Question

A 2190 kg car moving east at 10.5 m/s collides with a 3220 kg car moving east. The cars stick together and move east as a unit after the collision at a velocity of 4.74 m/s. a) What is the velocity of the 3220 kg car before the collision

Answer 1

To solve this problem it is necessary to apply the concepts related to the conservation of the Momentum describing the inelastic collision of two bodies. By definition the collision between the two bodies is given as:

$m_1v_1+m_2v_2 = (m_1+m_2)V_f$

Where,

$m_{1,2}$= Mass of each object

$v_{1,2}$= Initial Velocity of Each object

$V_f$= Final Velocity

Our values are given as

$m_1 = 2190Kg$

$v_1 =10.5m/s$

$m_2 = 3220kg$

$V_f = 4.74m/s$

Replacing we have that

$m_1v_1+m_2v_2 = (m_1+m_2)V_f$

$(2190)(10.5)+(3220)v_2 = (2190+3220)(4.74)$

$v_2 = 0.8224m/s$

Therefore the the velocity of the 3220 kg car before the collision was 0.8224m/s