1. PHYSICS 231 Lecture 12: Keeping momentum
Remco Zegers
Walk-in hour: Thursday 11:30-13:30 am
Helproom
PHY 231

2. The 3 most difficult exam problems
The Football player (28% correct)
Can out of the train (40% correct)
x-t and v-t graph & block-pulley (47% correct)
PHY 231

3. The football player 40 m/s 40o d Vertical direction
A football player throws a ball with an initial velocity of 40 m/s and an angle of 40o with respect to the field. At what distance from the player will the ball hit the field? Assume that the player’s length is negligible (i.e. the ball is thrown from a vertical height of 0.0 m).
40 m/s
40sin(40o)
40o
40cos(40o) d
Horizontal direction
Vertical direction
Y(t)=Y(0)+vy(0)t+½at2
0=0+40sin(40o)t-½gt2
25.7t-½gt2=0
(25.7-½gt)t=0 so
t=0 (start) or t=25.7/(½g)=5.2
X(t)=x(0)+vx(0)t+½at2
d=0+40cos(40o)t+0
d=31t
d=31*5.2=161.2 m
PHY 231

4. Can out of the train N Cancel! W E S
A train is moving with a speed of 25 km/h to the east. An environment-unfriendly passenger throws a can out of the window. The velocity with which he throws the can relative to the moving train is 25 km/h toward the back of the train the (west) and 10 km/h away from the train toward the south. To an onlooker standing on the ground (south of the track), what is the observed direction of motion of the can? N
Cancel! W
Train (25 km/h) E
velocity of can parallel of the train
(-25 km/h)
velocity of the can perpendicular to
the train (to the south) (10 km/h) S
PHY 231

5. X-t and v-t graph X x t t X v t t
PHY 231

6. Pulley
40 kg
A block of mass 40.0 kg sliding on a frictionless table is attached to another block of mass 10.0 kg by a string over a massless pulley (see figure). What is the acceleration of the bigger mass?
40 kg
10kg
10 kg
Use Newton’s second law for each object separately!
40 kg
10 kg
T=40a (only hor. direction)
-T+10g=10a
-T+98.1=10a
-40a+98.1=10a
98.1=50a so a=98.1/50=1.96 m/s2
PHY 231

7. Tail of chapter 5 If relieved from rest, what is
the velocity of the ball at the
lowest point?
30o
L=5m
(PE+KE)=constant
PErelease=mgh (h=5-5cos(30o))
=6.57m J
KErelease=0
PEbottom=0
KEbottom=½mv2
½mv2=6.57m so v=3.6 m/s h
PHY 231

8. A running person While running, a person dissipates about
0.60 J of mechanical energy per step per
kg of body mass. If a 60 kg person develops
a power of 70 Watt during a race, how
fast is she running (1 step=1.5 m long)
What is the force the person exerts on the
road?
W=Fx P=W/t=Fv
PHY 231

9. Chapter 6 Momentum & Collisions
When a bullet hits the wall, its velocity
is very much reduced. The wall does not
move, although the force on the ball
is the same as the force on the wall
(Newton’s 3rd law: Fwall-bullet=-Fbullet-wall).
Fwall-bullet=mbulletabullet
Fbullet-wall=mwallawall
Mbullet << Mwall
|abullet|>> awall
PHY 231

10. Is it only the mass??? Vbullet=100 m/s Vbullet=200 m/s
When the bullet gets stopped in the wall,
it deaccelerates from its initial velocity
to 0. So, its acceleration is vbullet/t,
with t some small time (independent of v).
Second law: Fwall-bullet=mbulletabullet=mbulletvbullet/t
The force also depends on the velocity of the bullet!
PHY 231

11. More general…and formal.
F=ma Newton’s 2nd law
F=mv/t a=v/t
F=m(vfinal-vinital)/t
Define p=mv p: momentum (kgm/s)
F=(pfinal-pinitial)/t
F=p/t
The net force acting on an object equals
the change in momentum (p) in a certain
time period (t).
Since velocity is a vector, momentum is also
a vector, pointing in the same direction as v.
PHY 231

12. Impulse F=p/t Force=change in (mv) per time period (t).
p=Ft The change in momentum equals
the force acting on the object times
how long you apply the force.
Definition: p=Impulse
What if the force is not constant within the time
period t? t s F1 F2 F3
p=Ft=(F1s+F2s+F3s)=
= t(F1s+F2s+F3s)/t
= tFaverage
p=Faverage t
PHY 231

13. Some examples
A tennis player receives a shot approaching him (horizontally) with 50m/s and returns the ball in the opposite direction with 40m/s. The mass of the ball is kg.
A) What is the impulse delivered
by the ball to the racket?
B) What is the work done by the
racket on the ball?
PHY 231

14. Child safety
A friend claims that it is safe to go on a car trip with your
child without a child seat since he can hold onto your 12kg
child even if the car makes a frontal collision (lasting 0.05s
and causing the vehicle to stop completely) at v=50 km/h
(about 30 miles/h). Is he to be trusted?
PHY 231