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Basphys LQ1

The document contains a series of physics problems and solutions related to vectors, forces, and motion, aimed at helping students prepare for the BASPHYS Quiz 1. It includes calculations involving vectors, gravitational forces, and kinematic equations. Additionally, it provides links to download further review materials for students' study needs.
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0% found this document useful (0 votes)
41 views10 pages

Basphys LQ1

The document contains a series of physics problems and solutions related to vectors, forces, and motion, aimed at helping students prepare for the BASPHYS Quiz 1. It includes calculations involving vectors, gravitational forces, and kinematic equations. Additionally, it provides links to download further review materials for students' study needs.
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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Society of Young Engineers Towards Achieving Excellence

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BASPHYS Quiz 1

8i + 5 j − 7k
8i − 5 j − 7k
8i − 5 j − k
8i + 5i − k

R
1
B 6

A 6
6
A+ B + R = 0 1
A + B = −R 6
A+ B = R 36
−( A + B ) = R

m
6
s
m
6
s
m
6
s

a = 3i + 2 j + 2k b = 2i − j + 3k

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BASPHYS Quiz 1

1.16m
1.30m

5500N
m a 5
4.0
s2 53.13
b = 5i − 12 j
m
g = 10 a
s2 3a + (2b − a)
1600N
a•b
3300N
4400N
7700N m
3.5
F m s
a m
200m 1.7
5m s

5a
25a
a
5
a
25
750m

8.5s
2.0844 10−7 N
60kg 8.5s
75kg

−11 m3
G = 6.67  10
kg s 2
1.20m
1.23m

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BASPHYS Quiz 1

m
20
s T2
25 T1

7.5kg
19

0.25

0.6kg
T1 T2

60 30

60 30

T1 T2

0.6kg

m
g = −10
s2

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BASPHYS Quiz 1

i j k
ab 3 2 2
2 −1 3

2 2 3 2 3 2
=i −j +k
−1 3 2 3 2 −1
A B = i[(2)(3) − (−1)(3)] − j[(3)(3) − (2)(2)] +
k[(3)(−1) − (2)(2)]
B = i (6 + 2) − j (9 − 4) + k (−3 − 4)
A a  b = 8i − 5 j − 7k

v 2 sin 2  1
A B R H= gM = g
2g 6
v 2 sin 2  v 2 sin 2  3v 2 sin 2 
R B HM = = =
2gM 1 g
A 2( g )
6
x=
H M = xH
3v 2 sin 2  v 2 sin 2 
= ( x)
g 2g
A+ B + R = 0 3 v 2 sin 2  2g
( 2 2 )=x
g v sin 
H M = 6H

a = 3i + 2 j + 2k
b = 2i − j + 3k

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BASPHYS Quiz 1

W = mg  5500 = m(10) a=5


 m = 550kg
ax ay
 F = ma (5cos53.13 )i (5sin 53.13 ) j
 T − mg = ma
 T = (550)(10) + (550)(4) 53.13 a = 3.00i + 4.00 j
 T = 7700 N

3a + (2b − a)
Gm1m2
Fg = 3(3.00i + 4.00 j ) + [2(5i − 12 j ) − (3.00i + 4.00 j )]
r2
9i + 12 j + (10i − 24 j − 3i − 4 j )
−11 m3
(6.67  10 )(60kg )(75kg ) 9i + 12 j + (7i − 28 j )
−7 kg s 2
2.0844  10 N = 9i + 7i + 12 j − 28 j
r2
m3 16i − 16 j
(6.67  10−11 )(60kg )(75kg )
kg  s 2
r2 = a b = ax bx + a y b y
m
2.0844  10−7 kg  2
s a b = (3  5) + [4  (−12)]
−11 m32 a b = 15 − 48
(6.67  10 )(60 kg )(75 kg )( s 2 )
kg  s 2

r2 = a b = −33
2.0844  10−7 kg  m
(6.67  10−11 m 2 )(60)(75)
r=
2.0844  10−7
a b =| a || b | cos 
a=5
ab
r = 1.20m 2 2 = cos
b= b +b x | a || b |
y

