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Head on Collision 
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#1
Feb2705, 10:09 PM

P: 5

Lost in this review problem:
Most of us know intuitively that in a headon collision between a large dump truck and a subcompact car, you are better off being in the truck than in the car. Why is this? Many people imagine that the collision force exerted on the car is much greater than that experienced by the truck. To substantiate this view, they point out that the car is crushed, whereas the truck is only dented. This idea of unequal forces, of course, is false. Newton's third law tells us that both objects experience forces of the same magnitude. The truck suffers less damage because it is made of stronger metal. But what about the two drivers? Do they experience the same forces? Suppose that each vehicle is initially moving at 6.40 m/s and that they undergo a perfectly inelastic headon collision. Each driver has mass 81.0 kg. Including the drivers, the total vehicle masses are 830 kg for the car and 3800 kg for the truck. The collision time is 0.150 s. (a) What force does the seat belt exert on the truck driver? (b) What force does the seat belt exert on the car driver? how do i go about solving this? Thanks in advance 


#2
Feb2705, 10:28 PM

Mentor
P: 22,243

This is a momentum problem. Find the final speed of the car/truck combination: m1v1+m2v2=m3v3. Then subtract that speed from the speeds of each driver to find the total change in speed for each. Divide by time to find the acceleration felt by each driver. Use f=ma to find the force.



#3
Feb2805, 06:33 AM

P: 453

Choose a positive direction, so that from the conservation of momentum:
[tex]m_tv_{0t}  m_cv_{0c} = (m_t + m_c)v_f[/tex], where we take the direction of the truck initial velocity as positive Then, since [tex]v_{0t}[/tex] and [tex]v_{0c}[/tex] have the same magnitudes, [tex](m_t  m_c)v_0 = (m_t + m_c) v_f[/tex] or [tex]v_f = \frac{m_t  m_c}{m_t + m_c}v_0[/tex] Because the mass of the truck is greater than that of the car, then numerator [tex]m_t  m_c[/tex] will be positive, meaning that the final velocity of the combined system (truck + car) will still be in the direction of the truck. Observe that the greater [tex]m_t[/tex] is in comparison with [tex]m_c[/tex], the closer will the final velocity be to the truck initial velocity! Therefore, the truck experiences less acceleration (same direction of velocity) than the car does (direction of velocity is reversed), and the same is true for the corresponding drivers. Hence a driver would rather prefer to be in the truck during the collision because a smaller acceleration would mean a smaller force as well (F = ma). 


#4
Feb2805, 09:44 AM

P: 5

Head on Collision
Thank you very much



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