Download Kinematics Multiples

Impulse
From Old AP’s
1 (1974) A force acting in a straight line on an object of mass 5 kg is plotted as a function
of time in the graph above. The change in the velocity of the object is:
a. .8 m/s.
b. 1.1 m/s.
c. 1.6 m/s.
d. 2.3 m/s.
e. 4.0 m/s.
*A. Impulse= Area under the curve = change in momentum:
 Fdt  mV
1
(base)( height )  (5kg)V
2
1
2
(2 sec onds )( 4 N )  (5kg)V
V  .8m / s
From Princeton Review Book
Approximat e the curve as a triangle.
2. A rubber ball of mass .08 kg is dropped from a height of 2 m and, after bouncing off
the floor, rises to its original height. If the impact time with the floor is measured to be
0.04 seconds, what average force did the floor exert on the ball?
a. .16 N.
b. 16 N.
c. 25 N.
d. 36 N.
e. 64 N.
*C Since the question is about impulse and momentum, we will use that relationship, but
first you must use kinematics or energy to find the speed of the ball before impact with
the floor:
Ug  KE
mgh  12 mv 2
(mass cancels)
v  2gh  2(9.8m / s 2 )( 2m)  6.26m / s
Now, use the Impulse/ch ange in momentum formula :
Ft  p
(note : this is actually only true if F is average force)
F(.04sec)  (.08kg)(6. 26m/s - (-6.26m/s) )
The ball reverses direction and thus sign of velocity
F  25 N
3. A ball of mass .5 kg, initially at rest, acquires a speed of 4 m/s immediately after being
kicked by a force of strength 20 N. For how long did this force act on the ball?
a. 0.01 s.
b. 0.02 s.
c. 0.1 s
d. 0.2 s
e. 1 s.
* C Use the impulse/change in momentum relationship:
F t  p  m(v f  vi )
(20 N )( t )  (.5kg)( 4m / s  0)
t  .1sec
4. A box with a mass of 2 kg accelerates in a straight line from 4 m/s to 8 m/s due to the
application of a force whose duration is 0.5 seconds. Find the average strength of this
force:
a. 2 N.
b. 4 N.
c. 8 N.
d. 12 N.
e. 16 N.
*E. Apply Impulse/momentum relationship:
F t  m(v f  vi )
F (.5 sec)  (2kg)(8m / s  4m / s)
F  16 N
5. A ball of mass m traveling horizontally with velocity v strikes a massive vertical wall
and rebounds back along its original direction with no change in speed. What is the
magnitude of the impulse delivered by the wall to the ball?
a. 0
b. (mv)/2
c. mv
d. 2mv
e. 4mv
* D.
Careful. The problem asks for Impulse, not force. The impulse is the change in
momentum. The ball initially had momentum + mv and ended with momentum (-mv).
Thus, the change in momentum was 2mv.
A block of mass 3 kg, initially at rest, is pulled along a frictionless, horizontal surface
with a force shown as a function of time, t, by the graph above.
6. (2004, 88%)
The acceleration of the block at t = 2 seconds is:
a. ¾ m/s2.
b. 4/3 m/s2.
c. 2 m/s2.
d. 8 m/s2.
e. 12 m/s2.
* B.
a
F (4 N )

 4/3 m/s 2
m (3kg)
(read the 4 N value off the graph)
7. The speed of the block at t = 2 seconds is:
a. 4/3 m/s.
b. 8/3 m/s
c. 4 m/s.
d. 8 m/s.
e. 24 m/s.
* A.
This is an impulse/mo mentum question :
 Fdt  m( v)
1
bh  (3 kg) (Vf  0)
2
1
(2 sec)( 4 N )  (3kg)V f
2
V f  4 / 3 m/s
8. (2004, 71%)
A 2 kg ball collides with the floor at an angle and rebounds at the same angle and speed
as shown above. Which of the following vectors represents the impulse exerted on the
ball by the floor?
* E. If you take a careful look at the initial and final velocity vectors of the ball, you will
note that the horizontal component of the velocity did not change. This means that there
was no net force in the horizontal direction. However, the vertical component of the
velocity reversed directions, which requires an upward force (to stop and then reverse the
direction of motion of the ball.) Even though the question asks for impulse and not force,
the direction of the impulse will be in the same direction as the force.
9.
(2004, 66%)
The momentum, p, of a moving object as a function of time t is given by the expression:
p  K t 3 , where K is a constant. The force causing this motion is given by the
expression:
a. 3 k t 2
c.
k t2
3
e.
k t4
4
b.
3k t2
2
d. k t4
*A
Remember the impulse/momentum relationship reworked is:
F = dp/dt = 3 k t2
(Also remember that you will always take the integral or the derivative of any function
they give you on the multiple choice.)
10.
(2004B, 18%)
A student obtains data on the magnitude of force applied to an object as a function of
time and displays the data on the graph above. The increase in momentum of the object
between t = 0 seconds and t = 4 seconds is most nearly:
a. 40 Ns
b. 50 Ns
c. 60 Ns
d. 80 Ns
e. More information is needed to determine this.
*C
The change in momentum is the area under the curve.
Use the trapezoid rule to calculate the area:
 h  h2 
 5N 25 N 
Area   1
 (b)  
(4 sec)  (15 N)(4 sec)  60 Ns
2
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