2. Section 7-3: Collisions & Impulse
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 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
During a collision, objects can be deformed due to the large forces involved.
Example: A tennis ball is hit by a racket as in the figure.

3. ∑F is time dependent & the collision time t is often very small
∑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 ≈ Fct
This approximation is the same as replacing the green area in the left figure by the green rectangle in the right figure.
Figure 9-9. Caption: Force as a function of time during a typical collision: F can become very large; Δt is typically milliseconds for macroscopic collisions.
Figure Caption: The average force acting over a very brief time interval gives the same impulse (FavgΔt) as the actual force.
Replace the green area at the left with the green rectangle at the right.

4. t is usually very small & Fc is time dependent
The true impulse
is the area under
the Fc vs. t curve.
t is usually very small & Fc is time dependent

5. J = p ≈ Fct 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

6. Opposite the person’s momentum
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

7. 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!!!

8. 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!!!

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

10. 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.
Figure 9-11.
Solution: Take the mass of the hand plus a reasonable portion of the arm to be 1 kg; if the speed goes from 10 m/s to zero in 1 cm the time is 2 ms. This gives a force of 5000 N.

11. 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 = N s.
Impulse on wall is in
opposite direction: N s.
vi = v sinθ
vf = - v sinθ