b = 52 + (−12) 2 −33 = cos


| 5 ||13 |
b = 25 + 144
−33
a b = 169 cos −1 ( ) =
65
b = 13  = 120.51

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BASPHYS Quiz 1

vbr = 3.5m / s vrs = 1.7m / s d = 200m

vrs d 1.7 d
=  =
vbr 200 3.5 200
 d = 97.1429  97.14m

m
vbs v0 = 0; a = −9.8 ; h = 750m
s2
0
vbs = (vbr ) 2 + (vrs ) 2 750m
 vbs = 3.5 + 1.72 2

m
 vbs = 3.8910  3.89
s
8.5s

 m 
x = vbr t v f = v0 + at = 0 +  −9.8 2  (8.5 s )
 200 = (3.5)t  s 
 t = 57.1429  57.14sec m
v f = −83.3
s

8.5s

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BASPHYS Quiz 1

1 1 m
d yf = d y 0 + v0t + at 2 = 0 + 0 +  −9.8 2  (8.52 s 2 )
2 2 s 
 m
v 2f = v02 + 2a  d y = 0 + 2  −9.8 2  (−750m)
 s 
d yf = −354.025 dis tan cey = 354.025m
m2 m
v = −9.8(−750) 2
2
f v f = −121.24
s s

8.5s
m
vo = 20  = 25
h8.5 s = h − dis tan cey = 750m − 354.025m s
h8.5 s = 395.975m

1
y = − gt 2 + (vo sin  )t
2

t
t
1
d fy = d y 0 + v0t + at 2
2
1 m
−750m = 0 + 0 +  −9.8 2  t 2 0m / s
2 s 
v y = − gt + vo sin 
−750 m s 2
 0 = −(9.81)(t ) + (20)(sin 25 )
= t2 t = 12.3718s
−4.9 m  t = 0.862sec

750m
d fy

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BASPHYS Quiz 1

1 x = (vo cos )t
y = − gt 2 + vo (sin  )t
2  x = (20)(cos 25 )(1.7232)
1
 y = − (9.81)(0.862)2 + (20)(sin 25 )(0.862)  x = 31.23m
2
 y = 3.641m  3.64m

vx = vo cos
 vx = (20)cos(25 )
 vx = 18.1261m / s
y
v y = − gt + v o sin 
 v y = −(9.81)(1.7232) + (20)sin(25 )
 v y = −8.4522m / s
1
y = − gt 2 + vo (sin  )t
2 v = (v x ) 2 + (v y ) 2
1
 0 = − (9.81)t 2 + (20)(sin 25 )t  v = (18.1261) 2 + (−8.4522) 2
2
 t = 1.72sec m
 v = 20
t = 0sec s

m = 7.5kg  = 19  = 0.25

y

F y =0
 N p − W sin  = 0
 N p = mg sin 
 N p = (7.5)(9.81)(cos19 )
 N p = 69.5665 N

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BASPHYS Quiz 1

T2
 Fx = ma Fx Fy
 W sin  − f = ma
−T1 cos60 T1 sin 60
 mg sin  −  N p = ma
T2 cos30 T2 sin 30
 (7.5)(9.81)(sin19 ) − (0.25)(69.5665) = (7.5)(a )
m
 a = 0.87 2 Fxnet = T2 cos30 − T1 cos60
s
 3 1
m = 0.6kg; = 60 ,30 0 = T2   − T1  
 2  2
 3
60 30
1
T1   = T2 
2
 T1 = T2 3 ( )
 2 
60 30 Fynet = T1 sin 60 + T2 sin 30 + Fweight
T1 T2  3 1
0 = T1   + T2   − 6.0 N
0.6kg  2  2
 3
Fweight = mg 6.0 N = T2 3  ( ) 1
 + T2  
2
 2 
6.0 N = 2T2 T2 = 3.0 N
T1 T2

T1
 m
Fweight = mg = (0.6kg )  −10 2  = −6.0 N
 s 

T1
T1 = T2 ( 3) T1 = 3.0 N ( 3)
T1 = 5.1962 N  5.20 N

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BASPHYS Quiz 1

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