Question

A car of mass $400kg$ and travelling at $72km{h}^{-1}$ crashes into a truck of mass $4000kg$ and travelling at $9km{h}^{-1}$ in the same direction. The car bounces back at a speed of $18km{h}^{-1}$. The speed of the truck after the impact is

• A. $9km{h}^{-1}$
• B. $18km{h}^{-1}$
• C. $27km{h}^{-1}$
• D. $36km{h}^{-1}$

Answer 1

Answer: (B)

Here, $m_1=400kg,m_2=4000kg$
$u_1=72km{h}^{-1}=72\times \frac{5}{18}m{s}^{-1}=20m{s}^{-1}$
$u_2=9km{h}^{-1}=9\times \frac{5}{18}=2.5m{s}^{-1}$
$v_1=-18km{h}^{-1}=-18\times \frac{5}{18}=-5m{s}^{-1}$
$v_2=?$
According to law of conservation of linear momentum, we get
$m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$
$400\times 20+4000\times 2.5=400\times (-5)+4000\times v_2$
$20+25=-5+10v_2$
$45=-5+10v_2$
$v_2=5m{s}^{-1}$
$=5\times \frac{18}{5}km{h}^{-1}$
$=18km{h}^{-1}$