A concave mirror creates a real, inverted image 16.0 cm from its surface. If the image is 3.5 times larger, how far away is the
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Part of the question is missing. Here it is:
<em>A 72 g autographed baseball slides off of a 1.3 m high table and strikes the floor a horizontal distance of 0.7m away from the table. The acceleration of gravity is 9.81 m/s2. What was the direction of the ball’s velocity just before it hit the floor? </em>
Answer:
Explanation:
The motion of the ball is a projectile motion, which consists of two separate motions:
- A horizontal motion at constant velocity
- A vertical motion at constant acceleration (free fall)
We start by analyzing the vertical motion, to find the time of flight of the ball. This can be done by using the suvat equation
where, choosing downward as positive direction:
s =1.3 m is the vertical displacement of the ball
u = 0 is the initial vertical velocity
is the acceleration of gravity
t is the time
Solving for t,
Now we can find the final vertical velocity of the ball, using:
And susbtituting t = 0.52 s, we find
It is important to keep in mind that the direction of this velocity is downward, since we chose downward as positive direction.
The horizontal velocity of the ball instead is constant; we know that the ball covers a horizontal distance of
d = 0.7 m
In a time of
t = 0.52 s
So, the horizontal velocity is
So now we can find the direction of the ball's velocity using:
And since the vertical direction is downward, this means that this velocity is below the horizontal, so the answer is
The problem is solved and the questions are answered below.
Explanation:
a. To calculate the speed of the 0.66 kg ball just before the collision
V₀ + K₀ = V₁ + K₁
= mgh₀ = 1/2 mv₁²
where, h= r - r cosθ
V =
V = 2.42 m/s
b. Calculate the speed of the 0.22 kg ball immediately after the collision
y = y₀ + Vy₀t - 1/2 gt²
0 = 1.2 - 1/2 gt²
t = 0.495 s
x = x₀ + Vx₀t
1.4 = 0 + vx₀ (0.495)
Vx₀ = 2.83 m/s
C. To Calculate the speed of the 0.66 kg ball immediately after the collision
m₁ v₁ = m₁ v₃ + m₂ v₄
(0.66)(2.42) = (0.66) v₃ + (0.22)(2.83)
V₃ = 1.48 m/s
D. To Indicate the direction of motion of the 0.66 kg ball immediately after the collision is to the right.
E. To Calculate the height to which the 0.66 kg ball rises after the collision
V₀ + k₀ = V₁ + k₁
1/2 mv₀² = mgh₁
h₁ = v₀²/2 g
= 0.112 m
F. Based on your data, No the collision is not elastic.
Δk = 1/2 m₁v₃² =1/2 m₂v₄² - 1/2 m₁v₁²
= 1/2 (0.66)(1.48)² + 1/2 (0.22)(2.83)² - 1/2 (0.66)(2.42)²
= - 0.329 J
Hence, kinetic energy is not conserved.