Question

car 1 is travelling at 6.8 m/s and has a mass 3,200 kg and collides with car 2, which is stopped. if the two cars collide and couple, and have final velocity of 4.1 m/s, what is the mass of car 2? round your answer to the nearest whole number. which equation do we use for this problem? o (m1v1 + m2v2)before=(m1 + m2)v f(after) o p(before)=p(after) o p = mv o (m1v1 + m2v2)before=(m1v1 + m2v2)after

Explanation:

Step1: Identify the conservation - of - momentum formula

For an in - elastic collision (where the two cars couple), the conservation of momentum formula is $(m_1v_1 + m_2v_2)_{before}=(m_1 + m_2)v_{after}$. Here, $m_1$ is the mass of car 1, $v_1$ is the initial velocity of car 1, $m_2$ is the mass of car 2, $v_2$ is the initial velocity of car 2, and $v_{after}$ is the final velocity of the combined cars.

Step2: Substitute the given values

We know that $m_1 = 3200$ kg, $v_1=6.8$ m/s, $v_2 = 0$ m/s (car 2 is stopped), and $v_{after}=4.1$ m/s. Substituting into the formula $(m_1v_1 + m_2v_2)_{before}=(m_1 + m_2)v_{after}$, we get $m_1v_1=(m_1 + m_2)v_{after}$ (since $v_2 = 0$). So, $3200\times6.8=(3200 + m_2)\times4.1$.

Step3: Expand and solve for $m_2$

First, expand the right - hand side: $3200\times6.8 = 3200\times4.1+m_2\times4.1$. Then, $21760=13120 + 4.1m_2$. Subtract 13120 from both sides: $21760−13120 = 4.1m_2$. So, $8640 = 4.1m_2$. Divide both sides by 4.1 to find $m_2=\frac{8640}{4.1}\approx2107$ kg.

Answer:

The mass of car 2 is 2107 kg.
The correct equation is $(m_1v_1 + m_2v_2)_{before}=(m_1 + m_2)v_{after}$