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Project on Impact – Football Helmet Collision March 13, 2019 Dynamics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Consider the Football Helmets as two particles coming together Dynamics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Y – Plane of Contact X – Line of Impact Oblique Impact Dynamics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. Project on Impact – Football Helmet Collision Problem Statement 1. The football player with helmet “A” has a velocity of vA = 6 ft/sec in the direction shown and his helmet collides with a stationary football player’s helmet, “B”, with a helmet of equal mass and diameter and same material. If e=0.6, determine the resulting magnitude of velocity of each football player’s helmet following impact. 2. What approx. force is exerted on helmet B if the delta time of impact is 0.01 sec? 3. What recommendations would you provide to lessen the impact of head injuries? Consider the Football Helmets as two particles vB =0 Line of Impact x vA = 6 ft/sec B y Plane of Impact B Oblique Impact WA = WB = 5 lb 30o A A Before Impact Dynamics, Fourteenth Edition R.C. Hibbeler During Impact Copyright ©2016 by Pearson Education, Inc. All rights reserved. Backup Material Dynamics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. IMPACT (Section 15.4) Impact occurs when two bodies collide during a very short time period, causing large impulsive forces to be exerted between the bodies. Common examples of impact are a hammer striking a nail or a bat striking a ball. The line of impact is a line through the mass centers of the colliding particles. In general, there are two types of impact: Central impact occurs when the directions of motion of the two colliding particles are along the line of impact. Oblique impact occurs when the direction of motion of one or both of the particles is at an angle to the line of impact. Dynamics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. CENTRAL IMPACT Central impact happens when the velocities of the two objects are along the line of impact (recall that the line of impact is a line through the particles’ mass centers). vA vB Line of impact Once the particles contact, they may deform if they are non-rigid. In any case, energy is transferred between the two particles. There are two primary equations used when solving impact problems. The textbook provides extensive detail on their derivation. Dynamics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. CENTRAL IMPACT (continued) In most problems, the initial velocities of the particles, (vA)1 and (vB)1, are known, and it is necessary to determine the final velocities, (vA)2 and (vB)2. So the first equation used is the conservation of linear momentum, applied along the line of impact. (mA vA)1 + (mB vB)1 = (mA vA)2 + (mB vB)2 This provides one equation, but there are usually two unknowns, (vA)2 and (vB)2. So another equation is needed. The principle of impulse and momentum is used to develop this equation, which involves the coefficient of restitution, or e. Dynamics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. CENTRAL IMPACT (continued) The coefficient of restitution, e, is the ratio of the particles’ relative separation velocity after impact, (vB)2 – (vA)2, to the particles’ relative approach velocity before impact, (vA)1 – (vB)1. The coefficient of restitution is also an indicator of the energy lost during the impact. The equation defining the coefficient of restitution, e, is e = (vB)2 – (vA)2 (vA)1 - (vB)1 If a value for e is specified, this relation provides the second equation necessary to solve for (vA)2 and (vB)2. Dynamics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. COEFFICIENT OF RESTITUTION In general, e has a value between zero and one. The two limiting conditions can be considered: • Elastic impact (e = 1): In a perfectly elastic collision, no energy is lost and the relative separation velocity equals the relative approach velocity of the particles. In practical situations, this condition cannot be achieved. • Plastic impact (e = 0): In a plastic impact, the relative separation velocity is zero. The particles stick together and move with a common velocity after the impact. Some typical values of e are: Steel on steel: 0.5 – 0.8 Wood on wood: 0.4 – 0.6 Lead on lead: 0.12 – 0.18 Glass on glass: 0.93 – 0.95 Dynamics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. IMPACT: ENERGY LOSSES Once the particles’ velocities before and after the collision have been determined, the energy loss during the collision can be calculated on the basis of the difference in the particles’ kinetic energy. The energy loss is  U1-2 =  T2 −  T1 where Ti = 0.5mi (vi)2 During a collision, some of the particles’ initial kinetic energy will be lost in the form of heat, sound, or due to localized deformation. In a plastic collision (e = 0), the energy lost is a maximum, although the energy of the combined masses does not necessarily go to zero. Why? Dynamics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. OBLIQUE IMPACT In an oblique impact, one or both of the particles’ motion is at an angle to the line of impact. Typically, there will be four unknowns: the magnitudes and directions of the final velocities. The four equations required to solve for the unknowns are: Conservation of momentum and the coefficient of restitution equation are applied along the line of impact (x-axis): mA(vAx)1 + mB(vBx)1 = mA(vAx)2 + mB(vBx)2 e = [(vBx)2 – (vAx)2]/[(vAx)1 – (vBx)1] Momentum of each particle is conserved in the direction perpendicular to the line of impact (y-axis): mA(vAy)1 = mA(vAy)2 and mB(vBy)1 = mB(vBy)2 Dynamics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved. PROCEDURE FOR ANALYSIS • In most impact problems, the initial velocities of the particles and the coefficient of restitution, e, are known, with the final velocities to be determined. • Define the x-y axes. Typically, the x-axis is defined along the line of impact and the y-axis is in the plane of contact perpendicular to the x-axis. • For both central and oblique impact problems, the following equations apply along the line of impact (x-dir.):  m(vx)1 =  m(vx)2 and e = [(vBx)2 – (vAx)2]/[(vAx)1 – (vBx)1] • For oblique impact problems, the following equations are also required, applied perpendicular to the line of impact (y-dir.): mA(vAy)1 = mA(vAy)2 and mB(vBy)1 = mB(vBy)2 Dynamics, Fourteenth Edition R.C. Hibbeler Copyright ©2016 by Pearson Education, Inc. All rights reserved.
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Running head: CASE ANALYSIS: THE HELMET PROBLEM

