Question

A 1.30 kg skateboard is coasting along the pavement at a speed of 6.64 m/s when a 0.680 kg cat drops from a tree vertically downward onto the skateboard. What is the speed of the skateboard-cat combination?

Answer 1

Answer:

$v_f=4.36\frac{m}{s}$

Explanation:

The principle of conservation of linear momentum states that if the sum of forces acting on a system is zero, its linear momentum remains constant over time. So, in this case we have:

$\Delta p=0\\p_i=p_f\\m_sv_i_s+m_cv_i_c=m_sv_f_s+m_cv_f_c$

The cat drops downward, so their initial speed horizontally is zero($v_i_c=0$). The skateboard and the cat collide and stick together, so their final speeds are the same($v_f_s=v_f_c=v_f$):

$m_sv_i_s=(m_s+m_c)v_f\\v_f=\frac{m_sv_i_s}{m_s+m_c}\\v_f=\frac{1.3kg(6.64\frac{m}{s})}{1.3kg+0.68kg}\\v_f=4.36\frac{m}{s}$