Question

A 300 kg cae traveling at 60-m/s rear ends a 600 kg truck plodding along at 10 m/s. The vehicles collide and bounce apart with the car moving forward at 15 m/s. What is the final velocity of the truck?

Answer 1

Answer:

32.5 m/s

Explanation:

The total momentum must be conserved before and after the collision:

$p_i = p_f\\m_c u_c + m_t u_t = m_c v_c + m_t v_t$

where

m_c = 300 kg is the mass of the car

m_t = 600 kg is the mass of the truck

u_c = 60 m/s is the initial velocity of the car

u_t = 10 m/s is the initial velocity of the truck

v_c = 15 m/s is the final velocity of the car

v_f is the final velocity of the truck

Solving for v_f, we find:

$v_f = \frac{m_c u_c + m_t u_t - m_c v_c}{m_t}=\frac{(300 kg)(60 m/s)+(600 kg)(10 m/s)-(300 kg)(15 m/s)}{600 kg}=32.5 m/s$