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Name Sept. 24, 2008 Quiz #2 — Fall 2008 Phys 2110 – Sec 3
30.0 m/s 40.0o
- A projectile is fired from a cliff at an initial speed of 30. 0 m s , directed up from the horizontal by 40. 0 ◦. It strikes the ground below 6.73 s later.
a) How high is the cliff?
What is the value of y at t = 6.37 s? We have
v 0 x = v 0 cos θ = 22. 98 m s and v 0 y = v 0 sin θ = 19. 28 m s
so the y equation gives
v = v 0 yt − 12 gt^2 = (19. 28 m s )(3.67 s) − 12 (9. (^8) sm 2 )(6.37)^2 = − 92 .2 m
which tells us that the cliff is 92 .2 m high.
b) How far did the projectile travel horizontally in its flight?
What is the value of x at t = 6.73 s? The x equation gives
x = v 0 xt + 0 = (22. 98 m s )(6.73 s) = 155 m
The projectile has traveled a horizontal distance of 155 m.
c) What was the speed of the projectile at impact?
At impact, the x component of the velocity is the same as the initial value: vx = 22. 98 m s. The y component is vy = v 0 y + ayt = (19. 28 m s ) + (− 9. 8 m s 2 )(6.73) = − 46. 67 m s
so the speed is
v =
√ v^2 x + v y^2 =
√ (22. 98 m s )^2 + (− 46. 67 m s )^2 = 52. 0 m s
The speed at impact is 52. 0 m s.
0.80 m
- A small mass moves at constant speed on a circle of radius 0.80 m such that it makes one revolution every 0.60 s.
a) Find the speed of the mass.
Mass travels a distance 2 πr in time T , so
v =
2 πr T
2 π(0.80 m) (0.60 s)
= 8. 38 m s
b) Find the acceleration of the mass; give its magnitude and direction.
Mass has a centripetal acceleration of magnitude
ac =
v^2 r
(8. 38 m s )^2 (0.80 m)
= 87. 7 m s 2
The direction of the acceleration is inward , i.e. toward the center of the circle.
2.0 kg
1.60 m/s^2
T
mg
- If we attach a string to a 2.0 kg mass and pull the mass upward so that its acceleration is 1. 6 m s 2 (upward), find the tension in the string.
Forces on the mass are the string tension T upward and gravity mg down- ward (added to the diagram at the right!). Newton's 2nd law gives
T − mg = ma =⇒ T = mg + ma = m(g + a)
Plug in the numbers: T = (2.0 kg)(9. 8 m s 2 + 1. 6 m s 2 ) = 22.8 N
Tension in the string is 22 .8 N.
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 = vx 0 + axt x = x 0 + vx 0 t + 12 axt^2 v x^2 = v x^20 + 2ax(x − x 0 ) x − x 0 = 12 (vx 0 + vx)t
vy = vy 0 + ayt y = y 0 + vy 0 t + 12 ayt^2 v y^2 = v y^20 + 2ay (y − y 0 ) y − y 0 = 12 (vy 0 + vy)t
g = 9. 80 m s 2 F = ma v =
2 πr T
ac =
v^2 r
Fspr = −kx
