Question

A car of mass \[1000{\text{ kg}}\] travelling at \[{\text{32 m/s}}\] clashes into the rear of a truck of mass \[{\text{8000 kg}}\] moving in the same direction with a velocity of \[{\text{4 m/s}}\] . After the collision, the car bounces with a velocity of \[{\text{8 m/s}}\]. The velocity of truck after the impact is \[{\text{________ m/s}}\]
A. \[{\text{8}}\]
B. \[{\text{4}}\]
C. \[{\text{6}}\]
D. \[{\text{9}}\]

Answer 1

Hint: In this question, a car crashes into the rear of a truck traveling in the same direction. After the collision the car moves in the opposite direction with a velocity of \[{\text{8 m/s}}\]. As collision is taking place, considering the car and the truck as a system we can say the momentum of the system before and after the collision is conserved. Thus, we can write the equation of momentum before the collision and the equation of momentum after the collision and equate.

Complete step by step answer:
We are given, the mass of the car, \[{{\text{m}}_1} = 1000{\text{ kg}}\]
The velocity of the car before the collision is \[{{\text{v}}_1} = 32{\text{ m/s}}\]
The mass of the truck, \[{{\text{m}}_2} = 8000{\text{ kg}}\]
Velocity of the truck before the collision is \[{{\text{v}}_2} = 4{\text{ m/s}}\]

After the collision, let the velocity of the car be \[{{\text{v}}_1}^1{\text{ m/s}}\].
Let the velocity of the truck be \[{{\text{v}}_2}^1{\text{ m/s}}\].
As the car after the collision bounces with a velocity of \[{\text{8 m/s}}\].
\[ \Rightarrow {{\text{v}}_1}^1{\text{ = }} - {\text{8 m/s}}\]

Considering the car and the truck as a system we can say the momentum of the system before and after the collision is conserved. Thus, the momentum of the system before collision is equal to the momentum of the system after the collision. Thus, equation of momentum conservation is,
\[{{\text{m}}_1}{{\text{v}}_1} + {{\text{m}}_2}{{\text{v}}_2} = {{\text{m}}_1}{{\text{v}}^1}_1 + {{\text{m}}_2}{{\text{v}}^1}_2\]

Substituting the values respectively, we get,
\[ \Rightarrow \left( {1000 \times 32} \right) + \left( {8000 \times 4} \right) = \left( {1000 \times - 8} \right) + 8000{{\text{v}}^1}_2\]
\[ \Rightarrow 32000 + 32000 = 8000{{\text{v}}^1}_2 - 8000\]
On solving we get,
\[ \Rightarrow 72000 = 8000{{\text{v}}^1}_2\]
On dividing we get,
\[ \therefore {{\text{v}}^1}_2 = 9{\text{ m/s}}\]
Thus, the velocity of the truck after the impact is \[9{\text{ m/s}}\] in the same direction as it was moving.

Therefore, the correct answer is option D.

Note:There are two types of collision- elastic collision and inelastic collision. In both cases the momentum of the system is conserved before and after the collision. In elastic collision, there is no loss of energy before and after the collision but for inelastic collision, there is a loss of energy. If both the colliding objects stick together then it is said to be a perfectly inelastic collision.