Case Analysis: The Helmet Problem
Name
Course
Date

1

CASE ANALYSIS: THE HELMET PROBLEM

2

Case Analysis: The Helmet Problem
Part 1
According to the information provided, the football player with helmet A has a velocity
of 6 ft/s and collides with the helmet of a stationary player (B). Considering how the collision is
inelastic, the following conditions must be met:


The total mechanical momentum of the two helmets must be the same than before the
collision



The total kinetic energy of the two helmets IS not the same after the collision

It is possible to write the law of conservation of the mechanical momentum as:
⃗ 𝐴,𝑜 + 𝑤𝐵 ∗ 𝑉
⃗ 𝐵,𝑜 = 𝑤𝐴 ∗ 𝑉
⃗ 𝐴,𝑓 + 𝑤𝐵 ∗ 𝑉
⃗ 𝐵,𝑓
𝑤𝐴 ∗ 𝑉
where wA and wB represent the mass of helmets A and B, VA,o and VB,o the initial speed of
helmets A and B, and VA,f and VB,f the final speed of helmets A and B.
According to the provided information, player B is stationary before the collision, which implies
that VB,o = 0 such that the previous equation is simplified into:
⃗ 𝐴,𝑜 = 𝑤𝐴 ∗ 𝑉
⃗ 𝐴,𝑓 + 𝑤𝐵 ∗ 𝑉
⃗ 𝐵,𝑓
𝑤𝐴 ∗ 𝑉
Moreover, since the two helmets have the same mass, it is possible to write this equation as:
⃗ 𝐴,𝑜 = 𝑤 ∗ (𝑉
⃗ 𝐴,𝑓 + 𝑉
⃗ 𝐵,𝑓 )
𝑤∗𝑉
which is equivalent to:
⃗ 𝐴,𝑜 = 𝑉
⃗ 𝐴,𝑓 + 𝑉
⃗ 𝐵,𝑓
𝑉
It is thus possible to write,
𝑉𝐴,𝑜𝑥 = 𝑉𝐴,𝑓𝑥 + 𝑉𝐵,𝑓𝑥
and
𝑉𝐴,𝑜𝑦 = 𝑉𝐴,𝑓𝑦 + 𝑉𝐵,𝑓𝑦

CASE ANALYSIS: THE HELMET PROBLEM

3

On the other hand, the coefficient of restitution, e, is:
𝑒=

𝑉𝐵,𝑓𝑥 − 𝑉𝐴,𝑓𝑥
= 0.6
𝑉𝐴,𝑜𝑥

Substituting from the previous equation,
𝑒=

𝑉𝐵,𝑓𝑥 − 𝑉𝐴,𝑓𝑥
= 0.6
𝑉𝐴,𝑓𝑥 + 𝑉𝐵,𝑓𝑥

Operating,
𝑉𝐵,𝑓𝑥 − 𝑉𝐴,𝑓𝑥 = 0.6 ∗ (𝑉𝐴,𝑓𝑥 + 𝑉𝐵,𝑓𝑥 )
𝑉𝐵,𝑓𝑥 − 𝑉𝐴,𝑓𝑥 = 0.6𝑉𝐴,𝑓𝑥 + 0.6𝑉𝐵,𝑓𝑥
0.4𝑉𝐵,𝑓𝑥 = 1.6 ∗ 𝑉𝐴,𝑓𝑥
𝑉𝐵,𝑓𝑥 = 4𝑉𝐴,𝑓𝑥
Besides, we know that the collision is oblique, since...


Anonymous
I was struggling with this subject, and this helped me a ton!

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