Section 7 3 collisions impulse
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Section 7-3: Collisions & Impulse. During a collision, objects can be deformed due to the large forces involved. Briefly consider the details of a collision Assume that a collision lasts a very small time t During the collision, the net force on the object is Newton’s 2 nd Law :

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Section 7-3: Collisions & Impulse

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Section 7 3 collisions impulse

Section 7-3: Collisions & Impulse

During a collision, objects can be deformed due to the large forces involved.

  • Brieflyconsider the details of a collision

    Assume that a collision lasts a very small time t

  • During the collision, the net force on the object is

    Newton’s 2nd Law:

    ∑F = p/tor p = (∑F)/t

    (momentum change of the object due to the collision)

    Define: p  Impulse Jthat the collision

    gives the object(change in momentum for the object!)

  • In the usual case: Either only one force is acting or we replace the left side by an average collision force:Fc = ∑F

    Impulse:J = p = Fc t

Example: A tennis ball is hit by a racket as in the figure.


Section 7 3 collisions impulse

∑Fis time dependent & the collision time tis often very small. So, not much error is made in replacing ∑Fwith an average force, Fc or Favg. The true Impulse in the collision is the area under the F vs. t curve (green in the left figure).In the approximation that Fc ≈∑F, the Impulseis approximately:

J = p ≈Fct

This approximation is the same as replacing the green area in the left figure by the green rectangle in the right figure.

Replace the green area at the left with the green rectangle at the right.


Section 7 3 collisions impulse

The true impulse

is the area under

the Fc vs. t curve.

t is usually very small & Fc is time dependent


Section 7 3 collisions impulse

It is often a good approximation to replace the

area under the Fc vs. t curve with the area of the

green rectangle. The approximate impulse is then

J = p ≈Fct


Example 7 6

Example 7-6

  • Advantage of bending knees when landing!

    a)m =70 kg, h =3.0 m

    Impulse: p = ?

    Ft= p = m(0-v)

    First, find v (just before

    hitting): KE + PE = 0

    m(v2 -0) + mg(0 - h) = 0

     v = 7.7 m/s

    Impulse: p = -540 N s

Just before he

hits the ground

Just after he

hits the ground

Opposite the person’s momentum


Section 7 3 collisions impulse

  • Advantage of bending knees when landing!

    Impulse: p = -540 N s

    m =70 kg, h =3.0 m, F = ?

    b) Stiff legged: v = 7.7 m/s to

    v = 0 in d = 1 cm (0.01m)!

    v = (½ )(7.7 +0) = 3.8 m/s

    Time t = d/v = 2.6  10-3 s

    F = |p/t| = 2.1  105 N

    (Net force upward on person)

    From free body diagram,

    F = Fgrd - mg  2.1  105 N

    Enough to fracture leg bone!!!


Section 7 3 collisions impulse

  • Advantage of bending knees when landing!

    Impulse: p = -540 N s

    m =70 kg, h =3.0 m, F = ?

    c) Knees bent: v = 7.7 m/s to

    v = 0 in d = 50 cm (0.5m)

    v = (½ )(7.7 +0) = 3.8 m/s

    Time t = d/v = 0.13s

    F = |p/t| = 4.2  103 N

    (Net force upward on person)

    From free body diagram,

    F = Fgrd - mg  4.9  103 N

    Leg bone does not break!!!


Example crash test

Example: Crash Test

  • Crash test: Car, m = 1500 kg, hits wall. 1 dimensional collision. +x is to the right. Before crash,v = -15 m/s. After crash, v = 2.6 m/s.Collision lasts Δt = 0.15 s. Find:Impulse car receives & average force on car.

    Assume: Force exerted by wall is large compared to other forces

    Gravity & normal forces are perpendicular & don’t effect the horizontal momentum

     Use impulse approximation

p1 = mv1 = -2.25 kg m/s, p2 = mv2 = 2.64 kg m/s

J = Δp = p2 – p1 = 2.64  104 kg m/s

(∑F)avg = (Δp/Δt) = 1.76  105 N


Section 7 3 collisions impulse

Example: Karate blow

Estimate the impulse & the average force delivered by a karate blow that breaks a board a few cm thick. Assume the hand moves at roughly 10 m/s when it hits the board.


Problem 17

Problem 17

  • Impulse:

    p = change in momentum  wall.

  • Momentum || to wall does not

    change.  Impulse will be  wall.

    Take + direction toward wall,

    Impulse = p = mv

    = m[(- v sinθ) - (v sinθ)]

    = -2mv sinθ = - 2.1 N s.

    Impulse on wall is in

    opposite direction: 2.1 N s.

vi = v sinθ

vf = - v sinθ


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