PHYS 101 - General Physics I Midterm Exam 1: (2.0) Cos (2.0) Sin
PHYS 101 - General Physics I Midterm Exam 1: (2.0) Cos (2.0) Sin
PHYS 101 - General Physics I Midterm Exam 1: (2.0) Cos (2.0) Sin
21 October 2017
1. The position of a point object is given by the expression r (2.0) cos( 2 t ) ˆi (2.0) sin( 2 t ) ˆj , where r is in
meters and time t is in seconds.
(a) (5 Pts.) What are the units of (2.0) and 2 in the expression for r ?
(b) (5 Pts.) What is the average velocity of the object during the time interval from t 0 to t 1 s ?
(c) (5 Pts.) What is the instantaneous velocity of the object as a function of time?
(d) (5 Pts.) What is the instantaneous acceleration of the object?
(e) (5 Pts.) Describe the motion of the object.
Solution:
(a) [2.0] m (meters), [𝜋⁄2] s −1 , (or Hz, rad/s).
(b) r(1) (2.0 m) ˆj , r(0) (2.0 m) ˆi, r r(1) r(0) (2.0 m) ˆi (2.0 m) ˆj ,
r
Since t 1 s , v av (2.0 m / s) ˆi (2.0 m / s) ˆj .
t
dr
(c) v (1.0 m / s) sin t ˆi (1.0 m / s) cos t ˆj .
dt 2 2
dv
(d) a (0.5 m / s 2 ) 2 cos t ˆi (0.5 m / s 2 ) 2 sin t ˆj .
dt 2 2
(e) Since r 2.0 m and v (1.0 m / s) are both constant, the object is in uniform circular motion.
2. A cannon is on a train car travelling with velocity 𝑢 = 2 m/s to the left. The cannon fires at an angle 𝜃 with the
horizontal as it passes a flag post fixed to the ground (see the figure). The cannonball leaves the cannon with speed
𝑣0 = 25 m/s (with respect to the cannon). (Gravitational acceleration is 𝑔 = 10𝑚/𝑠 2 , the height of the train car and
the cannon is negligible, and cos 𝜃 = 4⁄5.)
𝑦
(a) (7 Pts.) What is the velocity of the cannonball with respect to the
ground?
(b) (9 Pts.) How far away from the post does the cannonball hit the ⃗0
𝒗
ground?
⃗⃗
𝒈
(c) (9 Pts.) With |𝒗
⃗ 0 | fixed, the angle of the cannon can be changed to
change the range. For which value of 𝜃 is the distance from the post to the ⃗𝒖 𝜃
point hit on the ground is maximum? (You can express your answer in
terms of inverse trigonometric functions.) 𝑥
Solution:
(a) Velocity of the cannonball with respect to the train is v CT v 0 cos ˆi v 0 sin ˆj (20 m / s) ˆi (15 m / s) ˆj .
Since velocity of the train with respect to the ground is vTG (2 m / s) ˆi , velocity of the cannonball with respect to
the ground is
(b) Kinematical equations for the position of the cannonball with respect to the ground with the post at the origin is
1
x(t ) (18 m / s) t , y(t ) (15 m / s)t (10 m / s 2 )t 2 .
2
Therefore, the cannonball hits the ground when y 0 t 3 s after it is shot. During this time it covers the
horizontal distance
(c) For any value of the angle, the velocity of the cannonball with respect to the ground is
Maximum range means the angle between vCG and the horizontal is . Therefore,
4
25 sin 2
tan 1 25 sin 25 cos 2 cos sin .
25 cos 2 4 25
2
2
This gives (cos sin ) 2 1 2sin cos .
25
2
2
1 2 2
Since 2sin cos sin 2 , we find sin 2 1 arcsin 1 .
25 2 25
1 1 2
2
Another form of the solution is arccos
25 2 25
3. A small block of mass m rests on the sloping side of a triangular block of mass M which itself rests on a horizontal
table. A force F always acts horizontally on the small block as shown in the figure. Assume that all surfaces are
frictionless.
(a) (8 Pts.) Draw free-body diagrams of the two blocks.
⃗
𝑭 𝑚
(b) (8 Pts.) Determine the magnitude of the force F that must be applied to
m so that it remains in a fixed position relative to M (that is, m doesn’t move 𝑦 ⃗⃗
𝒈
𝑀
relative to the incline). 𝑥 𝜃
(c) (9 Pts.) If the force F is zero, what will be the acceleration of the
triangular block relative to the table?
Solution:
(a)
⃗ 𝑁2
𝑭
⃗ 𝑁1
𝑭 𝜃
𝑚 ⃗
𝑭 𝑀
𝑦
𝑥 𝜃 ⃗ 𝑁1
𝑭
⃗⃗
𝑚𝒈 x ⃗⃗
𝑀𝒈
(b) If the small block is to remain at rest relative to the triangular block, we should have ax a , a y 0 for both
blocks. Therefore, we have
F FN 1 sin ma , FN 1 cos mg 0
FN 1 sin Ma
m
for the triangular block. eliminating FN 1 , we get F ( M m)a and a g tan . Hence
M
m
F ( M m) g tan .
M
(c) If F is zero, acceleration of the two blocks will not be the same. So, we have
for the small block, and FN 1 sin Ma2 x for the triangular block. Eliminating FN 1 , we get
m
a2 x a1x and a1 y g a1x cot .
M
Furthermore, since the small block will be accelerating down the triangular block, its acceleration a R relative to the
triangular block should be along that direction, that is,
aRy
tan .
aRx
a1 y
Since a R a1 a2 , this means that we have tan . Solving these three equations, we find
a1x a2 x
cot m
a2 x g . An equivalent expression is a2 x g.
1 m csc M cot ( M m) tan
2
M
4. A block of mass m rests on a horizontal surface. The coefficients of kinetic
and static friction between the block and the horizontal surface are both . A
𝜃
force F acts on the block as shown in the figure. ⃗𝒈
⃗
⃗
𝑭
(a) (7 Pts.) Draw the free body diagram of the box.
𝑚
(b) (9 Pts.) Find the expression for the magnitude of the force needed to push 𝜇
the box at constant speed.
(c) (9 Pts.) Find the critical angle c above which the box becomes impossible to push.
Solution:
(a)
⃗𝑭𝑁
𝑦
𝑥 ⃗𝑓
𝑭 𝑚
x
𝜃
⃗
𝑭
⃗⃗
𝑚𝒈
(b) Since the box is pushed at constant speed, its acceleration is zero. So we have
mg
F .
cos sin
(c) The force becomes infinite when the expression in the denominator is zero. This means when C , where
1
cos C sin C 0 tan C
we need an infinite force to push the box. Therefore the box becomes impossible to push for C , where
1
C arctan .