(a) [5 pts] You are standing on a skateboard, initially at rest. A
friend throws a heavy ball towards you. You can either catch the object or
deflect the object back towards your friend (such that it moves away from you
with the same speed as it was originally thrown). What should you do to
MINIMIZE your speed on the skateboard?
(A) Catch the ball.
(B) Deflect the ball.
(C) Your final speed on the skateboard will be the same regardless
whether you catch the ball or deflect the ball.
ANS: (A)
(b) [5 pts] For general projectile motion, when the projectile is at
the highest pointof its trajectory,
(A) its acceleration is zero.
(B) its velocity is perpendicular to the acceleration.
(C) its velocity and acceleration are both zero.
(D) the horizontal component of its velocity is zero.
(E) the horizontal and vertical components of its velocity are zero.
ANS: (B)
(c) [5 pts] In an INELASTIC collision between two objects
(A) the momentum of each object is conserved.
(B) the kinetic energy of each object is conserved.
(C) both the momentum and kinetic energy of the system are conserved.
(D) the kinetic energy of the system is conserved, but the momentum
of the system is not conserved.
(E) the momentum of the system is conserved but the kinetic energy of
the system is not conserved.
ANS: (E)
(d) [5 pts] In a collision between two objects having unequal masses,
how does the magnitude of the impulse imparted to the lighter object by the
heavier one compare with the magnitude of the impulse imparted to the heavier
object by the lighter one?
(A) Both objects receive the same impulse.
(B) The lighter object receives a lighter impulse.
(C) The heavier object receives a lighter impulse.
(D) The answer depends on the ratio of the masses.
(E) The answer depends on the ratio of the speeds.
ANS: (A)
(e) [5 pts] Two blocks collide in a totally inelastic collision.
Intheinteractionregion (A) the system momentum is not conserved.
(B) the system kinetic energy is always conserved.
(C) the relative speed is never zero.
(D) both system momentum and kinetic energy are conserved.
(E) the system kinetic energy has to change regardless of the type of
collision.
ANS: (E)
The figure shows a velocity vs time curve describing a standard mass
mA colliding with a non-standard mass mB. There is a lot of
information contained in the graph. To avoid reading the velocities
incorrectly, I will give them to you.
vAi=+3.1 m/s
vAf=+1.4 m/s
vBi=+1.2 m/s
vBf=+2.0 m/s
(a) [7 pts] Calculate the relative speeds before and after collision
and state what kind of collision has occurred (totally elastic, partially
elastic, or totally inelastic).
vAi-vBi
=
3.1 m/s - 1.2 m/s
=
+1.9 m/s
vAf-vBf
=
1.4 m/s - 2.0 m/s
=
-0.6 m/s
The relative speeds are unequal indicating that the collision is partialy
elastic.
(b) [4 pts] Calculate the value of mass mB.
mB
mA
=
-
DvA
DvB
=
-
vAf-vAi
vBf-vBi
mB
=
-(1.0 kg)
� �
1.4 m/s - 3.1 m/s
2.0 m/s - 1.2 m/s
� �
=
2.125 kg � 2.1 kg
(c) [7 pts] Calculate the change in momentum for each mass and state
whether or not momentum is conserved.
DpA
=
pAf-pAi
=
(1.0 kg)(1.4 m/s - 3.1 m/s)
=
-1.7 kg·m/s
DpB
=
pBf-pBi
=
(2.125 kg)(1.4 m/s - 3.1 m/s)
=
+1.7 kg·m/s
Yes: momentum is conserved.
DpA = - DpB
(d) [7 pts] Calculate the coefficient of restitution, e, for this
collision. Does your e-value support your conclusion in part (a)?
e
=
vAf-vBf
vBi-vAi
=
1.4 m/s - 2.0 m/s
1.2 m/s - 3.1 m/s
=
+0.32
Yes. the result is consistent with part (a) which predicts a partially
elastic collision.
Mass m2 approaches mass m1 from the left. The collision
between them is totally elastic. The known quantities are
m2=2.0 kg
v1xi=3.0 m/s
v2xi=6.0 m/s
v1xf=2.5 v2xf
Mass m1 and the final velocities are unknown.
(a) [5 pts] Write down the two linear equations which can be used to
find the unknowns.
v1xi + v1xf
=
v2xi + v2xf
m1v1xi + m2v2xi
=
m1v1xf + m2v2xf
(b) [10 pts] Determine the values of the final velocities.
v1xi + v1xf
=
v2xi + v2xf
v1xi + 2.5 v2xf
=
v2xi + v2xf
3.0 m/s + 2.5 v2xf
=
6.0 m/s + v2xf
v2xf
=
2.0 m/s
v1xf
=
5.0 m/s
(c) [10 pts] Determine the value of mass m1.
m1v1xi + m2v2xi
=
m1v1xf + m2v2xf
m1(v1xi-v1xf)
=
m2(v2xf-v2xi)
m1
=
m2
� �
v2xf-v2xi
v1xi-v1xf
� �
=
(2.0 kg)
� �
2.0 m/s - 6.0 m/s
3.0 m/s - 5.0 m/s
� �
=
4.0 kg
An object is projected from a position 10.0 m above the ground at an
angle of 42.0 degrees. The object travels 100. m horizontally before hitting
the ground.
(a) [8 pts] Determine the x-component of the initial velocity.
yf
=
yi+Dxtanq-
g(Dx)2
2v2xi
0
=
10.0 m + 100.tan42.0o -
(9.80 m/s2)(100. m)2
2v2xi
vxi
=
22.1 m/s
(b) [5 pts] Determine the total travel time.
ttot
=
Dx
vxi
=
100. m
22.1 m/s
=
4.52 s
(c) [6 pts] How high did the object rise?
Find vi first.
vi
=
vxi
cosq
=
22.1 m/s
cos42.0o
=
29.8 m/s
Now, find Dy.
v2yf
=
v2yi -2gDy
0
=
v2yi -2gDymax
Dymax
=
[(29.8 m/s)sin42.0o]2
19.6 m/s2
=
20.3 m
(d) [6 pts] Calculate the speed of the object just before hitting the
ground.
Find the y-component of the object just before hitting the ground.
vyf
=
vyi-gttot
=
(29.8 m/s)sin42.0 - (9,80 m/s2)(4.52 s)
=
-24.4 m/s
Now, find the speed.
speed
=
�
v2xf+v2yf
=
[(22.1 m/s)2+(-24.4 m/s)2]1/2
=
32.9 m/s
FORMULAS
m1/m2
=
-Dv2/Dv1
v21
=
v1-v2
v12
=
v2-v1
xf
=
xi+vxit
yf
=
yi+vyit-gt2/2
vxi
=
vicosq
vyi
=
visinq
vyf
=
vyi-gt
y
=
yi+(x-xi)tanq- g(x-xi)2 /2vxi2
m1v1xi+m2v2xi
=
m1v1xf+m2v2xf
1
2
m1v21xi+
1
2
m2v22xi
=
1
2
m1v21xf+
1
2
m2v22xi
v1xi+v1xf
=
v2xi+v2xf
e
=
-
v2xf-v1xf
v2xi-v1xi
p
=
p1+p2
pfinal
=
pinitial
p1i+p2i
=
p1f+p2f
K
=
1
2
mv2
File translated from
TEX
by
TTH,
version 3.85. On 06 Oct 2015, 15:19.