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Name Sept. 20, 2006 Quiz #1 — Fall 2006
Phys 2010, NSCC
72 o
x
y
A
B
- Vector A has magnitude 4.2 and is directed at 72. 0 ◦ from the +x axis. Vector B points in the −y direction and has magnitude 3.00. Find the magnitude and direction of the sum of the two vectors.
Ax = 4.20 cos 72◦^ = 1. 298
Ay = 4.20 sin 72◦^ = 3. 994
And it is clear that: Bx = 0 By = − 3
So if C = A + B then
Cx = Ax + Bx = 1. 298 Cy = Ay + By = 0. 994
and C =
√ C x^2 + C y^2 = 1. 635
tan θ =
Cy Cx
= 0. 766 =⇒ θ = tan−^1 (0.766) = 37. 4 ◦
where θ is the direction of C measured from the +x axis.
Convert 3. (^12) mkg 2 to units of (^) cmg 2
(^12) mkg 2 = (3. (^12) mkg 2 ) ·
( 1000 g 1 kg
) ·
( (^) 1 m
100 cm
) 2 = 0. (^312) cmg 2
12.0 m/s (^) 3.0 m/s
- A cart rolling on a track undergoes a uniform t = 2.0^ s acceleration (deceleration, really). At is initial posi- tion it has a velocity of 12. 0 m s and after it has moved forward for 2.00 s its velocity is 3. 00 m s.
a) What is the acceleration of the cart?
a =
v − v 0 t
(3. 00 m s ) − (12. 0 m s ) (2.00 s)
= − 4. 5 m s 2
b) How far did it move forward in the 2.00 s?
A convenient equation to use is
x = 12 (v 0 + v)t = 12 ((12. 0 m s + 3. 0 m s )(2.00 s) = 15.0 m
- A projectile is shot straight upward from ground level with a speed of 55. 0 m s. How long does it take the projectile to reach maximum height?
With v 0 = 55. 0 m s v = 0 a = − 9. 80 m s 2
Use
v = v 0 + at =⇒ t =
v − v 0 a
t =
0 − 55. 0 m s (− 9. 80 m s 2 )
= 5.61 s
You must show all your work and include the right units with your answers!
Ax = A cos θ Ay = A sin θ A =
√ A^2 x + A^2 y tan θ = Ay/Ax
vx = v 0 x + axt x = v 0 xt + 12 axt^2 v x^2 = v 02 x + 2axx x = 12 (v 0 x + vx)t vy = v 0 y + ayt y = v 0 yt + 12 ayt^2 v^2 y = v 02 y + 2ayy y = 12 (v 0 y + vy)t
g = 9. (^80) sm 2 R =
2 v 02 sin θ cos θ g
Fnet = ma Weight = mg
