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. car 2 has a mass of . type the number in the first blank. type the units in the 2nd blank. (use these for your units: kg, m/s, kg*m/s)

Explanation:

Step1: Apply momentum - conservation formula

The law of conservation of momentum is $m_1v_1 + m_2v_2=(m_1 + m_2)v_f$. Here, $m_1 = 3200$ kg, $v_1=6.8$ m/s, $v_2 = 0$ m/s (car 2 is stopped), and $v_f = 4.1$ m/s. Substitute the values into the formula: $3200\times6.8+ m_2\times0=(3200 + m_2)\times4.1$.

Step2: Simplify the equation

The left - hand side simplifies to $3200\times6.8=21760$. So the equation becomes $21760=(3200 + m_2)\times4.1$.

Step3: Solve for $m_2$

First, divide both sides of the equation by 4.1: $\frac{21760}{4.1}=3200 + m_2$. $\frac{21760}{4.1}\approx5307.32$. Then, $m_2=\frac{21760}{4.1}-3200$. $m_2 = 5307.32-3200=2107.32$. Rounding to the nearest whole number, $m_2 = 2107$ kg.

Answer:

2107, kg