• Briefly consider the details of a collisionAssume 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/t or p = (∑F)/t(momentum change of the object due to the collision)

Define: p Impulse J that 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

Section 7-3: Collisions & Impulse

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

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

∑F is time dependent & the collision time t is often very small. So, not much error is made in replacing ∑F with 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 Impulse is approximately:

J = p ≈ FctThis 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.

The true impulse is the area under the Fc vs. t curve.

t is usually very small & Fc is time dependent

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 ≈ Fct

Example 7-6• Advantage of bending knees when landing!

a) m =70 kg, h =3.0 m

Impulse: p = ?

Ft= 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

• 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!!!

• 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 = 0.13 s

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• 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

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• Impulse:

p = change in momentum wall.• Momentum || to wall does not

change. Impulse will be wall.

Take + direction toward wall,

Impulse = p = mv

= 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