67 m/s

Step-by-step explanation:

The computation of the velocity of the arrow/ice combination after the collison is shown below:

Here the law of conservation is applied

`p_i = p_f\\\\m_av_a+m_b+v_b= (m_a + m_b)v`

Here

`m_a = 2.0 kg\\\\v_a = 10m/s\\\\m_b = 10 kg\\\\v_b = 0`

Now

`v = (m_av_a+ m_bv_b)/(m_a + m_b) \\\\= ((2.0kg) (10 m/s) + 0)/(2.0kg + 10.0kg)`

= 1.67 m/s

hence, the velocity is 1.67 m/s