Question

A 732 kg car stopped at an intersection is rear-ended by a 1720 kg truck moving with a speed of 15.5m/s. If the car was in neutral and its breaks were off, so the collisions are approximately elastic, find the final speed of both vehicles after the collisions.

Answer 1

Answer:

Truck= 6.25 m/s

Car= 21.7 m/s

Explanation:

For elastic collision, the truck's final velocity is given by

${v_1} = \frac{{\left( {{m_1} - {m_2}} \right)}}{{\left( {{m_1} + {m_2}} \right)}}{u_1}$

The car's final velocity is given by

${v_2} = \frac{{2{m_1}}}{{\left( {{m_1} + {m_2}} \right)}}{u_1}$

Where m and u denote masses and velocity respectively, subscripts 1 and 2 denote truck and car respectively.

Substituting 1720 Kg for mass of truck, 732 for mass of car and initial velocity as 15.5 m/s

$v_1=\frac{1720-732}{1720+732}\times 15.5=6.25 m/s$

$v_2=\frac {2\times 1720}{1720+732}\times 15.5=21.7 m/s$

Therefore, the final velocity of truck is 6.25 m/s while the car's final velocity is 21.7 m/s