Question

A 1050 kg automobile is at rest at a traffic signal. At the instant the light turns green, the automobile starts to move with a constant acceleration of 3.3 m/s2. At the same instant a 2350 kg truck, traveling at a constant speed of 8.6 m/s, overtakes and passes the automobile.
(a) How far is the center of mass of the automobile-truck system from the traffic light at t = 2.1 s?
(b) What is the speed of the center of mass of the automobile-truck system then?

Answer 1

Answer:

Part a)

$d = 14.7 m$

Part b)

$v_f = 8.08 m/s$

Explanation:

As we know that initial velocity of the center of mass of the system of truck and car is

$v_{cm} = \frac{m_1v_1 + m_2v_2}{m_1 + m_2}$

Here we know that

$m_1 = 1050 kg$

$v_1 = 0$

$m_2 = 2350 kg$

$v_2 = 8.6 m/s$

so we have

$v_{cm} = \frac{1050(0) + 2350(8.6)}{1050 + 2350}$

$v_{cm} = 5.94 m/s$

Similarly the acceleration of center of mass is given as

$a_{cm} = \frac{m_1a_1 + m_2a_2}{m_1 + m_2}$

$a_{cm} = \frac{1050(3.3) + 2350(0)}{1050 + 2350}$

$a_{cm} = 1.02 m/s^2$

Now we know that the center of mass of the system will move a distance

$d = v t + \frac{1}{2}at^2$

$d = 5.94(2.1) + \frac{1}{2}(1.02)(2.1)^2$

$d = 14.7 m$

Part b)

final speed of the center of mass is given as

$v_f = v_i + at$

$v_f = 5.94 + (1.02)2.1$

$v_f = 8.08 m/s$