Question

A lorry of mass 12,000kg travelling at a velocity of 2 m/s collides with a stationary car of mass 1500kg. The vehicles move together after the impact. Calculate their velocity.

Answer 1

Answer:

Approximately $1.8\; \rm m \cdot s^{-1}$. Assumption: the friction between the two vehicles and the ground is negligible.

Explanation:

The momentum $p$ of an object is the product of its mass $m$ and its velocity $v$:

$p = m \cdot v$.

Momentum should be conserved in this collision if there's no external force on these two vehicles. In other words, the sum of the momentum of the lorry and the car should be the same before and after the collision.

Mass of the lorry: $m_1 = 12000\; \rm kg$.

Velocity of the lorry right before the collision: $2\; \rm m \cdot s^{-1}$.

Therefore, the momentum of the lorry right before the collision would be $12000\; \rm kg \times 2\; \rm m \cdot s^{-1} = 24000\; \rm kg \cdot m\cdot s^{-1}$.

Mass of the car: $m_2 = 1500\; \rm kg$.

Velocity of the car before the collision: $0\; \rm m \cdot s^{-1}$.

Therefore, the velocity of the car before the collision would be $1500\; \rm kg \times 0\; \rm m \cdot s^{-1} = 0\; \rm kg \cdot m \cdot s^{-1}$.

Sum of the momentum of the lorry and the car right before the collision:

$24000\; \rm kg \cdot m \cdot s^{-1} + 0\; \rm kg \cdot m \cdot s^{-1} = 24000\; \rm kg \cdot m \cdot s^{-1}$.

Under the assumption that there is no external forces on the lorry and the car, the sum of the momentum of the lorry and the car right after the collision should also be $24000\; \rm kg \cdot m \cdot s^{-1}$.

Let $v\; \rm m \cdot s^{-1}$ be the velocity of the lorry and the car after the collision. (The velocity of the two vehicles should be the same because they were moving together.)

Right after the collision, the momentum of the lorry (with a mass of $12000\; \rm kg$) would be ${12000\; {\rm kg} \times v\; \rm m \cdot s^{-1}} = 12000\, v \; \rm kg \cdot m \cdot s^{-1}$.

Similarly, right after the collision, the momentum of the car (with a mass of $1500\; \rm kg$) would be ${1500\; \rm kg} \times v\; {\rm m \cdot s^{-1}} = 1500\, v\; \rm m \cdot s^{-1}$.

The sum of the velocities of the two vehicles right after the collision would be:

$\left(12000\, v \; {\rm kg \cdot m \cdot s^{-1}} + 1500\, v \; {\rm kg \cdot m \cdot s^{-1}}\right)$.

That quantity should match the momentum right before the collision. In other words:

${12000\; \rm kg \cdot m \cdot s^{-1}} = \left(12000\, v \; {\rm kg \cdot m \cdot s^{-1}} + 1500\, v \; {\rm kg \cdot m \cdot s^{-1}}\right)$.

Solve this equation for the velocity of the two vehicles after the collision:

$\displaystyle v = \frac{12000}{12000 + 1500}\approx 1.8$.

Therefore, the velocity of the two vehicles right after the collision would be approximately $1.8\; \rm m \cdot s^{-1}$.