A car of mass 400kg and travelling at 72 kmph crashes into a truck of mass 4000kg and travelling at 9 kmph, in the same direction. The car bounces back at a speed of 18 kmph. The speed of the truck after the impact is

The correct answer is 18 kmph.

Step-by-step Explanation:

To find the speed of the truck after the impact, we can apply the Law of Conservation of Linear Momentum. According to this law, the total linear momentum of a closed system before collision is equal to the total linear momentum of the system after collision.

Let us identify the given values:
Mass of the car ($m_c$) = 400 kg
Initial velocity of the car ($u_c$) = 72 kmph
Mass of the truck ($m_t$) = 4000 kg
Initial velocity of the truck ($u_t$) = 9 kmph (moving in the same direction as the car)
Final velocity of the car ($v_c$) = -18 kmph (since the car bounces back, its velocity becomes negative relative to its initial direction)
Final velocity of the truck = $v_t$

The equation for the conservation of linear momentum is:
$m_c{u}_c+m_t{u}_t=m_c{v}_c+m_t{v}_t$

Substitute the given values into the momentum equation:
$(400\times 72)+(4000\times 9)=(400\times (-18\left)\right)+(4000\times v_t)$

Now, calculate each term:
$28800+36000=-7200+4000v_t$
$64800=-7200+4000v_t$

Isolate the term with $v_t$ by adding 7200 to both sides of the equation:
$64800+7200=4000v_t$
$72000=4000v_t$

Divide by 4000 to solve for $v_t$:
$v_t=\frac{72000}{4000}$
$v_t=18$

Thus, the final speed of the truck after the impact is 18 kmph.