Question

1400-kg car going at 632 m/s in the positive z direction collides with a 2700-kg truck at rest the collision is totally inelastic and takes place over an interval of 0.203 s. assume that no brakes are applied during the collision and the car strikes the rear of the truck neglect the friction between the vehicles and the ground.
a) what is the average x component of the acceleration of the car during the collision?

Answer 1

Answer:

-2050.25 m/s2

Explanation:

Using the law of conservation of linear momentum, the sum of initial and final momentum are equal and since momentum,

p is the product of mass and velocity hence for this case

$m_cu_c+m_t_u_t=(m_c+m_t)v_{common}$where m represent mass, u and v represent initial and final velocities respectively, subscripts c and t represent car and truck respectively.

Substituting the given figures with consideration that the truck is initially at rest hence its initial velocity is zero then

(1400+2700)v=1400*632

V=215.80487804878048780487804878048780487804 m/s

Rounded off as 215.8 m/s

Acceleration, a is given as the rate of change of velocity per unit time.

The change in velocity is 215.8-632=-416.2

Time is given as 0.203 s

Acceleration is -416.2/0.203 s=−2050.246305418 m/s2

Rounded off, acceleration is -2050.25 m/s2