Question

A small car of mass 837 kg is parked behind a small truck of mass 1606 kg on a level road. The brakes of both the car and the truck are off so that they are free to roll with negligible friction. A 61 kg woman sitting on the tailgate of the truck shoves the car away by exerting a constant force on the car with her feet. The car accelerates at 1 m/s 2 . What is the acceleration of the truck? Answer in units of m/s 2 .

Answer 1

Answer:$a_t=0.502 m/s^2$

Explanation:

Given

Mass of car $\left ( m_c\right )=837 kg$

mass of small truck$\left ( m_t\right )=1606 kg$

mass of woman$\left ( m_w\right )=61 kg$

According to law of conservation of momentum

$\left ( m_t+m_w\right )v_t+m_cv_c=0$

$1667\times v_t=-837\times v_c$

where $v_t and v_c$ is the velocity of truck and car

and $v_t=u+a_t\times t$

$v_c=u+a_c\times t$

$a_c, a_t$ are the acceleration of car and truck

$v_t=a_t\times t$

$v_c=a_c\times t$

$v_c=t$

substitute the values of $v_t & v_c$

$1667\times a_t\times t=-837\times t$

$a_t=-0.502 m/s^2$