Download Impulsive Forces and Momentum

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Impulsive Force Model Worksheet 2:
Impulsive Forces and Momentum
1. Two objects, A & B, have identical velocities. Object A has 3 times the mass of object B.
a. Find the value of the ratio of momentum A to momentum B. Justify your answer.
Object A: 3mv Object B: mv Object A has 3 times the momentum of B
b. Find the value of the ratio of kinetic energy A to kinetic energy B. Justify your answer.
A 1.5mv 2

3
Object A: 21 (3m)v 2  1.5mv 2 Object B: 21 mv 2  0.5mv 2 Ratio:
B 0.5mv 2
EkA = 3 times EkB
2. Two objects, C & D, have the same momentum. Object C has ½ the mass of object D.
a. Find the value of the ratio of velocity C to velocity D. Justify your answer.
Object C must have 2 times the velocity of D for their momenta to be equal,  m2  2v  mv
b. Find the value of the ratio of kinetic energy C to kinetic energy D. Justify your answer.
C
mv 2
2
2
2
2
1 m
1

2
Object C: 2  2 (2v)  mv Object D: 2 mv  0.5mv Ratio:
D 0.5mv 2
EkC = 2 times EkD (The less massive has more kinetic energy when the momenta are equal.)
3. The following questions refer to the motion of a baseball.
a. While being thrown, a net force of 132 N acts on a baseball (mass = 140 g) for a period of
4.5 x 10-2 sec. What is the magnitude of the change in momentum of the ball?
F t  mv  F t  132N(0.045s)  5.9N s  5.9 kgs m
b. If the initial speed of the baseball is v = 0.0 m/s, what will its speed be when it leaves the
pitcher's hand?
5.9 kgs m
5.9 kgs m
mv  5.9 kgs m  v 

 42 ms
1kg
140g 1000
0.140
kg
g
c. When the batter hits the ball, a net force of 1150 N, opposite to the direction of the ball's initial
motion, acts on the ball for 9.0 x 10-3 s during the hit. What is the final velocity of the ball?
F t  mv  F t  1150N(0.0090s )  (0.140kg )v  v  74 ms
vi
v  v f  v i  v f  v  v i  v f  74 ms  42 ms  32 ms
d. How large is the force the ball exerts on the bat? Explain.
The ball exerts a force on the bat of 1150N. Ball on bat and bat on ball are a 3rd Law pair.
©Modeling Instruction - AMTA 2013
1
U9 Momentum - ws 2 v3.1
4. A rocket, weighing 4.36 x 104N, has an engine that provides an upward force of 1.2 x 105N. It
reaches a maximum speed of 860 m/s.
Fgases
a. Draw a force diagram for the rocket.
b. For how much time must the engine burn during the launch in order
to reach this speed?
Fg
4.35104 N
m
Fg
(860
)
N
s
10 kg
mv
g v
Fnet t  mv  t 


 49s
Fnet
1.2  105 N  4.35  104 N
7.65  104 N
5. A golf ball that weighs 0.45 N is dropped from a height of 1.0 m. Assume that the golf ball has a
perfectly elastic collision with the floor.
begin end
a. Construct a motion map for the golf ball from the time it is dropped
until it reaches its highest point of rebound.
For the motion map to the right the ball is in contact with the floor
and is not moving at that moment. There are other correct answers.
v
v
begin end
a a
a a
All a = g
except at
bottom
v
a a
v
b. Determine the time required for the ball to reach the floor.
y = -1.0m
2y
2(1.0m)
y  v yi t  21 at 2  t 

 0.45s
vyi = 0
a
10 sm2
a = g = -10 sm2
a
v
t = ?
c. What will the instantaneous momentum of the golf ball be immediately before it strikes the floor?
m  0.45N(
1kg
)  0.045kg
10N
p  mv  mgt  .045kg(10 sm2 )(.45s )  0.045kg(4.5 ms )  0.20 kgs m
d. What will be the change in momentum, (p) from the instant before the ball collides with the
floor until the instant after it rebounds from the floor? (Illustrate with a vector diagram.)
p  pf  pi  mv  0.20 kgs m  (0.20 kgs m )  0.40 kgs m
e. Suppose that the golf ball was in contact with the floor for 4.0 x 10-4s. What was the average
force on the ball while it was in contact with the floor?
Fnet t  mv  F 
0.40 kgsm
mv

 1000N  1.0  103 N
4
t
4.0  10 s
©Modeling Instruction - AMTA 2013
2
U9 Momentum - ws 2 v3.1