Answer to Question #125623 in Physics for Mwansa Kunda

Question

A 1000kg Toyota car is travelling north at 15m/s ⁄ when it collides with a 2000kg truck travelling east at 10m/s ⁄ . All occupants are wearing seat belts and there are no injuries, but the two vehicles are thoroughly tangled and move away from the impact point as one. The insurance officer then asks you to find the velocity of the wreckage just after impact. Using your PH-110 knowledge, demonstrate (using calculations) how you would arrive at the required information.

Answer 1

Let's write down the law of conservation of momentum in vector form:

$$m_1\mathbf{v_1} + m_2\mathbf{v_2} = (m_1 + m_2)\mathbf{u}$$

where $\mathbf{v_1}$ is the velocity of the car, $\mathbf{v_2}$ is the velocity of the truck before the collision and $\mathbf{u}$ is the velocity of the wreckage after the collision. $m_1 = 1000kg$ and $m_2 = 2000 kg$ are the masses of the car and truck respectively.

Expressing $\mathbf{u}$ from this equation, get:

$$\mathbf{u} = \dfrac{m_1\mathbf{v_1} + m_2\mathbf{v_2}}{m_1 + m_2}$$

The projection of this vector on the North-South direction is (since $\mathbf{v_2}$ does not have component in this direction):

$$u_N = \dfrac{m_1v_1}{m_1 + m_2}$$

Similarly, the projection of the East-West direction will be:

$$u_E = \dfrac{m_2v_2}{m_1 + m_2}$$

The module of the velocity of the wreckage just after impact will be:

$$u = \sqrt{u_N^2 + u_E^2} = \sqrt{(\dfrac{m_1v_1}{m_1 + m_2})^2 + (\dfrac{m_2v_2}{m_1 + m_2})^2} = \\ =\dfrac{1}{m_1 + m_2}\sqrt{m_1^2v_1^2 + m_2^2 v_2^2}$$

Substituting the numerical values, get:

$$u= \dfrac{1}{1000 + 2000}\sqrt{1000^2\cdot 15^2 + 2000^2 \cdot 10^2} \approx 8.33m/s$$

Answer. 8.33 m/s.