Download Solutions - Young Engineering and Science Scholars

California Institute of Technology
Young Engineers & Science Scholars 2010
Group Problems: Momentum—Solutions
Friday, July 2, 2010
Problem 1
A billiard ball rolls with velocity 1.20 m/s in the +y direction and strikes an identical ball initially
at rest. The first (initially rolling) ball is deflected so that its velocity has a component of 0.80 m/s
in the +y direction and a component of 0.56 m/s in the +x direction.
(a) Draw a diagram of before and after.
Before:
6
y
After:
6
v2,f
7v1,f
QQm k
m
2 1
y
m
2
6
1.20 m/s
m
1
-
-
x
x
(b) Write a momentum conservation equation for the collision.
Answer:
m~v1,i = m~v1,f + m~v2,f
i.e.
v1y,i = v1y,f + v2y,f ,
0 = v1x,f + v2x,f
(c) What is final velocity and speed of the ball that was initially at rest?
Answer:
We’re given that v1y,i = 1.2 m/s, v1y,f = 0.8 m/s, v1x,f = 0.56 m/s. Thus
v2y,f , = v1y,i − v1y,f = 0.4 m/s
v2x,f , = −v1x,f = −0.56 m/s
So,
~v2,f = (−0.56 m/s)î + (0.4 m/s)ĵ
v2,f =
p
0.562 + 0.42 m/s = 0.69 m/s
Problem 2
The average force exerted by a bat on a baseball during the time of contact, 0.002 s, is 6600 N. The
mass of a baseball is 0.145 kg and its speed is 75 mph just before the bat collides with it.
(a) What is the impulse of the ball hitting the bat?
Answer: Let the positive x direction be the direction in which the bat moves. Impulse
is the average force times time the force is applies, so J~ = (6600 N) × (0.002 s) × (−î)
= (−13.2 kg m / s)î . By Newton’s third law, the impulse of the ball on the bat is the same
as that of the bat on ball, but in the opposite direction.
1
(b) What is the velocity of the ball when it leaves the bat? Assume the ball leaves in the same
direction from which it came.
Answer: Conservation of momentum implies
J~bat on ball = (+13.2 kg m/s)î = p~ball,f − p~ball,i = (0.145 kg) (vball,f − (−75 mph)) î.
Convert m/s to mph:
13.2 kg
m 1 mi 3600 s
m
= 13.2 kg
= 29.5 kg mi/hr
s
s 1609 m 1 hr
Thus
vball,f =
29.5 kg mi/hr
− 75 mph = 129 mph
0.145 kg
Problem 3
A dog that jumps into the interior of an ice boat is moving with a (horizontal) velocity of 32 km/h
just before it enters the boat. The boat had already been moving (in the same forward, horizontal
direction as the dog) with a speed of 5 km/h. The dog has a mass of 14 kg. The boat plus boater
weigh 160 kg.
(a) Assuming the ice is frictionless (it almost is!), what is the speed of the boat after the collision?
Answer: Conservation of momentum implies
mboat vboat,i + mdog vdog,i = (mboat + mdog )vboat+dog
thus
vboat+dog =
160 × 5 + 14 × 32
mboat vboat + mdog vdog
=
km/h = 7.2 km/h
mboat + mdog
160 + 14
(b) What is the percentage of energy lost in the collision?
Answer: We’ll calculate the percentage of mechanical kinetic energy lost.
KEi =
1
1
2
2
mboat vboat
+ mdog vdog
= 9168 kgkm2 /h2
2
2
1
2
mboat+dog vboat+dog
= 4476 kgkm2 /h2
2
So the percentage of energy lost is
KEf
1−
× 100% = 51%
KEi
KEf =
(c) Where does the energy go?
Answer:
The energy goes into heat, acoustic energy, and internal energy of the boat (things on the boat
slide around...).
Problem 4 A 10 gram bullet is fired into a stationary block of wood that has a mass of 5 kg. The
speed of the bullet just before entry into the wood block is 500 m/s.
(a) What is the speed of the block just after the bullet has become embedded?
Answer: Use conservation of momentum:
mb vb = (mb + mw )vw+b
⇒
vw+b =
2
mb vb
0.010 × 500
=
m/s = 0.998 m/s
mb + mw
5 + 0.010
(b) How far will the block slide on a surface with coefficient of friction of 0.50? (Assume the
bullet’s path and the surface is essentially perpendicular to the force of gravity.)
Answer: The deceleration of the block due to friction is given by
mb+w~as = −µFN î = −µmb+w g î
⇒ as,x = −µg.
The distance the wooden block travels (in the poitive x direction) as a function of time, from
the point when the bullet is embedded is
1
x = v0 t + as,x t2
2
The block stops sliding once its velocity hits zero. The velocity is given by
v = v0 + as,x t
so the time at which it stops sliding is
t=−
v0
.
as,x
Plug this back into the distance equation to find how far the block slides:
x=−
1
v2
1 v02
1 v02
0.9982
v02
+ as,x 20 = −
=
=
m = 10.2 cm
as,x
2
as,x
2 as,x
2 µg
2 × 0.5 × 9.81
3
Brain Teaser Solution:
Part I: Answer is c: Both would be equally effective in stopping Mighty Mike. If you
calculate the momentum of both Speedy Gonzalez and Ponderous Poncho, they have
equal momentum!
Part II: Answer is a: Even though Speedy Gonzalez is small, the kinetic energy he has
is much bigger than Ponderous Poncho and would thus have more energy to break
Mighty Mike’s bones.
If we calculate the kinetic energy of both we see that
Speedy Gonzalez Kinetic Energy: KE SG =
1
ft
(100lb)(16 ) 2 = 12800 lb*ft/s
2
s
1
ft
(400lb)(4 ) 2 = 3200 lb*ft/s
2
s
In fact, Speedy Gonzalez has 4 times more energy than Ponderous Poncho despite being
300 lb heavier than Speedy Gonzalez.
Ponderous Poncho: KE PP =