Physics Practice Problems and Solutions

This article provides solutions to physics practice problems related to topics such as distance, velocity, acceleration, Kepler's laws of planetary motion, and force. The solutions are explained step-by-step and are useful for engineering and science students. References to relevant textbooks are also provided.

Suggested Solutions
1. (a) (i)
(ii) to estimate the distance between Oxford and Didcot, we need to find the area under the
graph since displacement can be obtained from integrating the velocity function.
S= ( 1
2 ×200 × 36 ) + ( 1
2 × ( 36+ 40 ) × 40) + ( 40 ×120 ) + ( 1
2 × ( 40+38 ) × 40 ) + ( 1
2 ×200 ×38 )
S ≈ 15280 m

(b)
2. (a) (i)
U x=45 cos 50=28.93 m/s
U y =45 sin 50=34.47 m/ s
(ii) The x and y components of the balls acceleration
10m

In horizontal projection U x is always constant. This implies that ax=0 m/ s2
Vertically, the ball is experiencing gravitational acceleration since no external force acting on
it and thus a y=−9.8 m/s2
(b) (i) To find the time of flight, we need to find the time when the ball reaches at 10m above the
ground
10=U y t + 1
2 g t2 we need ¿ find the value of t .
Now 10=34.47 t−0.5 ( 9.8 ) t2
⇒ 4.9t2−34.47 t +10=0
Which can be solved using the quadratic formula
t=−b ± √b2−4 ac
2 a
t=34.47 ± √ 34.472−4 (10) ¿ ¿¿
t=6.7 secs As Required
(ii) at Hmax, V y=0⇒ V y−U y=a y t ⇒ 0−34.47=−9.8 t
solving for t , we have t = 34.47
9.8 =3.52 sec
⇒ at Hmax , t=3.52 sec
⇒ x=34.47 ( 3.52 ) −0.5 ( 9.8 ) ( 3.522 )=60.63 m
(iii) the ball lands when t=6.7 sec; at Hy =10m,
V x=28.93 m
s ( horizontal velocity is constant ) ⇒ V y=U y +¿
⇒ V y=34.47−9.8 ( 6.7 )=−31.19 m/ s
3. Kepler’s laws of planetary motion
To Jeremiah
Hallo pal, how are you? Johannes Kepler in attempt to describe planetary motion, he devised
three laws called the Kepler’s Laws. The first law states that planets move in an elliptical orbit,
with the sun being one focus of the ellipse. The law means that the distance between the earth and
the sun changes constantly as the earth revolves around the sun.
Second law states that the radius of the vector joining the planets to the sun sweeps out equal
areas in equal times as the planet travels around the ellipse. That is to say, the movement is
faster when the vector radius is shortest and slowest when the vector radius is long.
The third law states that the ratio of the squares of the orbital period for two planets is equal to
the ratio of the cubes of their mean orbit radius. This third law means that the length of time for a

planet to move around the sun increases with increase of the planets orbit radius. Hope you enjoy
the planetary motion study.
Thanks
4. (a) T = 2 π
ω = 2 π
2760 =2.28 ×1 0−3 s
(b) V ( x )= dx
dt = Aω cos ( ωt +φ )
but Max velocity occurs whenthe cos β=1
⇒ V max = Aω=0.0016× 2760=4.416 m/s
( c ) a= dv
dt =−A ω2 sin ( ωt +φ )
¿−A ω2 cos ( π
2 − ( ωt + π
2 ) )
¿−A ω2 cos ωt=−0.0016 ×27602
¿−12188.16 m/s2
(d) X ( t=0 ) =A sin π
2 = ( 1.6 ×1 0−3 ) sin π
2 =1.6 ×1 0−3 m
V ( t=0 ) = Aω cos ( ωt + π
2 )=1.6 ×1 0−3 ×2760 cos π
2 =0
a ( t=0 ) =− A ω2 cos ωt=−1.6 ×1 0−3 ×276 02=−12.19 m/ s2
5. (a) (i) x- component of Resultant force
Rx= ( 8.0 ×1 07 + 4.0× 1 07 )=1.2 ×1 08 N
y- component of Resultant force
Ry = ( 7.0 ×1 07 +−2.0 ×1 07 ) =5.0× 107 N
(ii) Resultant force F=2
√ ( Rx
2+ R y
2 ) =1.3 ×1 08 N
θ=tan−1
( R y
Rx )=tan−1
( 5
12 )=22.62 °

(b) F=G m1 m2
d2
⟹ d= √ 6.67 ×1 0−11 × 7.5× 1012 ×6.0 ×1 024
1.3 × 108
¿ 4.81 ×1 06 km
( c) (i) a= F
m= 1.3 ×1 08
7.5 ×1 012 =1.73× 10−5 ms−2 in the same direction as F
(ii) At U=0 , t = V
a = 1
1.73 ×1 0−5 =5.77 × 1014 seconds
⟹ 5.77 × 1014
3600 =16 hours 1 minute32 seconds
≅ 16 hours ∎
References:
1. Browne, Michael E. (July 1999). Schaum's outline of theory and problems of physics for
engineering and science (Series: Schaum's Outline Series). McGraw-Hill Companies.
p. 58. ISBN 978-0-07-008498-8.
2. Holzner, Steven (December 2005). Physics for Dummies. Wiley, John & Sons,
Incorporated. p. 64. ISBN 978-0-7645-5433-9.