PHYS 101 – General Physics I Midterm Exam 1

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

v CG  v CT  vTG  (20 m / s) ˆi  (15 m / s) ˆj  ( 2 m / s) ˆi  (18 m / s) ˆi  (15 m / s) ˆj .

(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

x(3)  (18 m / s)(3 s)  54 m .

(c) For any value of the angle, the velocity of the cannonball with respect to the ground is

v CG  ( v 0 cos   u ) ˆi  v 0 sin  ˆj  (25 cos  2) ˆi  (25 sin ) ˆj m / s .


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

for the small block, and

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

 FN1 sin   ma1x , FN1 cos  mg  ma1 y

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

F cos   Ff  0 and FN  mg  F sin   0 .

We also know that Ff   FN . So Ff  F cos    (mg  F sin  ) . Solving for F , we find

 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  .
