Question

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

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

Answer 1

Answer: (B)

Here, $m_{1}=400\, kg , m_{2}=4000\, kg$
$u_{1}=72\, km\, h ^{-1}=72 \times \frac{5}{18} m s ^{-1}=20\, m\, s ^{-1}$
$u_{2}=9 \,km\, h ^{-1}=9 \times \frac{5}{18}=2.5\, m \,s ^{-1}$
$v_{1}=-18 \,km\, h ^{-1}=-18 \times \frac{5}{18}=-5 m s ^{-1}$
$v_{2}=?$
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+10 v_{2} $
$ 45=-5+10 v_{2} $
$ v_{2} =5 \,m\, s ^{-1}$
$=5 \times \frac{18}{5} \,km\, h ^{-1}$
$=18\, km\, h ^{-1} $