Question

need explanation plz

m1 01 m2 3. Consider the above image depicting an inelastic collision between two masses ml - 5kg, and m2 10kg. Masses ml and m2 combine into a single particle after their collision and move with a velocity v_f at an angle q 2. Suppose that q 1 70 degrees, vl - 10m/s and v2 10 m/s. a. What are the values of v _f and q_2?

Answer 1

Solution:

Given that
Given that rightarrow M_1 = 5 kg m_2 = 10 kg v_1 = 10 m/s theta_1 = 70 degree v_2 = 10 m/s theta = ? After collision They move with common vector rightarrow By law of conservation of momentum Initial momentum = final momentum m_1 v_1 i^bar + m_2 v_2 cos theta_1 i^bar + m_2 v_2 sin theta_1 j^bar = (m_1 + m_2)^2 f cos theta_2 i^bar + (m_1 + m_2) v_f sin theta_2 for Horizontal components m_1 v_1 + m_2 v_2 cos theta = (m_1 + m_2) cos theta_2 v_f 5(10) + (10 times 10 times cos 70 degree) = (5 + 10) cos theta_2 v_f v_f cos theta_2 = 5.61 m/s equation (1) \r\n rightarrow for vertical components of velocity m_2 v_2 sin theta_1 = (m_1 + m_2) sin degree v_f v_f sin theta_2 = 10 times 10 times sin 70 degree/(10 + 5) v_f sin theta_2 = 6.26 m/s rightarrow (2) from eq (1) & eq (2) we get