Collision Force

Classical Mechanics Level 3

A car having a mass of $2 \text{ Mg}$ strikes a smooth, rigid sign post with an initial speed of $30 \text{ km/h}$ . To stop the car, the front end horizontally deforms $0.2 \text{ m}$ . If the car is free to roll during the collision, determine the average horizontal collision force causing the deformation.

F_{\text{avg}} = 347 \text{ kN}

F_{\text{avg}}= 4500 \text{ kN}

F_{\text{avg}} = 5400 \text{ kN}

F_{\text{avg}} = 9000 \text{ kN}

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Samuel Wong
Mar 25, 2014

The kinetic energy of the car:

$E=\frac{1}{2}\times m\times v^{2}$

$E=68890J$

This is also equal to the work done on the post, which is the Force times distance.

So, 68890=0.2*F

$F=\boxed{347KN}$

In given case collision is inelastic. How can you equate entire kinetic energy with work done?

puneeth vandrapu - 7 years, 2 months ago

please understand work energy theorem clearly, then u can do this

Venkat Krishna - 7 years, 2 months ago

initial velocity is 8.33 m/s final velocity is 0m/s by 3rd equation of kinematics accln comes out to be 173.472 so F=ma therefore 2x173.47225=347 N

Sagnick Mukherjee - 7 years, 2 months ago

it was easy though 347 KN

Abdun Fabaswir - 7 years, 2 months ago

Naveen Dubey
Mar 28, 2014

Apply equation of motion V^2=u^2-2as (v-final velocity = 0, U initial velocity = 50/6 m/s, S=0.2 mtr) we get a= 173.666 Striking force F= Ma ==> F = 2000 Kg X 173.666 = 347 N

Acceleration = initial velocity squared / 2 * distance, Force = Mass * Acceleration, which gives, Force = 347.222 Newton, but I dont understand how some of us have got 347 kilo Newton

Sriharsha Venkatesh - 7 years, 2 months ago

Collision Force

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